Measure Theory

T.K.SUBRAHMONIAN MOOTHATHU

Contents

1. Introduction: Riemann vs Lebesgue 2 2. Small subsets of Rd 2 3. About functions behaving nicely outside a small set 9 4. σ-algebras and measurable spaces 17 5. Measures and outer measures 22 6. Lebesgue measure on Rd 24 7. Pre-measure → outer measure → measure 30 8. Measurable functions 36 9. Approximating measurable functions with simple functions 40 10. Limit theorems for measurable functions 42 11. Integration with respect to a measure 46 12. Properties of the integral, and limit theorems 48 13. Lp spaces 53 14. Product spaces and product measures 56 15. Fundamental Theorem of Calculus (FTC) 59 16. Measures on metric spaces 61 17. Measure Theory and Functional Analysis 65

Prerequisites: Real Analysis and basic Metric Space Theory.

Convention: Axiom of choice and continuum hypothesis will be assumed throughout.

Topics for self-reading: Banach-Tarski paradox, Littlewood’s three principles.

Our approach: We will do measure theory ﬁrst, and then will do integration theory. We will also discuss the interaction of measure theory with topology in Rd.

Some textbooks for further reading: 1. R.G. Bartle, The elements of integration and Lebesgue measure, 1995. 2. D.L. Cohn, Measure Theory, 2nd Edn., 2013. 1 2 T.K.SUBRAHMONIAN MOOTHATHU

3. H.L. Royden, Real Analysis, 3rd Edn., 1988. 4. E.M. Stein and R. Shakarchi, Real Analysis (Princeton Lectures in Analysis, 3), 2005.

1. Introduction: Riemann vs Lebesgue

Measure theory helps us to assign numbers to certain sets and functions - to a measurable set we may assign its measure, and to an integrable function we may assign the value of its integral. Lebesgue integration theory is a generalization and completion of Riemann integration theory. In Lebesgue’s theory, we can assign numbers to more sets and more functions than what is possible in Riemann’s theory. If we are asked to distinguish between Riemann integration theory and Lebesgue integration theory by pointing out an essential feature, the answer is perhaps the following. Riemann integration theory 7→ finiteness. Lebesgue integration theory 7→ countable infiniteness. Riemann integration theory is developed through approximations of a finite nature (eg: one tries to approximate the area of a bounded subset of R2 by the sum of the areas of finitely many rectangles), and this theory works well with respect to finite operations - if we can assign numbers to

ﬁnitely many sets A1,...,An and ﬁnitely many functions f1, . . . , fn, then we can assign numbers to

A1 ∪· · ·∪An, f1 +···+fn, max{f1, . . . , fn}, etc. The disadvantage of Riemann integration theory is that it does not behave well with respect to operations of a countably infinite nature - there may not ∪ ∑ ∞ ∞ { ∈ N} be any consistent way to assign numbers to n=1 An, n=1 fn, limn→∞ fn, sup fn : n , etc. even if we can assign numbers to the sets A1,A2,..., and functions f1, f2 .... Lebesgue integration theory rectifies this disadvantage to a large extent. In Riemann integration theory, we proceed by considering a partition of the domain of a function, where as in Lebesgue integration theory, we proceed by considering a partition of the range of the function - this is observed as another difference. Moreover, while Riemann’s theory is restricted to the Euclidean space, the ideas involved in Lebesgue’s theory are applicable to more general spaces, yielding an abstract measure theory. This abstract measure theory intersects with many branches of mathematics and is very useful. There is even a philosophy that measures are easier to deal with than sets.

2. Small subsets of Rd

It is possible to think about many mathematical notions expressing in some sense the idea that a subset Y ⊂ Rd is a small set (or a big set) with respect to Rd. We will discuss this a little as a warm-up. We will also use this opportunity to introduce Lebesgue outer measure. Suppose you have a certain notion of smallness or bigness for a subset of Rd. Then there are some natural questions. Two sample questions are: MEASURE THEORY 3

1. If Y ⊂ Rd is big, is Rd \ Y small? ∪ ⊂ Rd ∞ 2. If Y1,Y2,... are small, is n=1 Yn small? For instance, consider the following two elementary notions. Saying that Y ⊂ Rd is unbounded is one way of saying Y is big, and saying that Y ⊂ Rd is a finite set is one way of saying Y is small. Note that the complement of an unbounded set can also be unbounded and a countable union of finite sets need not be finite. So here we have negative answers to the above two questions.

Exercise-1: Find an uncountable collection {Yα : α ∈ I} of subsets of R such that Yα’s are pairwise disjoint, and each Yα is bounded neither above nor below.

To discuss some other notions of smallness, we introduce a few deﬁnitions.

Deﬁnitions: (i) We say Y ⊂ Rd is a discrete subset of Rd if for each y ∈ Y , there is an open set U ⊂ Rd such that U ∩ Y = {y}. For example, {1/n : n ∈ N} is a discrete subset of R and Zd is a discrete subset of Rd.

(ii) A subset Y ⊂ Rd is nowhere dense in Rd if int[Y ] = ∅, or equivalently if for any nonempty open set U ⊂ Rd, there is a nonempty open set V ⊂ U such that V ∩ Y = ∅. If f : R → R is a continuous map, then its graph G(f) := {(x, f(x)) : x ∈ R} is nowhere dense in R2 (∵ G(f) is closed and does not contain any open disc). Also, {A ∈ Rd×d : det(A) = 0} is nowhere dense in Rd×d.

(iii) A subset Y ⊂ Rd is of ﬁrst category in Rd if Y can be written as a countable union of nowhere dense subsets of Rd; otherwise, Y is said to be of second category in Rd. For example, Y = Q × R ∪ R2 is of ﬁrst category in since Y can be written as the countable union Y = r∈Q Yr, where 2 Yr := {r} × R is nowhere dense in R .

(iv) (The following definition can be extended by considering ordinal numbers, but we consider only non-negative integers). For Y ⊂ Rd and integer n ≥ 0, define the nth derived set of Y inductively as Y (0) = Y , Y (n+1) = {limit points of Y (n) in Rd}. We say Y ⊂ Rd has derived length n if Y (n) ≠ ∅ and Y (n+1) = ∅; and we say Y has infinite derived length if Y (n) ≠ ∅ for every integer n ≥ 0. For example, Q has infinite derived length (since Q = R), and {(1/m, 1/n): m, n ∈ N} has derived length 2. ∏ d d (v) We say A ⊂ R is a d-box if A = Ij, where Ij’s are bounded intervals. The d-dimensional ∏ j=1 d | | × × − volume of a d-box A is V old(A) = j=1 Ij . For example, V ol3([1, 4) [0, 1/2] ( 1, 3]) = 6.

Exercise-2: If U ⊂ Rd is a nonempty open set, then U can be written as a countable union of pairwise disjoint d-boxes (these d-boxes may not be open). [Hint: Let B be the collection of all open d-boxes whose vertices have rational coordinates. For each x ∈ U, there is B ∈ B with 4 T.K.SUBRAHMONIAN MOOTHATHU ∪ x ∈ B ⊂ U. Hence U can be written as a countable union U = Bn with Bn ∈ B. Now note that ∪ n \ n Bn+1 j=1 Bj can be written as a ﬁnite union of pairwise disjoint d-boxes.]

(vi) The d-dimensional Jordan outer content µ∗ [Y ] of a bounded subset Y ⊂ Rd is deﬁned as ∑ J,d ∪ ∗ { k ∈ N ⊂ k } µJ,d[Y ] = inf n=1 V old(An): k , and An’s are d-boxes with Y n=1 An .

(vii) The d-dimensional Lebesgue outer measure µ∗ [Y ] of an arbitrary set Y ⊂ Rd is deﬁned as ∑ L,d ∪ ∗ { ∞ ⊂ ∞ } µL,d[Y ] = inf n=1 V old(An): An’s are d-boxes with Y n=1 An . ∗ ≤ ∗ ⊂ Rd ∗ ∗ [101] We have that µL,d[Y ] µJ,d[Y ] for any bounded set Y , and µL,d[A] = µJ,d[A] = V old(A) for any d-box A ⊂ Rd.

∪ ∪ k ∞ Proof. Any finite union n=1 An of d-boxes can be extended to an infinite union n=1 An of d-boxes without changing the total volume by taking An’s to be singletons for n > k. This observation ∗ ≤ ∗ ∗ yields that µL,d[Y ] µJ,d[Y ]. It is easy to see µJ,d[A] = V old(A) if A is a d-box. It remains ∗ to show µ [A] ≥ V old(A) when A is a d-box. First suppose A is closed. Then A is compact L,d ∪ d ∞ by Heine-Borel. Let ε > 0 and let A1,A2,... ⊂ R be d-boxes such that A ⊂ An and ∑ n=1 ∞ ∗ ∈ N ⊂ n=1 V old(An) < µL,d[A] + ε. For each n , let Bn be an open d-box with An Bn and n V old(Bn) < V old(An)+ε/2 . Then {Bn : n ∈ N} is an open cover for the compact set A. Extracting ∑ ∑ ≤ k ≤ ∞ n ∗ a finite subcover, we have V old(A) n=1 V old(Bn) n=1(V old(An) + ε/2 ) < µL,d[A] + 2ε. ∗ Thus µL,d[A] = V old(A) for closed d-boxes. Now if B is an arbitrary d-box and ε > 0, then there ⊂ − ∗ ≤ ∗ is a closed d-box A B with V old(B) ε < V old(A) = µL,d[A] µL,d[B].

Other basic properties of Lebesgue outer measure and Jordan outer content are given below.

∗ ∅ [102] (i) µL,d[ ] = 0. ∗ ≤ ∗ ⊂ ⊂ Rd (ii) [Monotonicity] µL,d[X] µL,d[Y ] if X Y . (iii) [Translation-invariance] µ∗ [Y + x] = µ∗ [Y ] for every Y ⊂ Rd and every x ∈ Rd. L,d L,d ∪ ∑ ⊂ Rd ∞ ∗ ≤ ∞ ∗ (iv) [Countable subadditivity] If Y1,Y2,... and Y = n=1 Yn, then µL,d[Y ] n=1 µL,d[Yn]. ∗ ⊂ Rd (v) µL,d[Y ] = 0 for every countable set Y . ⊂ Rd ∗ (vi) For any Y , we have µL,d[Y ] ∑∞ ∪∞ = inf{ V old(An): An’s are closed d-boxes with Y ⊂ An} ∑n=1 ∪ n=1 { ∞ ⊂ ∞ } = inf n=1 V old(An): An’s are open d-boxes with Y n=1 An . ⊂ Rd ∗ { ∗ ⊂ Rd} (vii) For any Y , we have µL,d[Y ] = inf µL,d[U]: Y U and U is open in . ∗ Rd ∞ (viii) µL,d[ ] = . ⊂ Rd {∥ − ∥ ∈ ∈ } ∗ ∪ (ix) If X,Y are such that dist(X,Y ) := inf x y : x X, y Y > 0, then µL,d[X Y ] = ∗ ∗ µL,d[X] + µL,d[Y ]. MEASURE THEORY 5

Proof. (i), (ii) and (iii) are clear. To prove (iv), without loss of generality we may assume ∑ ∪ ∞ ∗ ∞ µ [Yn] < ∞. Given ε > 0, there exist d-boxes A(n, k) such that Yn ⊂ A(n, k) and ∑n=1 L,d ∪ ∪ k=1 ∞ ∗ n ∞ ∞ V old(A(n, k)) < µ [Yn] + ε/2 . Then Y ⊂ A(n, k) and we have the estimate ∑k=1 ∑ L,d ∑ ( ) n=1∑ k=1 ∞ ∞ ≤ ∞ ∗ n ∞ ∗ n=1 k=1 V old(A(n, k)) n=1 µL,d[Yn] + ε/2 = ( n=1 µL,d[Yn]) + ε. Now (v) follows from (iv) since singletons have Lebesgue outer measure zero (or we can see it directly by noting that singletons are d-boxes with zero volume). The first part of (vi) is clear since any d-box and its closure have equal volume. To get the second part, note that if A1,A2,... are d-boxes and ε > 0, there exist open d-boxes B1,B2,... such that An ⊂ Bn and V old(Bn) < n V old(An) + ε/2 . We may deduce (vii) using part (vi) and Exercise-2. And (viii) follows from [101] and [102](ii). ∗ ∪ ≤ ∗ ∗ Now we prove (ix). From countable subadditivity, we have µL,d[X Y ] µL,d[X] + µL,d[Y ]. To prove the other inequality, we may assume µ∗ [X ∪ Y ] < ∞. Let δ = dist(X,Y ). Given ε > 0, ∪L,d ∑ ∪ ⊂ ∞ ∞ ∗ ∪ find d-boxes A1,A2,... such that X Y n=1 An and n=1 V old(An) < µL,d[X Y ] + ε. By partitioning the d-boxes into smaller d-boxes and throwing away the unnecessary ones, we may assume that diam[An] < δ and (X ∪ Y ) ∩ An ≠ ∅ for every n ∈ N. Let Γ = {n ∈ N : X ∩ An ≠ ∅} ∪ ′ ′ and Γ = {n ∈ N : Y ∩ An ≠ ∅}. Then N = Γ ∪ Γ is a disjoint union, X ⊂ ∈ An, and ∪ ∑ ∑ ∑ n Γ ⊂ ∗ ∗ ≤ ∞ Y n∈Γ′ An. Hence µL,d[X] + µL,d[Y ] n∈Γ V old(An) + n∈Γ′ V old(An) = n=1 V old(An) < ∗ ∪ µL,d[X Y ] + ε. ∗ ∅ [103] (i) µJ,d[ ] = 0. ∗ ≤ ∗ ⊂ Rd (ii) [Monotonicity] µJ,d[X] µJ,d[Y ] if X Y are bounded subsets of . ∗ ∗ ⊂ Rd ∈ Rd (iii) [Translation-invariance] µL,d[Y +x] = µL,d[Y ] for every bounded set Y and every x . ⊂ Rd ∗ ∪ ≤ ∗ ∗ (iv) [Finite subadditivity] If X,Y are bounded subsets, then µJ,d[X Y ] µJ,d[X] + µJ,d[Y ]. ∗ ⊂ Rd (v) µJ,d[Y ] = 0 for every finite set Y . (vi) For any bounded set Y ⊂ Rd, we have µ∗ [Y ] ∑ J,d ∪ k k = inf{ V old(An): k ∈ N, and An’s are closed d-boxes with Y ⊂ An} ∑n=1 ∪ n=1 k k = inf{ V old(An): k ∈ N, and An’s are open d-boxes with Y ⊂ An} ∑n=1 n=1 ∪ { k ∈ N ⊂ k } = inf n=1 V old(An): k , and An’s are pairwise disjoint d-boxes with Y n=1 An . (vii) If X,Y ⊂ Rd are bounded sets with dist(X,Y ) := inf{∥x − y∥ : x ∈ X, y ∈ Y } > 0, then ∗ ∪ ∗ ∗ µJ,d[X Y ] = µJ,d[X] + µJ,d[Y ]. ⊂ Rd ∗ ∗ (viii) For any bounded set Y , we have µJ,d[Y ] = µJ,d[Y ].

Proof. Statements (i) to (vii) can be proved as in [102]. To prove (viii), use the ﬁrst expression for ∗ µJ,d[Y ] in (vi) and note that a ﬁnite union of closed sets is closed.

Qd ∩ d ∗ ̸ ∗ Example: Let Y = [0, 1] . Note that µL,d[Y ] = 0 = 1 = µL,d[Y ]. But we have µJ,d[Y ] = ∗ µJ,d[Y ] = 1 by [103](viii). So the Jordan outer content of a bounded countable set need not be 6 T.K.SUBRAHMONIAN MOOTHATHU

∗ ∗ zero. This example also shows that µL,d[Y ] < µJ,d[Y ] is possible for a bounded set, and that the Jordan outer content does not satisfy countable subadditivity for bounded sets (since the Jordan d \ ∗ d outer content of a singleton is zero). If X = [0, 1] Y , then µJ,d[X] = 1 since X = [0, 1] and ∗ ∗ ̸ ∗ ∪ hence µJ,d[X] + µJ,d[Y ] = 2 = 1 = µJ,d[X Y ]. This says that the assumption dist(X,Y ) > 0 is necessary in [103](viii). .

Some ways of saying that Y ⊂ Rd is a small set: (i) Y is a countable set. (ii) Y is a discrete subset of Rd. (iii) Y is contained in a vector subspace of Rd of dimension ≤ d − 1. (iv) Y is nowhere dense in Rd. (v) Y is of ﬁrst category in Rd. (vi) Y has ﬁnite derived length. ∗ (vii) Y is a bounded set with µJ,d[Y ] = 0. ∗ (viii) µL,d[Y ] = 0.

It is good to investigate various possible implications between pairs of notions given above. Some are discussed below.

Exercise-3: If Y is a discrete subset of Rd, then Y is countable. [Hint: Let B = {B(x, 1/n): x ∈ Qd, n ∈ N}. Then, B is countable and for each y ∈ Y , there is B ∈ B such that B ∩ Y = {y}.]

Exercise-4: If Y is contained in a vector subspace W of Rd with dim(W ) ≤ d − 1, then Y is a nowhere dense subset of Rd.[Hint: W is closed in Rd (∵ ﬁx a basis for W and argue with the coeﬃcients of each basis vector separately) and W does not contain any open ball of Rd.]

Exercise-5: For any Y ⊂ Rd, the set Y \ Y (1) is discrete and hence countable. In particular, every uncountable subset of Rd has a limit point in Rd.[Hint: Let y ∈ Y \Y (1). If B(y, 1/n)∩Y contains a point other than y for every n ∈ N, then y ∈ Y (1), a contradiction.]

Exercise-6: If Y ⊂ Rd has ﬁnite derived length, then Y is countable. [Hint: Suppose Y (n) ≠ ∅ and Y (n+1) = ∅. Then Y (n) = Y (n) \ Y (n+1) must be countable by Exercise-5. Go backwards inductively.]

⊂ Rd ∗ Exercise-7: Every nonempty open set U contains a bounded countable set Y with µJ,d[Y ] > 0. ⊂ Rd ∗ Exercise-8: If Y is a bounded subset with finite derived length, then µJ,d[Y ] = 0. [Hint: Suppose that Y (n) ≠ ∅ and Y (n+1) = ∅, and let ε > 0. Find a compact set K ⊂ Rd with Y ⊂ K. Note that Y (n) must be finite using the compactness of K. Find finitely many open d-boxes of ∪ (n) (n−1) \ k A1,...,Ak of total volume < ε/2 covering Y . Then, observe that Y j=1 Aj is finite by MEASURE THEORY 7 the compactness of K. Next, find finitely many open d-boxes of total volume < ε/22 covering ∪ (n−1) \ k × { } ⊂ R2 Y j=1 Aj. Proceed backwards like this.] [Converse is not true: consider [0, 1] 0 .] ∗ Rd−1 × { } ⊂ Rd−1 − Exercise-9: µL,d[ 0 ] = 0. [Hint: Let B1,B2,... be countably many (d 1)- d−1 boxes covering R such that V old−1(Bn) = 1 for every n ∈ N. Given ε > 0, let An = Bn × ∪ ∑ ∑ − n+1 n+1 Rd−1 × { } ⊂ ∞ ∞ ∞ n [ ε/2 , ε/2 ]. Then 0 n=1 An and n=1 V old(An) = n=1 ε/2 = ε.] [With more effort we can prove the following: if Y is contained in a vector subspace W of Rd with ≤ − ∗ dim(W ) d 1, then µL,d[Y ] = 0.] ∪ Rd ∗ ∈ N ∗ ∞ Exercise-10: If Y1,Y2,... are such that µL,d[Yn] = 0 for every n , then µL,d[ n=1 Yn] = 0.

Now we discuss some interesting examples.

[104] Let K ⊂ [0, 1] be the middle-third Cantor set. Then, K is an uncountable, nowhere dense ∗ ∗ compact set with µJ,1[K] = µL,1[K] = 0. Moreover, K has no isolated points.

Proof. We recall the construction of K. Let K0 = [0, 1], K1 = [0, 1/3] ∪ [2/3, 1], K2 = [0, 1/9] ∪ n [2/9, 1/3]∪[2/3, 7/9]∪[8/9, 1], and so on. That is, Kn is the disjoint union of 2 closed subintervals n of [0, 1], each having length 1/3 , and Kn+1 is obtained from Kn by removing the middle-third open intervals from each of these 2n closed intervals. The middle-third Cantor set K is defined as ∩∞ K = n=0 Kn. Being the intersection of compact sets, K is compact. Since the maximal length n of an interval contained in Kn is (1/3) , K does not contain any open interval, and hence K is ⊂ ∗ ≤ n ∗ nowhere dense. Also, since K Kn, the above description yields µJ,1[K] (2/3) . So µJ,1[K] = 0 and hence µ∗ [K] = 0 also. L,1 ∑ { ∞ n ∈ { }} It may be verified that K = n=1 xn/3 : xn 0, 2 . That is, K is precisely the set of those x ∈ [0, 1] whose ternary expansion (i.e., base 3 expansion) x = 0.x1x2 ··· contains only 0’s and 2’s. Hence K is bijective with {0, 2}N which is uncountable. We show K has no isolated point. Let x ∈ K and let U be a neighborhood of x. Choose n large n enough so that one of the 2 closed intervals constituting Kn, say Jn, satisfies x ∈ Jn ⊂ U. Let y ∈ Jn \{x} be an end point of Jn. This end point is never removed in the later construction, so y ∈ Km for every m ≥ n. Thus y ∈ K ∩ (U \{x}).

Remark: It may be noted that for x ∈ K, the base 3 expansion x = 0.x1x2 ··· is eventually constant iﬀ x is an end point of a removed open interval. This helps to see that K contains points other than the end points of the (countably many) removed open intervals.

The following theorem is relevant while considering big and small sets in a topological sense.

[105] [Baire Category Theorem] Let (X, ρ) be a complete metric space and let Un ⊂ X be open ∩ ∩ ∈ N ∞ ∞ ̸ ∅ and dense in X for n . Then, n=1 Un is also dense in X. In particular, n=1 Un = . 8 T.K.SUBRAHMONIAN MOOTHATHU ∩ ⊂ ∩ ∞ ̸ ∅ Proof. Let V X be a nonempty open set. It suﬃces to show V ( n=1 Un) = . Since U1 is open and dense, U1 ∩ V is a nonempty open set. Let B1 be an open ball in X such that B1 ⊂ U1 ∩ V and diam[B1] < 1. Since U2 is open and dense, B1 ∩ U2 ≠ ∅. Let B2 ⊂ X be an open ball with

B2 ⊂ B1∩U2 and diam[B2] < 1/2. In general, let Bn+1 ⊂ X be an open ball with Bn+1 ⊂ Bn∩Un+1 and diam[Bn+1] < 1/(n+1). If xn is the center of the ball Bn, then we note that for every n, m ≥ k we have xn, xm ∈ Bk and hence ρ(xn, xm) ≤ diam[Bk] < 1/k. So (xn) is a Cauchy sequence. Since

(X, ρ) is complete, there is x ∈ X such that (xn) → x. Now, for any n, we have xm ∈ Bn for m ≥ n ∩ ∩ ∈ ∈ ∞ ⊂ ∩ ∞ and hence x Bn. Thus x n=1 Bn V ( n=1 Un).

Remarks: (i) By considering the complements of Un’s in the above, we get the following conclusion: if (X, ρ) is a complete metric space, then X cannot be written as a countable union of nowhere dense (closed) subsets of X. That is, X is of second category in itself. (ii) Since Rd is a complete metric space with respect to the Euclidean metric, Rd cannot be written as a countable union of nowhere dense (closed) subsets of Rd. (iii) From a topological point of view, a ﬁrst category subset is considered as a small set and a dense Gδ subset is considered as a big set because of Baire Category Theorem. However, a set that is topologically big (small) need not be measure theoretically big (small); we will see this in [106] below. (iv) The uncountability of the middle-third Cantor set can be proved with the help of Baire Category Theorem also, see the exercise below.

Exercise-11: Let (X, ρ) be a complete metric space without isolated points. Then X must be uncountable. [Hint: If X is countable, write X = {x1, x2,...} and let Un = X \{xn}. Then Un ∩ { } ∵ ∅ is open, and Un is dense since xn is not open ( xn is not an isolated point). But n Un = , a contradiction.]

Exercise-12: It is not possible to write Qd as a countable intersection of open subsets of Rd.[Hint: If ∩ d ∞ d d d possible, let Q = Un, where Un is open in R . Write Q = {x1, x2,...} and let Vn = R \{xn}. ∩ n∩=1 ∞ ∩ ∞ ∅ Then, ( n=1 Un) ( n=1 Vn) = , contradicting Baire Category Theorem.]

We observe in the following the distinction between topological bigness (smallness) and measure theoretical bigness (smallness).

⊂ Rd ∗ [106] (i) For every ε > 0, there is a dense open set U such that µL,d[U] < ε. ⊂ Rd ∗ (ii) There is a dense Gδ subset Y with µL,d[Y ] = 0. ⊂ Rd ∗ ∞ ∗ Rd \ (iii) There is an Fσ set X of ﬁrst category with µL,d[X] = and µL,d[ X] = 0. (iv) For every closed d-box A and every ε > 0, there is a nowhere dense compact set K ⊂ Rd such ⊂ ∗ − that K A and µL,d[K] > V old(A) ε.

d Proof. (i) Write Q = {x1, x2,...}. For each n ∈ N, let An be an open d-box with xn ∈ An and ∪ n ∞ V old(An) < ε/2 . Put U = n=1 An. MEASURE THEORY 9 ∩ ⊂ Rd ∗ ∞ (ii) Let Un be a dense open subset with µL,d[Un] < 1/n and put Y = n=1 Un. (iii) Let Y be as in (ii) and take X = Rd \ Y . (iv) Let U be as in (i) and let K = A \ U.

The next result shows that the Lebesgue outer measure does not satisfy ﬁnite additivity (and hence it does not satisfy countable additivity), even though it satisﬁes countable subadditivity.

[107] Let X = Qd ∩ [0, 1]d. Then, there is a subset Y ⊂ [0, 1]d satisfying the following: (i) The translations Y + x are pairwise disjoint for x ∈ X. ∪ ∗ n (ii) There exist ﬁnitely many distinct elements x1, . . . , xn ∈ X such that µ [ (Y + xi)] ≠ ∑ L,d i=1 n ∗ i=1 µL,d[Y + xi].

Proof. Define an equivalence relation on [0, 1]d by the condition that a ∼ b iff a − b ∈ Qd. By the axiom of choice, we can form a set Y ⊂ [0, 1]d whose intersection with each equivalence class is a singleton. (i) We verify that (Y + r) ∩ (Y + s) = ∅ for any two distinct r, s ∈ X. If (Y + r) ∩ (Y + s) ≠ ∅ for r, s ∈ X, then there are a, b ∈ Y such that a + r = b + s. Now we have a − b = s − r ∈ Qd, and hence a ∼ b. So we must have a = b by the definition of Y , and then necessarily r = s. (ii) If z ∈ Rd, there is r ∈ Qd such that z − r ∈ [0, 1]d. Then there is y ∈ Y such that y ∼ z − r and ∪ ′ ∈ Qd ′ − ′ Rd so there is r such that y + r = z r or z = y + r + r . This shows that = r∈Qd (Y + r). By [102](iii) and [102](viii), we conclude that µ∗ [Y ] > 0. Let δ = µ∗ [Y ] and n ∈ N be such that L,d ∑ L,d d n ∗ d nδ > 2 . Choose distinct elements x1, . . . , xn ∈ X. Then µ [Y + xi] = nδ > 2 again by i=1 L,d ∪ ⊂ d ∗ n ≤ d translation invariance. On the other hand, Y +X [0, 2] and therefore µL,d[ i=1(Y +xi)] 2 .

Remark: The construction above is due to Vitali, and hence the set Y is called a Vitali set.

3. About functions behaving nicely outside a small set

There are a few classical results in Analysis with conclusion of the following form: “... the function has nice behavior outside a small set”. We will consider some such results here. We know that a function that is the pointwise limit of a sequence of continuous functions may not be continuous. For instance, f : [0, 1] → R given by f(1) = 1 and f(x) = 0 for x < 1 is n the pointwise limit of (fn), where fn : [0, 1] → R is fn(x) = x . However, the set of points of discontinuity of the limit function must be a small set often, see [108] below.

Definition: Let X,Y be metric spaces and let f : X → Y be a function. Then the oscillation ω(f, x) of f at a point x ∈ X is defined as ω(f, x) = limδ→0+ diam[f(B(x, δ))]. Clearly, f is continuous at x iff ω(f, x) = 0. 10 T.K.SUBRAHMONIAN MOOTHATHU

Exercise-13: Let X,Y be metric spaces and let f : X → Y be a function. Then the set {x ∈ ∩ } ∞ X : f is continuous at x is a Gδ subset of X.[Hint: The given set is equal to n=1 Un, where

Un = {x ∈ X : ω(f, x) < 1/n}, and Un is open.]

Exercise-14: There does not exist any function f : Rd → R such that {x ∈ Rd : f is continuous at x} = Qd.[Hint: Use Exercise-12 and Exercise-13.] [Remark: However, for f : R → R deﬁned as f(0) = 1, f(x) = 0 for x ∈ R \ Q, and f(p/q) = 1/q for p ∈ Z \{0}, q ∈ N, and p, q are relatively prime, satisﬁes {x ∈ R : f is continuous at x} = R \ Q.]

[108] Let (X, ρ1) be a complete metric space, (Y, ρ2) be an arbitrary metric space, and let (fn) be a sequence of continuous functions from X to Y , converging pointwise to a function f : X → Y .

Then the set {x ∈ X : f is continuous at x} is a dense Gδ subset of X.

Proof. Let ε > 0 and Dε = {x ∈ X : ω(f, x) ≥ ε}. We know from the hint of Exercise-13 that Dε is a closed set. We claim that Dε is nowhere dense in X. Let U ⊂ X be a nonempty open set. We have to ﬁnd a nonempty open set V ⊂ U such that Dε ∩ V = ∅.

For n ∈ N, let Kn = {x ∈ X : ρ2(fn(x), fj(x)) ≤ ε/4 for every j ≥ n}. Then Kn is a closed set ∪ ∞ ≤ and X = n=1 Kn. The continuity of the distance function ρ2 implies that ρ2(fn(x), f(x)) ε/4 for every x ∈ Kn. Let U1 ⊂ X be a nonempty open set with U1 ⊂ U. Since (U1, ρ1) is a complete metric space, there is n ∈ N such that U2 := int[Kn ∩ U1] ≠ ∅. Let V ⊂ U2 be a nonempty open set with diam[fn(V )] ≤ ε/4. For any x, y ∈ V , we have ρ2(f(x), f(y)) ≤ ρ2(f(x), fn(x)) +

ρ2(fn(x), fn(y))+ρ2(fn(y), f(y)) ≤ ε/4+ε/4+ε/4 = 3ε/4. Hence diam[f(V )] ≤ 3ε/4 and therefore

ω(f, x) ≤ 3ε/4 for every x ∈ V . This shows Dε ∩ V = ∅, proving our claim. ∪∞ The claim implies that D := n=1 D1/n is an Fσ set of ﬁrst category in X. This completes the proof since {x ∈ X : f is continuous at x} = X \ D, and X is a complete metric space.

We know that the derivative of a diﬀerentiable real function need not be continuous. However, we can say the following.

[109] Let f : R → R be diﬀerentiable. Then there exists a sequence (gn) of continuous functions from R to R converging to f ′ pointwise. Consequently, {x ∈ R : f ′ is continuous at x} is a dense

Gδ subset of R.

Proof. Since f is diﬀerentiable, f is continuous. Deﬁne gn(x) = [f(x + 1/n) − f(x)]/(1/n) and use [108].

Now we will show that a monotone real function (increasing or decreasing) is continuous and diﬀerentiable at most of the points. MEASURE THEORY 11

[110] Let −∞ ≤ a < b ≤ ∞ and let f :(a, b) → R be a monotone function. Then, Y = {x ∈ (a, b): f is discontinuous at x} is a countable set (possibly empty).

Proof. Suppose f is increasing. If x ∈ Y , then necessarily f(x−) < f(x+), and we may choose a rational number between f(x−) and f(x+). This gives a one-one map from Y to Q.

To talk about the points of diﬀerentiability of a monotone real function, we need a technical lemma due to Vitali.

Deﬁnition: A collection Γ of nondegenerate intervals is a Vitali cover for a set X ⊂ R if for each ε > 0, the subcollection {I ∈ Γ : 0 < |I| < ε} is also a cover for X.

⊂ R ∗ ∞ [111] [Vitali’s covering lemma] Let X be such that µL,1[X] < and let Γ be a collection of intervals forming a Vitali cover for X. Then, ∪ ∈ ∗ \ (i) There are countably many pairwise disjoint intervals I1,I2,... Γ such that µL,1[X n In] = 0.

(ii) For every ε > 0, there exist ﬁnitely many pairwise disjoint intervals I1,...,Ik ∈ Γ with the ∪ ∗ \ k property that µL,1[X n=1 In] < ε.

∗ ∗ Proof. Write µ = µL,1 for simplicity. (i) With out loss of generality assume that every I ∈ Γ is a (nondegenerate) closed interval. By [102](vi), choose an open set U ⊂ R such that X ⊂ U and µ∗[U] < ∞. Every x ∈ X has a neighborhood contained in U. Hence Γ′ = {I ∈ Γ: I ⊂ U} is also a vitali cover for X. We will ′ ∗ choose the intervals In inductively. Let δ0 = sup{|J| : J ∈ Γ } (note that δ0 ≤ µ [U] < ∞) and let ′ I1 ∈ Γ be any interval with |I1| > δ0/2. Suppose that we have chosen pairwise disjoint intervals ∪ ∪ ′ n n I1,...,In ∈ Γ . If X ⊂ Ii, then we are done. Else, any x ∈ X \ Ii is at a positive i=1∪ i=1 n {| | ∈ ′ ∩ ∅ ≤ ≤ } distance from the closed set i=1 Ii. Let δn = sup J : J Γ and Ii J = for 1 i n . ∗ ′ Then 0 < δn ≤ µ [U] < ∞. Let In+1 ∈ Γ be an interval with |In+1| > δn/2. We will show that the sequence (In) does the job. ∑ ′ ∗ Observation: For every J ∈ Γ , there is n ∈ N such that In ∩ J ≠ ∅ (∵ |In| ≤ µ [U] < ∞ so that

(|In|) → 0, and hence there is n ∈ N such that |In| < |J|/2). ∪ \ ∞ ∗ Let Y = X n=1 In and ε > 0. We claim that µ [Y ] < ε. Let cn be the midpoint of In and let ′ Yn ⊂ R be the closed interval with midpoint cn and |Yn| = 6|In| (this Yn may not be in Γ ). Let ∑∞ k ∈ N be so that |In| < ε/6. If x ∈ Y , then in particular x does not belong to the closed ∪ n=k+1 k ∈ ′ ∈ ∩ ∅ ≤ ≤ set n=1 In. Choose J Γ with x J and In J = for 1 n k. By our observation above,

Im ∩ J ≠ ∅ for some m ≥ k + 1. Let m be the smallest such number. Then |J| ≤ δm−1 < 2|Im| ∪∞ and hence |x − cm| ≤ |J| + |Im| ≤ 3|Im|. Therefore, x ∈ Ym. We have shown that Y ⊂ Yn. ∑ ∑ n=k+1 ∞ | | ≤ ∞ | | ∗ Since n=k+1 Yn 6 n=k+1 In < ε, we have proved that µ [Y ] < ε. 12 T.K.SUBRAHMONIAN MOOTHATHU

Now, note that the argument given above actually shows that for every ε > 0, there is k ∈ N ∪ ∗ \ k such that µ [X n=1 In] < ε. Hence we have established (ii) also.

When a mathematical problem is diﬃcult, it is a good idea to divide the problem into many subcases and to treat each case separately. If f :(a, b) → R is a function, then the four Dini derivatives of f at a point x ∈ (a, b) are deﬁned as follows. f(x + h) − f(x) D+f(x) = lim sup [upper right derivative] h→0+ h f(x + h) − f(x) D+f(x) = lim inf [lower right derivative] h→0+ h f(x + h) − f(x) D−f(x) = lim sup [upper left derivative] h→0− h f(x + h) − f(x) D−f(x) = lim inf [lower left derivative]. h→0− h [ ] f(x + h) − f(x) f(x + h) − f(x) Remark: Here, lim sup := lim sup , and similarly the others. h→0+ h y→0+ 0

Example: Let f :(−1, 1) → R be f(0) = 0 and f(x) = x sin(1/x) for x ≠ 0. Then, D+f(0) = 1 = − D f(0) and D+f(0) = −1 = D−f(0) so that f is not diﬀerentiable at 0.

[112] [Lebesgue’s diﬀerentiation theorem] Let −∞ ≤ a < b ≤ ∞, let f :(a, b) → R be a monotone { ∈ } ∗ function and let Y = x (a, b): f is not diﬀerentiable at x . Then µL,1[Y ] = 0.

Proof. Since (a, b) can be written as a countable union of bounded open intervals, we may as well assume (a, b) itself is bounded in view of Exercise-10. Assume f is increasing and write ∗ ∗ ∪ { ∈ + } µ = µL,1. By the remark above, Y = Y1 Y2, where Y1 = x (a, b): D−f(x) < D f(x) and − ∗ Y2 = {x ∈ (a, b): D+f(x) < D f(x)}. We will only show that µ [Y1] = 0; the case of Y2 is similar. 2 + Let Γ = {(r, s) ∈ Q : r < s}, let X(r, s) = {x ∈ (a, b): D−f(x) < r < s < D f(x)} and note ∪ ∗ ∈ that Y1 = (r,s)∈Γ X(r, s). Hence it suﬃces to show there µ [X(r, s)] = 0 for every (r, s) Γ, by Exercise-10. Fix (r, s) ∈ Γ, write X = X(r, s) and let ε > 0 be arbitrary. Choose an open set U ⊂ (a, b) such that X ⊂ U and µ∗[U] < µ∗[X] + ε by [102](vi).

Since D−f < r on X, for each x ∈ X and δ > 0 we can find a nondegenerate closed interval I(x, δ) = [x − α, x] ⊂ U such that 0 < α < δ and f(x) − f(x − α) < rα. Then Γ = {I(x, δ): x ∈ X, δ > 0} is a Vitali cover for X. By Vitali’s lemma, we can find finitely many pairwise disjoint ∪ ∈ ∗ \ k | | intervals I1,...,Ik Γ such that µ [X n=1 In ] < ε. MEASURE THEORY 13 ∪ k ⊂ ∗ − ∗ ≤ ∗ ∗ Let V = n=1 int[In]. Then, V is open, V U, and µ [X] ε < µ [V ] µ [U] < µ [X] + ε. Let X′ = V ∩ X. Since D+f > s on X, and hence on X′, for each y ∈ X′ and δ > 0 we can find a nondegenerate closed interval J(y, δ) = [y, y + β] ⊂ V (hence J(y, δ) ⊂ In for some n ∈ {1, . . . , k}) such that 0 < β < δ, and f(y +β)−f(y) > sβ. Then Γ = {J(y, δ): y ∈ X′, δ > 0} is a Vitali cover ′ for X . Again by Vitali’s lemma, we can find finitely many pairwise disjoint intervals J1,...,Jm ∈ Γ ∪ ∑ ∗ ′ \ m | | m | | ≥ ∗ ′ − ≥ ∗ − such that µ [X j=1 Jj ] < ε. Then j=1 Jj µ [X ] ε µ [X] 2ε.

Write In = [xn −αn, xn] and Jj = [yj, yj +βj]. For each n ∈ {1, . . . , k}, let Dn = {j ∈ {1, . . . , m} :

Jj ⊂ In}. Then {1, . . . , m} is the disjoint union of Dn’s. ∑ Note that ∈ (f(yj + βj) − f(yj)) ≤ f(xn) − f(xn − αn) for each n ∈ {1, . . . , k} since f j Dn ∑ ∑ m k is increasing. Summing over n, we get (f(yj + βj) − f(yj)) ≤ (f(xn) − f(xn − αn)), ∑ ∑ ∑ j=1 ∑ n=1 m k m | | k | | and hence j=1 sβj < n=1 rαn, or s( j=1 Jj ) < r( n=1 In ). From the earlier estimates we conclude that s(µ∗[X] − 2ε) < r(µ∗[X] + ε). Since ε > 0 was arbitrary and r < s, we must have µ∗[X] = 0.

The conclusion of [112] can be extended to more general class of real functions.

Deﬁnition: If f :[a, b] → R is a function and P = {a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b} is a ∑ b n | − | b partition of [a, b], let Va (f, P ) = i=1 f(ai) f(ai−1) . Deﬁne the total variation of f as Va (f) = { b } b ∞ sup Va (f, P ): P is a partition of [a, b] . We say f is of bounded variation if Va (f) < . It is easy to see that if f is of bounded variation, then f is bounded (∵ if x ∈ [a, b], take P = {a ≤ x ≤ b} to | − | ≤ b see that f(x) f(a) Va (f)).

Remark: Imagine a particle moving along the graph of f from the point (a, f(a)) to the point (b, f(b)) (the particle has to make a jump wherever there is a discontinuity in the graph). Loosely speaking, f is of bounded variation iff the total distance travelled by the particle along the graph of f is finite (∵ the total variation of f is just the vertical variation; but since the horizontal distance b − a is finite, the above interpretation is justified).

→ R b | − | Examples: (i) If f :[a, b] is a monotone function, then Va (f, P ) = f(b) f(a) for any b | − | ∞ partition P of [a, b] and hence Va (f) = f(b) f(a) < . So f is of bounded variation. (ii) Suppose f :[a, b] → R is Lipschitz continuous (this happens if f is C1) with Lipschitz constant b ≤ − ∞ λ > 0. Then, it may be seen that Va (f) λ(b a) < and hence f is of bounded variation.

Non-example: A (uniformly) continuous function f :[a, b] → R need not be of bounded variation. Let f : [0, 1] → R be the (uniformly) continuous function deﬁned as f(0) = 0 and f(x) = x sin(1/x) if x ∈ (0, 1]. Let ak = 2/kπ ∈ [0, 1] for k ∈ N. Observe that |f(a2k)−f(a2k−1)| = |0−a2k−1| = a2k−1. ∑ ∈ N { ≤ ≤ ≤ · · · ≤ ≤ } 1 ≥ m | − Let m and P = 0 a2m a2m−1 a1 1 . Then V0 (f, P ) k=1 f(a2k) 14 T.K.SUBRAHMONIAN MOOTHATHU ∑ ∑ | m m − −1 → ∞ → ∞ 1 ∞ f(a2k−1) = k=1 a2k−1 = (2/π) k=1(2k 1) as m . Hence V0 (f) = , and thus f is not of bounded variation. This example also shows that bounded ; bounded variation.

Remark: If f, g :[a, b] → R are of bounded variation, r, s ∈ R, and h :[a, b] → R is de- b ≤ | | b | | b ∞ { → R ﬁned as h = rf(x) + sg(x), then Va (h) r Va (f) + s Va (g) < . Hence f :[a, b] : f is of bounded variation } is a real vector space (in fact, it is a normed space with the norm ∥ ∥ | | b f = f(a) + Va (f)).

Exercise-15: If f, g :[a, b] → R are of bounded variation, then fg is of bounded variation. [Hint:

Let M > 0 be such that |f|, |g| ≤ M. Now, subtracting and adding the term f(ai)g(ai−1), note that |(fg)(ai) − (fg)(ai−1)| ≤ |f(ai)||g(ai) − g(ai−1)| + |f(ai) − f(ai−1)||g(ai−1)| and hence b ≤ b b Va (fg) M(Va (f) + Va (g)).] → R ∈ b c b Exercise-16: If f :[a, b] is a function and c [a, b], then Va (f) = Va (f)+Vc (f). [Hint: If P1 is c b b ∪ ≤ b a partition of [a, c] and P2 is a partition of [c, b], then Va (f, P1)+Vc (f, P2) = Va (f, P1 P2) Va (f). Conversely, if P is a partition of [a, b], ﬁrst reﬁne it by inserting c and then divide into partitions b ≤ c b ≤ c b P1 of [a, c] and P2 of [c, b]. Check that Va (f, P ) Va (f, P1) + Vc (f, P2) Va (f) + Vc (f).]

[113] A function f :[a, b] → R is of bounded variation iﬀ there exist monotone functions g, h : [a, b] → R such that f(x) = g(x) − h(x) for every x ∈ [a, b]. Consequently, for any function → R ∗ { ∈ } f :[a, b] of bounded variation, we have µL,1[ x [a, b]: f is not diﬀerentiable at x ] = 0.

Proof. Suppose f = g −h, where g, h are monotone. Since g, h are of bounded variation, f is also of bounded variation since the collection of functions of bounded variation on [a, b] is a vector space. → R x Conversely assume that f is of bounded variation and deﬁne g :[a, b] as g(x) = Va (f). Then g is monotone increasing. Let h = g − f, and consider points x < y in [a, b]. By Exercise-16, we y have g(y) − g(x) = Vx (f) ≥ |f(y) − f(x)| ≥ f(y) − f(x), and therefore h(y) ≥ h(x). Thus h is also monotone increasing. Clearly, f = g − h.

The result below and related discussions are left as a reading assignment (see page 108 of H.L. Royden, Real Analysis, 3rd Edn.).

[114] Let [a, b] be a compact interval. Then for a function f :[a, b] → R, we have the following implications: f is Lipschitz continuous ⇒ f is absolutely continuous ⇒ f is of bounded variation. ∗ Consequently, if f is either Lipschitz continuous or absolutely continuous, then µL,1[Y ] = 0, where Y = {x ∈ [a, b]: f is not diﬀerentiable at x}.

Remark: However, there is a limit to these type of results; there are continuous functions f : [a, b] → R which are not diﬀerentiable at any point (see Theorem 7.18 of W. Rudin, Principles of Mathematical Analysis). MEASURE THEORY 15

Now we mention a characterization of Riemann integrable functions in terms of small sets. For simplicity, we restrict ourselves to dimension one, even though the corresponding result is true in higher dimensions as well. If f :[a, b] → R is a bounded function and if P = {a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b} is a partition of [a, b], let Mi = sup{f(x): ai−1 ≤ x ≤ ai} and mi = inf{f(x): ai−1 ≤ x ≤ ai}. The upper and lower Riemann sums with respect to the partition ∑ ∑ n − n − P are deﬁned as U(f, P ) = i=1 Mi(ai ai−1) and L(f, P ) = i=1 mi(ai ai−1). A bounded function f :[a, b] → R is said to be Riemann integrable if for every ε > 0 there is a partition P of [a, b] such that U(f, P ) − L(f, P ) < ε. The following characterization says that a Riemann integrable function is not very diﬀerent from a continuous function.

[115] Let f :[a, b] → R be a bounded function and let Y = {x ∈ [a, b]: f is not continuous at x}. ∗ Then, f is Riemann integrable iﬀ µL,1[Y ] = 0.

Proof. Let ω(f, x) be the oscillation of f at x defined earlier. ∪ ⇒ ∞ { ∈ ≥ } ∗ : Since Y = k=1 Yk, where Yk = x [a, b]: ω(f, x) 1/k , it suffices to show µL,1[Yk] = 0 for every k ∈ N. Fix k ∈ N and let ε > 0. Let P = {a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b} be a partition of [a, b] with U(f, P ) − L(f, P ) < ε/2k. Let Ai = (ai−1, ai) and Γ = {1 ≤ i ≤ n : Yk ∩ Ai ≠ ∅}. ∪ ′ ′′ ′ Note that Mi − mi ≥ 1/k for i ∈ Γ. Write Yk = Y ∪ Y , where Y = Yk ∩ ( ∈ Ai) and k k ∑ k i Γ∑ ′′ Y = Yk ∩{a0, a1, . . . , an}. We have ε/2k > U(f, P )−L(f, P ) ≥ ∈ (Mi−mi)|Ai| ≥ 1/k ∈ |Ai| k ∑ i Γ i Γ ′′ and hence ∈ |Ai| < ε/2. And since Y is a finite set, there are finitely many intervals B1,...,Bm i Γ ∪ ∑ k ∪ ∪ ∑ ′′ m m m such that Y ⊂ Bj and |Bj| < ε/2. Thus Yk ⊂ [ ∈ Ai] ∪ [ Bj] and ∈ |Ai| + ∑ k j=1 j=1 i Γ j=1 i Γ m | | ∗ j=1 Bj < ε. Since ε > 0 was arbitrary, µL,1[Yk] = 0. ⇐: Let ε > 0 be given. We have to find a partition P of [a, b] such that U(f, P ) − L(f, P ) < ε. Let λ = sup{|f(x)| : x ∈ [a, b]} and let ε′ = ε/[2λ + 2(b − a)]. For each x ∈ [a, b] \ Y , choose an open interval A(x) ⊂ R containing x such that |f(x) − f(z)| < ε′ for every z ∈ [a, b] ∩ A(x) by continuity. ∪ ∑ ⊂ ∞ ∞ | | ′ Also choose countably many open intervals Bm such that Y m=1 Bm and m=1 Bm < ε .

Then {A(x): x ∈ [a, b] \ Y } ∪ {Bm : m ∈ N} is an open cover for the compact set [a, b]. Extract a

ﬁnite subcover {A(xj) : 1 ≤ j ≤ p}∪{Bm : 1 ≤ m ≤ q}. The end points inside [a, b] of these ﬁnitely many intervals determine a partition P = {a0 = a ≤ a1 ≤ · · · ≤ an−1 ≤ an = b} of [a, b]. Observe that for each i ∈ {1, . . . , n}, we have [ai−1, ai] ⊂ A(xj) for some j ∈ {1, . . . , p}, or [ai−1, ai] ⊂ Bm for ′ some m ∈ {1, . . . , q}. Let Γ = {1 ≤ i ≤ n :[ai−1, ai] ⊂ A(xj) for some j} and Γ = {1, . . . , n}\ Γ. ∑ ∑ ′ Note that Mi−mi ≤ 2ε if i ∈ Γ. Hence U(f, P )−L(f, P ) ≤ ∈ (Mi−mi)(ai−ai−1) + ∈ ′ (Mi− ∑ ∑ i Γ∑ ∑ i Γ − ≤ ′ − − ≤ ′ n − q | | mi)(ai ai−1) 2ε i∈Γ(ai ai−1)+2λ i∈Γ′ (ai ai−1) 2ε i=1(ai ai−1)+2λ m=1 Bm < 2ε′(b − a) + 2λε′ = ε. 16 T.K.SUBRAHMONIAN MOOTHATHU

A corollary of [115] is that any bounded function f :[a, b] → R with at most countably many points of discontinuity (in particular, any continuous function) is Riemann integrable. The higher dimensional generalization of [115] can be stated as follows.

[115′] Let A ⊂ Rd be a d-box, let f : A → R be a bounded function and let Y be the set { ∈ } ∗ x A : f is not continuous at x . Then, f is Riemann integrable iﬀ µL,d[Y ] = 0.

Example: Let f : [0, 1] → R be f(0) = 0 and f(x) = sin (1/x) for x ≠ 0. Even though the graph of f has inﬁnitely many ups and downs (in fact, f is not of bounded variation), f is Riemann integrable since f is bounded and is discontinuous only at one point, namely 0.

Exercise-17: Let f :[a, b] → R be a bounded function. If f is either monotone or of bounded variation, then f is Riemann integrable. [Hint: Use [110], [113], and [115].]

Exercise-18: If f, g;[a, b] → R are Riemann integrable, then max{f, g}, min{f, g}, and af + bg (where a, b ∈ R) are also Riemann integrable. [Hint: The set of discontinuities of max{f, g} is contained in {x : f is not continuous at x} ∪ {x : g is not continuous at x}.]

1 Deﬁnition: Let X be a set and A ⊂ X. The indicator function 1A : X → R of the subset A is  1, if x ∈ A, deﬁned as 1A(x) =  0, if x ∈ X \ A. Example: We discuss an example that illustrates the main drawback of Riemann integration theory.

Write [0, 1] ∩ Q = {r1, r2,...}, let fn : [0, 1] → R be the indicator function of {r1, . . . , rn}, and let f : [0, 1] → R be the indicator function of [0, 1] ∩ Q. We have 0 ≤ f1 ≤ f2 ≤ · · · ≤ f ≤ 1 and the sequence (fn) converges to f pointwise. Each fn is Riemann integrable since fn is discontinuous only at ﬁnitely many points. But f is discontinuous at every point of [0, 1], and the Lebesgue outer measure of [0, 1] is positive. Hence f is not Riemann integrable by [115]. Thus even the pointwise limit of a uniformly bounded, monotone sequence of Riemann integrable functions need not be Riemann integrable.

Remark: Let f : [0, 1] → R be the indicator function of [0, 1] ∩ Q. Since f is not continuous at any point, it is not possible to realize f as the pointwise limit of a sequence of continuous functions from [0, 1] to R, in view of [108].

Remark: Let (fn) be a sequence of continuous functions from [a, b] to R converging pointwise to a function f :[a, b] → R, and let Y = {x ∈ [a, b]: f is not continuous at x}. From [108] we know that Y is an Fσ set of ﬁrst category in [a, b]. But Y can have positive outer Lebesgue measure by

1 The indicator function of a set A is also called the characteristic function of A and is also denoted as χA. MEASURE THEORY 17

[106]. Hence f may not be Riemann integrable. Thus even the pointwise limit of a sequence of continuous functions may not be Riemann integrable (of course, we did not give an example).

Remark: Lebesgue integration theory is developed not just for the sake of making the indicator function of [0, 1] ∩ Q integrable. The limit theorems in Lebesgue’s theory allow us to integrate the pointwise limit of a sequence of integrable functions, and to interchange limit and integration, under very mild hypothesis. Moreover, the powerful tools in Lebesgue’s theory make many proofs simpler (eg: the proof of the change of variable theorem in d-dimension), and provide us with new ways of dealing with functions (eg: Lp spaces). Also, as we will see later, in Lebesgue’s theory we have a more satisfactory version of the Fundamental Theorem of Calculus (describing diﬀerentiation and integration as inverse operations of each other).

4. σ-algebras and measurable spaces

A d-box in Rd has a well-deﬁned d-dimensional volume. We may ask whether it is possible to deﬁne the notion of a d-dimensional volue for all subsets of Rd. Of course, we would like to have consistency conditions such as monotonicity and countable additivity.

Question: Can we have a function µ : P(Rd) → [0, ∞] such that d (i) µ[A] = V old(A) if A ⊂ R is a d-box, (ii) [Monotonicity] µ[A] ≤ µ[B] for subsets A, B of Rd with A ⊂ B, ∪ ∑ ∞ ∞ Rd (iii) [Countable additivity] µ[ n=1 An] = n=1 µ[An] if An’s are pairwise disjoint subsets of ? ∗ Remark: From [107] we know that the Lebesgue outer measure µL,d does not satisfy countable ∗ additivity. The key observation of Lebesgue’s theory is that µL,d will satisfy all the three conditions ∗ A ⊂ P Rd stated above if we restrict µL,d to a slightly smaller collection ( ) by discarding some pathological subsets of Rd. In order to describe the structure of this smaller collection A, it is convenient to proceed in an abstract manner, which we do below.

Deﬁnition: Let X be a nonempty set. A collection A ⊂ P(X) of subsets of X is said to be a σ-algebra on X if the following hold: (i) ∅,X ∈ A. (ii) A ∈ A ⇒ X \ A ∈ A. ∪ ∈ A ⇒ ∞ ∈ A (iii) A1,A2,... n=1 An . If A is a σ-algebra on X, then (X, A) is called a measurable space.

Example: {∅,X} and P(X) are trivial examples of σ-algebras on any nonempty set X. The following are some σ-algebras on Rd (verify): d d A1 = {A ⊂ R : A or R \ A is countable}, 18 T.K.SUBRAHMONIAN MOOTHATHU

d d d A2 = {A ⊂ R : A or R \ A is of ﬁrst category in R }, A { ⊂ Rd ∗ ∗ Rd \ } 3 = A : µL,d[A] = 0 or µL,d[ A] = 0 . d d d d A4 = {A ⊂ R : [0, 1] ⊂ A or [0, 1] ⊂ R \ A}.

Exercise-19: If A is a σ-algebra on a set X show that (i) A \ B,A∆B ∈ A if A, B ∈ A, (ii) ∩ ∞ ∈ A ∈ A n=1 An if A1,A2,... .

Exercise-20: Let A = {A ⊂ Rd : A is a countable (possibly ﬁnite or empty) union of d-boxes}. Is A a σ-algebra on Rd?[Hint: Let d = 1. Consider Q and R \ Q, or the middle-third Cantor set and its complement.]

d d d d Exercise-21: Are the following σ-algebras on R : A1 = {A ⊂ R : A or R \ A is open in R } and d d d A2 = {A ⊂ R : A or R \ A is dense in R }? ∩ A A { ⊂ ∈ A } Exercise-22: If α’s are σ-algebras on a set X, then α α := A X : A α for every α is also a σ-algebra on X.

Deﬁnition: Let X be a nonempty set and C ⊂ P(X) be a collection of subsets of X.A σ-algebra A on X is said to be generated by C if A is the smallest σ-algebra on X containing C. Here, A exists and is unique since A is precisely the intersection of all σ-algebras on X containing C (note that there is at least one σ-algebra on X containing C, namely P(X)).

Deﬁnition: Let X be a metric space. Then the σ-algebra on X generated by the collection of all open subsets of X is called the Borel σ-algebra on X, and is denoted as B(X) (or just B, if X is clear from the context). The subsets of X belonging to B(X) are called Borel subsets of X. For example, open subsets, closed subsets, Gδ subsets and Fσ subsets of X are Borel subsets of X.

[116] [Characterizations of the Borel σ-algebra on Rd] Consider the following collections of subsets of Rd: d C1 = {A ⊂ R : A is closed}, d C2 = {A ⊂ R : A is compact}, d C3 = {A ⊂ R : A is closed d-box}, d C4 = {A ⊂ R : A is an open d-box}, d C5 = {A ⊂ R : A is a d-box}, d C6 = {A ⊂ R : A is an open ball}, −1 d C7 = {f (W ): f : R → R is continuous and W ⊂ R is open}. d d If Ai is the σ-algebra on R generated by Ci for 1 ≤ i ≤ 7, then Ai = B(R ) for 1 ≤ i ≤ 7.

∪∞ Proof. Clearly A ⊂ A ⊂ A = B(Rd). Since (a, b) = [a + 1/n, b − 1/n] (where n is chosen 3 2 1 n=n0 0 so that a + 1/n0 ≤ b − 1/n0), it follows that any open d-box is a countable union of closed d-boxes, MEASURE THEORY 19 ∩ ∩ A ⊂ A ∞ − ∞ − and therefore 4 3. Since [a, b] = n=1(a 1/n, b + 1/n), [a, b) = n=1(a 1/n, b), and ∩∞ (a, b] = n=1(a, b+1/n), we deduce that any d-box is a countable intersection of open d-boxes, and d hence A4 = A5. Since any open set in R can be written as a countable union of open d-boxes as d d well as a countable union of open balls, we have A4 = A6 = B(R ). Thus Ai = B(R ) for 1 ≤ i ≤ 6. d d By the definition of continuity, we have A7 ⊂ B(R ). If U ⊂ R is an open set different from Rd, let A = Rd \ U and define f : Rd → R as f(x) = dist(x, A) := inf{∥x − a∥ : a ∈ A}. Then f is continuous, and A = f −1(0) because A is closed. Now, U = f −1(R \{0}) and R \{0} is open in R. d Hence B(R ) ⊂ A7, completing the proof.

Topological remarks: (i) If X is a separable metric space, then any base or subbase for the topology of X will generate the Borel σ-algebra B(X). (ii) In the above characterization we used implicitly the fact that Rd is second countable and locally compact. If a metric space X fails to be second countable or locally compact, then the σ-algebra generated by all compact subsets of X will only be a proper sub-collection of B(X). For example, try to ﬁgure out what happens for the spaces (R, discrete metric) (which is not second countable), and (Q, Euclidean metric) (which is not locally compact).

Exercise-23: Show that B(R) is generated by each of the following collections: {(a, ∞): a ∈ R}, {[a, ∞): a ∈ R}, {(−∞, b): b ∈ R}, {(−∞, b]: b ∈ R}, {(a, b): a < b and a, b ∈ Q}.

Next our aim is to determine the cardinality of B(Rd). We need some set-theoretic preparation.

Deﬁnition: An order ≤ on a set X is a partial order if (i) x ≤ x for every x ∈ X, (ii) x ≤ y and y ≤ x ⇒ x = y for every x, y ∈ X, (iii) x ≤ y and y ≤ z ⇒ x ≤ z for every x, y, z ∈ X. We say (X, ≤) is a totally ordered set if ≤ is a partial order and any two elements of X are comparable. We say (X, ≤) is a well-ordered set if (X, ≤) is totally ordered and any nonempty subset Y ⊂ X has a least element in Y .

Examples: (i) Let X be the collection of all nonempty subsets of N. Deﬁne an order ≤ on X as A ≤ B iﬀ the minimum of A is less than or equal to the minimum of B. Then this is not a partial order since the second condition fails. (ii) If X is any nonempty set, then P(X) with inclusion as order is partially ordered, but in general not totally ordered. (iii) R with the usual order is totally ordered, but not well-ordered since the subset (0, 1) does not contain a least element. (iv) N with the usual order is well-ordered.

Well-ordering principle (equivalent to the axiom of choice): Any nonempty set admits a well- ordering.

Now we describe the construction of some ordinal numbers. Start with an uncountable set X such that card(X) = card(R), and let ≤ be a well-ordering on X. Let θ denote the least element of 20 T.K.SUBRAHMONIAN MOOTHATHU

X. By adding one extra element to X if necessary, we may also assume that (X, ≤) has a largest ′ element, say θ . For each β ∈ X, let Lβ = {α ∈ X : α < β} be the left section of β in X. Let ′ Y = {β ∈ X : Lβ is uncountable}. Then Y ≠ ∅ since θ ∈ Y . So Y has a least element, say Ω.

Then LΩ is uncountable, but Lβ is countable for every β < Ω. Here, Ω is called the ﬁrst uncountable ordinal, and each β ∈ LΩ is called a countable ordinal number since each β ∈ LΩ represents the type of a countable well-ordered set through Lβ.

Fact: If A ⊂ LΩ is a nonempty countable set, then A has a least upper bound in LΩ.[Proof : If ∪ \ ̸ ∅ \ B = β∈A Lβ, then B is countable and hence LΩ B = . The least element of LΩ B is the least upper bound of A.]

If α ∈ LΩ, then the least element of the nonempty set {β ∈ LΩ : α < β} will be denoted as α +1.

Note that there are no elements between α and α + 1 in LΩ. On the other hand, given β ∈ LΩ, there may or may not exist α ∈ LΩ such that α + 1 = β. For example, if β ∈ LΩ is the least upper bound of the countable set {θ, θ + 1, θ + 2,...} (where recall that θ is the least element of LΩ), then there is no α ∈ X with α + 1 = β. We say β ∈ LΩ is a limit ordinal if there is no α ∈ LΩ with α + 1 = β.

Exercise-24: (i) If card(X) ≤ card(R), then card(XN) ≤ card(R). (ii) If card(J) ≤ card(R) and ∪ ≤ R ∈ ≤ R card(Xβ) card( ) for each β J, then, card( β∈J Xβ) card( ). [Hint: (i) Assume X = (0, 1). N N Deﬁne a one-one map f : (0, 1) → (0, 1) as follows. If x = (xn) ∈ (0, 1) and if xn = 0. xn,1xn,2 ··· , then f(x) = 0. x1,1x1,2x2,1x1,3x2,2x3,1 ··· . (ii) Let g : R → J and hβ : R → Xβ be surjections. Then ∪ R2 → R2 R f : β∈J Xβ deﬁned as f(x, y) = hg(y)(x) is a surjection, and card( ) = card( ).]

[117] card(B(Rd)) = card(R).

Proof. We will use transﬁnite induction (i.e., induction with respect to ordinal numbers) by using

LΩ described above. Recall that we denoted the least element of LΩ by the symbol θ. To start the d induction process, let Aθ = {U ⊂ R : U is open}. Let β ∈ LΩ and assume that we have defined Aα ∪ ∈ A A ∈ for every α Lβ. If β is a limit ordinal, define β = α<β α. If β = α + 1 for some α LΩ, let ′ d d d A = {A ⊂ R : R \ A ∈ Aα}, and Aβ = {A ⊂ R : A is a countable union of members from Aα ∪ α ∪ A′ } A ∈ A A α . This defines β for every β LΩ. Finally, put = β<Ω β. From our construction, it is clear that A ⊂ B(Rd). We verify that A is a σ-algebra on Rd. It suffices to check only the third property. So consider ∈ A ∈ ∈ A ∈ N A1,A2,... . Then there are β1, β2,... LΩ such that An βn for every n . By the

Fact mentioned above, the countable set {βn : n ∈ N} has a least upper bound, say δ in LΩ. Then ∪ ∈ A ∈ N ∞ ∈ A ⊂ A A ⊂ B Rd An δ for every n and hence n=1 An δ+1 . Thus is a σ-algebra on containing all open subsets of Rd. Hence A = B(Rd). MEASURE THEORY 21

Now, it suﬃces to show that card(A) = card(R). Since there is an open ball of radius 1 centered d at each point of R , we have card(A) ≥ card(Aθ) ≥ card(R). So it suﬃces to establish that ∪ A ≤ R R A A card( ) card( ). Since card(LΩ) = card( ) and = β<Ω β, it is enough to show that card(Aβ) ≤ card(R) for each β ∈ LΩ in view of Exercise-24. Let D be the collection of all open balls in Rd with rational radius and center in Qd. Then D is countable, and any open set U ⊂ Rd can be written as a countable union of members of N D. Hence card(Aθ) ≤ card(D ) = card(R). Let β ∈ LΩ and suppose we have proved that ∪ A ≤ R A A card( α) card( ) for every α < β. If β is a limit ordinal, then β = α<β α is a countable union and hence card(Aβ) ≤ card(R). If there is α with α+1 = β, then any A ∈ Aβ can be written ∪ ∞ ∈ A ∪ A′ A A ∪ A′ N as A = n=1 An with An α α. This gives a one-one map from β into ( α α) . Hence A ≤ A ∪ A′ N ≤ R card( β) card(( α α) ) card( ), again by Exercise-24. This completes the proof.

[118] [Corollary] For any uncountable set Y ⊂ Rd, there is A ⊂ Y such that A is not a Borel subset of Rd.

Proof. We have card(B(Rd)) = card(R) = card(Y ) < card(P(Y )).

N Rd { ⊂ Rd ∗ } N Rd Deﬁnition: Let ( ) = A : µL,d[A] = 0 . The members of ( ) are called Lebesgue null sets. The σ-algebra L(Rd) on Rd generated by B(Rd) ∪ N (Rd) is called the Lebesgue σ-algebra on Rd, and members of L(Rd) are called Lebesgue measurable subsets of Rd.

[119] card(N (Rd)) = card(L(Rd)) = card(P(Rd)) > card(R). Hence, N (Rd) * B(Rd) $ L(Rd).

⊂ ∗ ≤ Proof. Let K be the middle-third Cantor set. Then, for any subset A K, we have µL,1[A] ∗ ∗ P ⊂ N Rd ⊂ L Rd P µL,1[K] = 0. So µL,d[A] = 0 also. This shows that (K) ( ) ( ). And card( (K)) = card(P(R)) since K is an uncountable subset of R.

[120] [Translation invariance] (i) A + x ∈ B(Rd) for every A ∈ B(Rd) and x ∈ Rd. (ii) A + x ∈ N (Rd) for every A ∈ N (Rd) and x ∈ Rd. (iii) A + x ∈ L(Rd) for every A ∈ L(Rd) and x ∈ Rd.

Proof. First let us mention a general principle that will be used at many places. To establish that the members of a certain σ-algebra D on a set X satisfies a certain property P , it suffices to do the following: show that the collection {A ⊂ X : A satisfies property P } is a σ-algebra, and then find a suitable collection C ⊂ P(X) generating D and show that every member of C satisfies the property P . Let A = {A ⊂ Rd : A + x ∈ B(Rd) for every x ∈ Rd}. It is easy to check that A is a σ-algebra containing all d-boxes. And recall that the collection of all d-boxes generates B(Rd). This proves 22 T.K.SUBRAHMONIAN MOOTHATHU

(i). Next, statement (ii) is a consequence of the translation invariance property of the Lebesgue outer measure, and (iii) follows from (i) and (ii) by applying the principle mentioned above.

We will give other characterizations of the Lebesgue measurable sets shortly, and we will also show that L(Rd) ≠ P(Rd).

5. Measures and outer measures

Deﬁnition: Let X be a nonempty set. A function µ∗ : P(X) → [0, ∞] is called an outer measure if the following are satisﬁed: (i) µ∗[∅] = 0, (ii) [Monotonicity] A ⊂ B ⊂ X ⇒ µ∗[A] ≤ µ∗[B], ∪ ∑ ∗ ∞ ≤ ∞ ∗ ⊂ (iii) [Countable subadditivity] µ [ n=1 An] n=1 µ [An] for every A1,A2,... X. ∗ Rd ∗ Example: µL,d is an outer measure on . But the outer Jordan content µJ,d is not an outer measure on [0, 1]d since it does not satisfy countable subadditivity.

Remark: Note that we do not need the concept of a σ-algebra to deﬁne an outer measure.

Deﬁnition: Let (X, A) be a measurable space. A function µ : A → [0, ∞] is called a measure on (X, A) if the following are satisﬁed: (i) µ[∅] = 0, ∪ ∑ ∞ ∞ ∈ A (ii) [Countable additivity] µ[ n=1 An] = n=1 µ[An] if A1,A2,... are pairwise disjoint. If µ is a measure on (X, A), the triple (X, A, µ) is called a measure space.

Remark: Note that countable additivity implies ﬁnite additivity (by taking An = ∅ for all large n), and ﬁnite additivity implies monotonicity for a measure.

Deﬁnition: Let (X, A, µ) be a measure space. A set A ∈ A is said to be a µ-null set if µ[A] = 0. We say (X, A, µ) is complete if any subset of any µ-null set also belongs to A. Sometimes we will just say that µ is complete to mean (X, A, µ) is complete.

Examples: (i) Let (X, A) be a measurable space and let µ be the counting measure on (X, A) defined as µ[A] = ∞ if A is infinite and µ[A] = card(A) if A is finite. Then (X, A, µ) is a complete measure space. (ii) Let (X, A) be a measurable space, let x ∈ X, and define a measure µ on (X, A) as µ[A] = 1 if x ∈ A, and µ[A] = 0 if x∈ / A. Here µ is called the dirac measure concentrated at x. d d If (X, A) = (R , B(R )), then µ is not complete. (iii) Let µ1, µ2 be measures on a measurable space

(X, A) and let c1, c2 ∈ [0, ∞). Then the linear combination µ := c1µ1 + c2µ2 is also a measure on

(X, A). If at least one of µ1, µ2 is complete and the corresponding ci ≠ 0, then µ is also complete. MEASURE THEORY 23

Exercise-25: [Completion of a measure space] Let (X, A, µ) be a measure space, let Nµ = {B ⊂ X : e there is C ∈ A such that B ⊂ C and µ[C] = 0}, and let A be the σ-algebra generated by A ∪ Nµ.

Let A1 = {A ∪ B : A ∈ A and B ∈ Nµ}, A2 = {A∆B : A ∈ A and B ∈ Nµ},

A3 = {A \ B : A ∈ A and B ∈ Nµ}, and A4 = {(A ∪ B1) \ B2 : A ∈ A and B1,B2 ∈ Nµ}. e e e Show that Ai = A for 1 ≤ i ≤ 4 and µ has a natural extension µe to A. Here, (X, A, µe) is called the completion of (X, A, µ). [Hint: Let A ∈ A, B ∈ Nµ, and let C ∈ A be so that B ⊂ C and µ[C] = 0. ′ ′ ′ ′ ′′ If B = B \ A, then B ∈ Nµ and A ∪ B = A∆B . If A = A \ C and B = [A ∩ (C \ B)] ∪ [B \ A], ′ ′′ ′ ′′ then A ∈ A, B ∈ Nµ and A∆B = A ∪ B . So A1 = A2. Similarly, A2 = A3 = A4. Next, use the equalities of Ai’s to show Ai is a σ-algebra.]

Remark: We know that it is often advantageous to work with a complete metric space rather than arbitrary metric spaces (to ensure, for example, that the limits of sequences do not go out of the space). Similarly, we will see in our course that it is advantageous to work with a complete measure space rather than an arbitrary measure space.

Now we would like to establish an important result which says that given an outer measure µ∗ on a set X, there is a σ-algebra A on X such that the restriction of µ∗ to A is a complete measure. The key idea is to take A as the collection of those subsets A of X for which the ‘boundary’ between A and X \ A is not very bad so that if an arbitrary subset E is divided into two pieces by intersecting with A and X \ A, then the outer measures of the two pieces add up to give the outer measure of E. Before proving the result, it is convenient to introduce another structure on a set X.

Deﬁnition: Let X be a nonempty set. A collection A ⊂ P(X) is an algebra on X if (i) ∅,X ∈ A, (ii) A ∈ A ⇒ X \ A ∈ A, (iii) A, B ∈ A ⇒ A ∪ B ∈ A.

Remark: Let A be a collection of subsets of a set X satisfying the first two conditions in the definition above. Then note that A satisfies condition (iii) iff A is closed under finite intersections. ∩ ∪ A \ n \ n Therefore, if is an algebra, then sets of the form A0 A1, j=0 Aj,A0 [ j=1 Aj], etc. belong to

A for every A0,A1,...,An ∈ A.

Example: Let X ⊂ Rd be a d-box and let A = {A ⊂ X : A is a ﬁnite union of d-boxes}. Then A is an algebra on X (but not a σ-algebra on X).

Exercise-26: Let A be an algebra on a set X. If A is closed under countable union of disjoint members, then A is a σ-algebra on X.[Hint: If A1,A2,... ∈ A, let B1 = A1 and Bn+1 = ∪ ∪ ∪ \ n ∈ A ∞ ∞ ∈ A An+1 [ i=1 Ai]. Then Bn’s are disjoint, and Bn . So n=1 An = n=1 Bn .]

We write Ac = X \ A below. 24 T.K.SUBRAHMONIAN MOOTHATHU

[121] [Caratheodory’s restriction theorem] Let X be a nonempty set and µ∗ : P(X) → [0, ∞] be an ∗ ∗ ∗ c ∗ outer measure. Let A = {A ⊂ X : µ [E] = µ [A ∩ E] + µ [A ∩ E] for every E ⊂ X} and µ = µ |A. Then A is a σ-algebra on X and µ is a complete measure on (X, A).

Proof. Note that A can also be represented as A = {A ⊂ X : µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ X} since µ∗ is subadditive. Step-1 : First our aim is to show A is an algebra on X. Clearly, ∅,X ∈ A, and A is closed under complementation. Now, let A, B ∈ A and note that (A ∩ B)c is the disjoint union of the three sets Ac ∩ B,A ∩ Bc,Ac ∩ Bc. Hence for any E ⊂ X, we have µ∗[E] ≥ µ∗[B ∩ E] + µ∗[Bc ∩ E] = µ∗[A∩B ∩E]+µ∗[Ac ∩B ∩E]+µ∗[A∩Bc ∩E]+µ∗[Ac ∩Bc ∩E] ≥ µ∗[(A∩B)∩E]+µ∗[(A∩B)c ∩E] by the subadditivity of µ∗. Hence A ∩ B ∈ A for every A, B ∈ A. Thus A is an algebra on X. Step-2 : We claim that if A, B ∈ A are disjoint, and E ⊂ X, then µ∗[(A ∪ B) ∩ E] = µ∗[A ∩ E] + µ∗[B∩E]. This claim is true because µ∗[(A∪B)∩E] = µ∗[A∩[(A∪B)∩E]]+µ∗[Ac ∩[(A∪B)∩E] = ∗ ∗ µ [A ∩ E] + µ [B ∩ E]. Inductively, we have the following: if A1,...,Ak ∈ A are pairwise disjoint ∪ ∑ ⊂ ∗ k ∩ k ∗ ∩ and E X, then µ [( n=1 An) E] = n=1 µ [An E]. Step-3 : To show A is a σ-algebra on X, it is enough to check that A is closed under countable union of pairwise disjoint members, in view of Exercise-26. Let A1,A2,... ∈ A be pairwise disjoint sets, ∪ ∪ ∞ k let A = An and Bk = An. Note that Bk ∈ A since A is an algebra. For any E ⊂ X, we n=1 n=1 ∑ ∗ ∗ ∗ c k ∗ ∗ c have µ [E] = µ [Bk ∩E]+µ [B ∩E] ≥ µ [An ∩E]+µ [A ∩E] by Step-2 and by the inclusion k n∑=1 c ⊂ c → ∞ ∗ ≥ ∞ ∗ ∩ ∗ c ∩ ≥ ∗ ∩ ∗ c ∩ A Bk. Letting k , we get µ [E] n=1 µ [An E] + µ [A E] µ [A E] + µ [A E] by the countable subadditivity of µ∗. Thus A ∈ A and so A is a σ-algebra on X. ∪∞ Step-4 : Let A1,A2,... ∈ A be pairwise disjoint and A = An. Then, taking E = A in the ∑ n=1 ∑ ∗ ≥ ∞ ∗ ∗ ∞ ∗ ∗ argument of Step-3, we have µ [A] n=1 µ [An], and therefore µ [A] = n=1 µ [An] since µ is countably subadditive. This shows that µ∗ restricted to A is a measure. ∗ Step-5 : Finally we show that the measure µ |A is complete. Let E ⊂ X, let A ∈ A be such that µ[A] = 0 and let B ⊂ A. Then µ∗[B ∩ E] + µ∗[Bc ∩ E] ≤ µ∗[A ∩ E] + µ∗[E] = 0 + µ∗[E] = µ∗[E] and hence B ∈ A.

6. Lebesgue measure on Rd

∗ Rd A Rd Since µL,d is an outer measure on , we can use [121] get a σ-algebra on such that the ∗ A A L Rd restriction of µL,d to is a complete measure. We show below that = ( ). ∗ ∗ ⊂ Rd [122] [Characterization of Lebesgue measurable sets] Let µ = µL,d. Then, for a subset A , the following are equivalent. (i) A ∈ L(Rd), where L(Rd) is the σ-algebra generated by B(Rd) ∪ N (Rd). (ii) A satisﬁes Caratheodory’s condition, i.e., µ∗[E] = µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ Rd. MEASURE THEORY 25

(iii) For every ε > 0, there is an open set U ⊂ Rd such that A ⊂ U and µ∗[U \ A] < ε. d ∗ (iv) There is a Gδ set G ⊂ R such that A ⊂ G and µ [G \ A] = 0. (v) For every ε > 0, there is a closed set K ⊂ Rd such that K ⊂ A and µ∗[A \ K] < ε. d ∗ (vi) There is an Fσ set F ⊂ R such that F ⊂ A and µ [A \ F ] = 0. (vii) ∀ ε > 0, ∃ open U ⊂ Rd and closed K ⊂ Rd such that K ⊂ A ⊂ U and µ∗[U \ K] < ε. d d ∗ (viii) There exist a Gδ set G ⊂ R and an Fσ set F ⊂ R such that F ⊂ A ⊂ G and µ [G \ F ] = 0.

Proof. (i) ⇒ (ii): Let A = {A ⊂ Rd : µ∗[E] = µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ Rd}. Since µ∗ is an outer measure, A is a σ-algebra on Rd by [121]. Let C = {A ⊂ Rd : A is a d-box} ∪ N (Rd). It is clear that C generates L(Rd). Hence it suffices to show C ⊂ A. If A ∈ N (Rd), then µ∗(E) ≥ µ∗(Ac ∩ E) = µ∗(A ∩ E) + µ∗(Ac ∩ E) for every E ⊂ Rd and hence A ∈ A. Thus N (Rd) ⊂ A. So it suffices to show that every d-box A belongs to A. Consider a d-box A and an arbitrary set E ⊂ Rd. We have to show µ∗[E] ≥ µ∗[A∩E]+µ∗[Ac ∩E] (∵ the other inequality is true by subadditivity). If µ∗[E] = ∞, we are done. So assume µ∗[E] < ∞ ∪ ∑ ⊂ ∞ ∞ ∗ ∗ and let ε > 0. Then there are d-boxes Bn such that E n=1 Bn and n=1 µ [Bn] < µ [E] + ε by c the definition. If n is such that Bn intersects both A and A , then we may partition Bn into finitely many d-boxes so that each d-box of the partition is contained in either A or Ac (this is possible since

A is a d-box). After this sort of reﬁnement, we may assume that each Bn is contained in either A or ∪ c ′ c c A . Let Γ = {n ∈ N : Bn ⊂ A} and Γ = {n ∈ N : Bn ⊂ A }. Then A ∩ E ⊂ ∈ Bn and A ∩ E ⊂ ∪ ∑ ∑ ∑ n Γ ∗ ∞ ∗ ∗ ∗ ≥ ∗ ∩ ∗ c ∩ n∈Γ′ Bn. Hence µ [E]+ε > n=1 µ [Bn] = n∈Γ µ [Bn]+ n∈Γ′ µ [Bn] µ [A E]+µ [A E]. Since ε > 0 is arbitrary, µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩ E] and thus A ∈ A. (ii) ⇒ (iii): Let ε > 0. Let A = {B ⊂ Rd : µ∗[E] = µ∗[B ∩ E] + µ∗[Bc ∩ E] for every E ⊂ Rd}. We know that A is a σ-algebra by [121]. From the argument above, we also know that all d-boxes belong ∪ d ∞ to A. Write R = Bn, where each Bn is a d-box, and let An = A ∩ Bn. Then, An ∈ A for n=1 ∪ ∈ N ∈ A A ∞ ∗ ∞ every n (since A, Bn and since is a σ-algebra). Note that A = n=1 An and µ [An] < d ∗ ∗ n for each n ∈ N. Choose an open set Un ⊂ R such that An ⊂ Un and µ [Un] < µ [An] + ε/2 by ∗ ∗ ∗ c ∗ ∗ [102](vii). Since An ∈ A, we have µ [Un] = µ [An ∩ Un] + µ [A ∩ Un] = µ [An] + µ [Un \ An]. ∪ n ∗ n ∞ ∗ Hence µ [Un \ An] < ε/2 . Put U = Un. Then, U is open, A ⊂ U, and µ [U \ A] ≤ ∑ ∑ n=1 ∪ ∞ ∗ \ ∞ n \ ⊂ ∞ \ ∗ n=1 µ [Un An] < n=1 ε/2 = ε since U A n=1(Un An) and µ is countably subadditive. ∩ ⇒ ⊂ Rd ⊂ ∗ \ ∞ (iii) (iv): Let Un be an open set such that A Un and µ [Un A] < 1/n. Let G = n=1 Un. ∗ ∗ ∗ Then A ⊂ G, and µ [G \ A] ≤ µ [Un \ A] < 1/n for every n ∈ N so that µ [G \ A] = 0. (iv) ⇒ (i): We have A = G \ (G \ A), G ∈ B(Rd) and G \ A ∈ N (Rd). Hence A ∈ L(Rd). Thus (i)-(iv) are equivalent. The implications (i)+(iii) ⇒ (v) ⇒ (vi) ⇒ (i) can be proved similarly. Hence statements (i)-(vi) are equivalent. It is easy to see that (iii)+(v) ⇔ (vii), and (iv)+(vi) ⇔ (viii). This completes the proof. 26 T.K.SUBRAHMONIAN MOOTHATHU

Remark: For sets A ⊂ B and ε > 0, the statements µ∗[B \ A] ≤ ε and µ∗[B] ≤ µ∗[A] + ε are not equivalent (problems: it can happen that µ∗[A] = ∞; even otherwise µ∗ is not ﬁnitely additive).

∗ L Rd Deﬁnition: Let µL,d be the restriction of µL,d to ( ). Then µL,d is a complete measure on the d d measurable space (R , L(R )) by [121] and [122]. We call µL,d the d-dimensional Lebesgue measure.

d d d d Remark:(R , L(R ), µL,d) is the completion of the measure space (R , B(R ), µL,d). This can be d d seen as follows. Let µ = µL,d and Nµ = {A ⊂ R : A ⊂ B for some B ∈ B(R ) with µ[B] = 0}. d d Then Nµ ⊂ N (R ). To get the other inclusion, note that if A ∈ N (R ), then by [122](iv) there is d d a Gδ subset (hence a Borel subset) G ⊂ R such that A ⊂ G and µ[G] = 0. Thus Nµ = N (R ). d d d Now apply Exercise-25. In particular, if A ∈ L(R ), then there exist Bi ∈ B(R ) and Yi ∈ N (R ) such that A = B1 ∪ Y1 = B2 \ Y2 = B3∆Y3 = (B4 ∪ Y4) \ Y5. But we have better representations d d d d from [122]. If A ∈ L(R ), then there exist Gδ set G ⊂ R , Fσ set F ⊂ R and Y1,Y2 ∈ N (R ) such that A = G \ Y1 = F ∪ Y2.

′ d d [123] [Uniqueness of Lebesgue measure] Let µ = µL,d and let µ be a measure on (R , L(R )) such that µ′[A] = µ[A] for every d-box A. Then µ′ = µ.

Proof. If U ⊂ Rd is open, then U can be written as a countable union of pairwise disjoint d-boxes and hence µ′[U] = µ[U]. If A ∈ L(Rd), then using [122] we get µ[A] = inf{µ[U]: A ⊂ U and U open} = inf{µ′[U]: A ⊂ U and U open} ≥ µ′[A]. If A ∈ L(Rd) is a bounded set contained in a d-box B, then µ[A] ≥ µ′[A] and µ[B \ A] ≥ µ′[B \ A] so that by additivity we obtain µ[A] = µ′[A]. For the ∪ ∈ L Rd ∞ general case, note that any A ( ) can be written as a countable disjoint union A = n=1 An d where An ∈ L(R ) are bounded.

Exercise-27: Let (X, A, µ) be a measure space and let An ∈ A for n ∈ N. ∪∞ (i) If A1 ⊂ A2 ⊂ · · · , then µ[ An] = limn→∞ µ[An]. n=1 ∩ ⊃ ⊃ · · · ∞ ∈ N ∞ (ii) If A1 A2 , and if µ[An] < for all large n , then µ[ n=1 An] = limn→∞ µ[An].

(iii) Give an example to say that the ﬁniteness assumption ‘µ[An] < ∞’ is necessary in (ii). ∪∞ [Hint: (i) Let B1 = A1 and Bn+1 = An+1 \ An. Then Bn’s are pairwise disjoint and Bn = ∪ n=1 ∞ ∞ n=1 An. Now use countable additivity. (ii) Assume µ[A1] < and apply the ﬁrst part to the sets Bn := A1 \ An. (iii) Consider the Lebesgue measure on R and let An = (n, ∞).]

[124] [Properties of Lebesgue measure] Let µ = µL,d. Then we have the following: d (i) µ[A] = V old(A) for every d-box A ⊂ R . (ii) µ[K] < ∞ for every compact set K ⊂ Rd. (iii) [Translation invariance] µ[A + x] = µ[A] for every A ∈ L(Rd) and x ∈ Rd. (iv) ∀ A ∈ L(Rd) and ε > 0, ∃ open U ⊂ Rd and closed K ⊂ Rd with K ⊂ A ⊂ U and µ[U \K] < ε. MEASURE THEORY 27

d d d (v) ∀ A ∈ L(R ), ∃ a Gδ set G ⊂ R and an Fσ set F ⊂ R such that F ⊂ A ⊂ G and µ[G \ F ] = 0. (vi) [Outer regularity] µ[A] = inf{µ[U]: U ⊂ Rd is open and A ⊂ U} for every A ∈ L(Rd). (vii) [Inner regularity] µ[A] = sup{µ[K]: K ⊂ Rd is compact and K ⊂ A} for every A ∈ L(Rd).

Proof. (i) is clear; (ii) follows from (i) since every compact subset is contained in some d-box; and ∗ (iii) is a consequence of the translation invariance of µL,d and [120]. Statements (iv) and (v) are just taken from [121]; and (vi) is a consequence. So we just have to prove (vii). d d Given A ∈ L(R ), let An = A ∩ B(0, n). Let ε > 0 and let Kn ⊂ R be a closed set so that ∪ ⊂ \ n+1 n Kn An and µ[An Kn] < ε/2 . Replacing Kn with i=1 Ki if necessary, we may assume that

K1 ⊂ K2 ⊂ · · · . Note that Kn is compact since Kn is a closed subset of the compact set B(0, n). ∪ ∑ ∞ \ ≤ ∞ n+1 If F = n=1 Kn (where F may not be compact), then we see that µ[A F ] n=1 ε/2 = ε/2. Note that µ[A] = µ[F ] + µ[A \ F ]. If µ[A] = ∞, then we must have µ[F ] = ∞, and hence limn→∞ µ[Kn] = µ[F ] = ∞ = µ[A] by Exercise-27(i). If µ[A] < ∞, then µ[F ] + ε/2 ≥ µ[A], and therefore µ[Kn] + ε > µ[A] for all large n ∈ N, again by Exercise-27(i).

Remark: Let X be a metric space. A measure µ defined on (X, B(X)) is said to be regular if for every A ∈ B(X) we have µ[A] = inf{µ[U]: U ⊂ X is open and A ⊂ U} = sup{µ[K]: K ⊂ X is compact and K ⊂ A}. The last two statements of the above result say that the Lebesgue measure is regular. More generally, the following fact is true. Let X be a locally compact, second countable metric space and let µ be a measure on B(X) having the property that µ[K] < ∞ for every compact K ⊂ X. Then µ is regular. Remark: Since B(Rd) ⊂ L(Rd), now we have a well-defined notion of d-dimensional volume for each Borel subset of Rd. However, there are subsets of Rd for which the concept of a d-dimensional volume is not well-defined, as the following result shows.

[125] [Existence of non-measurable set] There is a subset Y ⊂ Rd which is not Lebesgue measurable; that is, L(Rd) ≠ P(Rd).

∗ P Rd ∗ Proof. We know that µL,d is not ﬁnitely additive on ( ) by [107]. But the restriction of µL,d to L Rd ∗ L Rd L Rd ̸ P Rd ( ) is a measure and hence µL,d is countably additive on ( ). Therefore, ( ) = ( ). In fact, we may check that the Vitali set Y constructed in [107] does not belong to L(Rd).

There are other methods to produce sets which are not Lebesgue measurable.

Deﬁnition: A subset B ⊂ Rd is called a Bernstien set if for every uncountable closed set F ⊂ Rd, we have B ∩ F ≠ ∅ and Bc ∩ F ≠ ∅.

[126] (i) There exists a Bernstien set B ⊂ Rd. (ii) If B ⊂ Rd is a Bernstien set, then B/∈ L(Rd). 28 T.K.SUBRAHMONIAN MOOTHATHU

d d (iii) If A ∈ L(R ) is with µL,d[A] > 0, then there is Y ⊂ A such that Y/∈ L(R ).

Proof. Write µ = µL,d. (i) Let F be the collection of all uncountable closed subsets of Rd. We have card(F) = card(R).

So we may index the members of F using the well-ordered set LΩ (where Ω is the ﬁrst uncountable ordinal) as F = {Fβ : β ∈ LΩ}. We will choose two distinct points xβ, yβ ∈ Fβ for each β ∈ LΩ by transﬁnite induction. Let θ denote the least element of LΩ and let xθ, yθ ∈ Fθ be two distinct points. Let β ∈ LΩ and assume that we have chosen points xα, yα ∈ Fα for every α < β. Then

{xα : α < β} ∪ {yα : α < β} is countable since Lβ is countable. Therefore, we may choose two distinct points xβ, yβ from the uncountable set Fβ \ [{xα : α < β} ∪ {yα : α < β}]. This completes c the induction step. Finally put B = {xβ : β ∈ LΩ}. Then xβ ∈ B ∩ Fβ and yβ ∈ B ∩ Fβ for every

β ∈ LΩ so that B is a Bernstien set. (ii) Let if possible B ∈ L(Rd). Then µ[B] > 0 or µ[Bc] > 0. Since both B and Bc are Bernstien sets, we may assume µ[B] > 0. Then by [124], there is a closed set F ⊂ Rd such that F ⊂ B and µ[F ] > 0. Now, F must be uncountable since µ[F ] > 0. This is a contradiction to the deﬁnition of Bernstien set since Bc ∩ F = ∅. (iii) If possible, let P(A) ⊂ L(Rd). Then, in particular, A ∩ B,A ∩ Bc ∈ L(Rd). So we must have µ[A∩B] > 0 or µ[A∩Bc] > 0. This leads to a contradiction as in the proof of part (ii) since neither A ∩ B nor A ∩ Bc contains an uncountable closed subset of Rd.

Remark: Thus for A ⊂ Rd, we observe that (i) P(A) ⊂ B(Rd) iﬀ A is countable, and (ii) P(A) ⊂ L(Rd) iﬀ A ∈ N (Rd). Below we discuss some more properties of the Lebesgue measure.

d Exercise-28: Let A ⊂ L(R ) be a collection of pairwise disjoint sets such that µL,d[A] > 0 for every

A ∈ A. Then A is countable. [Hint: For k, n ∈ N, let Ak,n = {A ∈ A : µL,d[A ∩ B(0, n)] > 1/k}. ∪ ∪ A A ∞ ∞ A Then k,n is ﬁnite and = k=1 n=1 k,n.]

d Exercise-29: Let µ = µL,d, and A ∈ L(R ) be with µ[A] > 0. Then for every 0 < λ < 1, there is an open d-box B ⊂ Rd such that µ[A ∩ B] > λµ[B]. [Hint: Replacing A with A ∩ B(0, r) for ∪∞ some r > 0, assume µ[A] < ∞. Choose open d-boxes B1,B2,... such that A ⊂ Bn and ∑ ∪ ∑ ∑ n=1 ∞ ∞ ∩ ∞ ∞ ∩ λ n=1 µ[Bn] < µ[A]. Since A = n=1(A Bn), we get λ n=1 µ[Bn] < n=1 µ[A Bn], so

λµ[Bn] < µ[A ∩ Bn] for some n ∈ N.]

d [127] [Steinhaus’ theorem] If A ∈ L(R ) is with µL,d[A] > 0, then A − A contains an open neighborhood of 0 ∈ Rd, where A − A := {x − y : x, y ∈ A}. ∏ d Proof. Write µ = µL,d. Let 1/2 < λ < 1 and let B = j=1(aj, bj) be an open d-box such that µ[A∩B] > λµ[B], chosen by Exercise-29. Replacing A with A∩B, we may assume that A ⊂ B and ∏ d − µ[A] > λµ[B]. Let r > 0 be so that for C := j=1(aj r, bj + r) we have µ[C] < 2λµ[B]. We claim MEASURE THEORY 29 ∏ d − ⊂ − ∩ ̸ ∅ that U := j=1( r, r) A A. To prove the claim, we need to show that A (A + z) = for every z ∈ U. Note that for each z ∈ U, both A and A + z are contained in C and µ[A] + µ[A + z] = 2µ[A] > 2λµ[B] > µ[C]. Another proof is as follows. Since µ is inner regular, there is compact K ⊂ A with µ[K] > 0. By the outer regularity of µ (applied to K), ﬁnd an open neighborhood U of K with 2µ[K] > µ[U]. Since the compact set K is contained in the open set U, there is δ > 0 such that B(x, δ) ⊂ U for every x ∈ K. That is, K + B(0, δ) ⊂ U. If z ∈ B(0, δ), then both K and K + z are contained in U and µ[K] + µ[K + z] = 2µ[K] > µ[U]; hence K ∩ (K + z) ≠ ∅. So B(0, δ) ⊂ K − K ⊂ A − A.

d Exercise-30: Let µ = µL,d, and A ∈ L(R ) be with µ[A] > 0. Then for every 0 < β < µ[A], there is ∪ d d ∞ B ∈ L(R ) such that B ⊂ A and µ[B] = β.[Hint: Let ε > 0 and write R = Cn, a pairwise ∪ n=1 ∈ N k ∩ disjoint union of d-boxes with µ[Cn] < ε. Then there is k such that β < µ[ n=1(A Cn)] < β+ε.

Using this idea, inductively choose Lebesgue measurable sets B0 = A ⊃ B1 ⊃ B2 ⊃ · · · such that ∩∞ β < µ[Bm] < β + 1/m and put B = m=0 Bm.]

d d Deﬁnition: Let µ = µL,d. If A ∈ L(R ) and x ∈ R , deﬁne the Lebesgue density of A at x as µ[A ∩ B(x, r)] dens(A, x) = lim → , if the limit exists. r 0 µ[B(x, r)] [128] [Lebesgue density theorem] If A ∈ L(Rd), then there is Y ∈ N (Rd) such that dens(A, x) = 1 for every x ∈ A \ Y .

Proof. Write µ = µL,d. We may assume A is bounded (∵ consider A ∩ B(0, n) for each n ∈ N separately). First suppose d = 1. Let (bn) be a sequence in (0, 1) converging to 1.

Let Yn = {x ∈ A : for every δ > 0, there is r ∈ (0, δ) such that µ[A ∩ B(x, r)] < bnµ[B(x, r)]} ∪ ∞ ∈ \ and Y = n=1 Yn. Clearly, dens(A, x) = 1 for every x A Y . Thus it suﬃces to show µ[Yn] = 0 for every n ∈ N. Fix n ∈ N and consider Yn. Given ε > 0, choose an open set U such that Yn ⊂ U ∗ and µ[U] < µ [Yn] + ε (at present we do not know whether Yn is measurable). For each x ∈ Yn and δ > 0, there is an open interval I(x, δ) ⊂ U of the form (x − r, x + r) such that r < δ and

µ[A ∩ I(x, δ)] < bn|I(x, δ)|. So Γ = {I(x, δ): x ∈ Yn and δ > 0} is a Vitali cover for Yn. Hence by Vitali’s covering lemma ([111]), there exist ﬁnitely pairwise disjoint intervals I1,...,Ik ∈ Γ ∪ ∪ ∑ ∗ k ∗ ∗ k k ∗ such that µ [Yn \ Ij] < ε. Hence µ [Yn] − ε < µ [Yn ∩ ( Ij)] ≤ µ [Yn ∩ Ij] ≤ ∑ j=1∑ j=1 j=1 k ∗ ∩ k | | ≤ ∗ ∗ − j=1 µ [A Ij] < bn j=1 Ij < bnµ[U] bn(µ [Yn] + ε), and hence µ [Yn] < (1 + bn)ε/(1 bn). ∗ Since ε > 0 is arbitrary, µ [Yn] = 0, or µ[Yn] = 0. About the general case: Vitali’s covering lemma (which we stated as result [111] for d = 1) can be generalized to dimension d by replacing intervals with balls, and then the above proof can be imitated to establish Lebesgue density theorem in dimension d. 30 T.K.SUBRAHMONIAN MOOTHATHU

Remark: The above result says that if A ∈ L(Rd), then for almost every x ∈ Rd, either A or Ac has full density at x.

Reading assignment: [See Theorem 7.7 of W. Rudin, Real and Complex Analysis, 3rd Edn.] Let 1 d µ = µL,d and f ∈ L (µ). Then µ-almost every x ∈ R is a Lebesgue point of of f in the sense that 1 ∫ lim → |f(y) − f(x)|dµ(y) = 0. r 0 µ[B(x, r)] B(x,r)

7. Pre-measure → outer measure → measure

Deﬁnition: Let (X, A, µ) be a measure space. We say (i) µ is a probability measure if µ[X] = 1,

(ii) µ is a finite measure if µ[X] < ∞, and (iii) µ is a σ-finite measure if there exist X1,X2,... ∈ A ∪ ∞ ∞ ∈ N such that X = n=1 Xn and µ[Xn] < for every n . For example, µL,d restricted to d d d d ([0, 1] , B([0, 1] )) is a probability measure, µL,d restricted to ([0, 2] , B([0, 2] )) is a finite measure d d (but not a probability measure), and µL,d on to (R , L(R )) is a σ-finite measure (but not a finite measure). If X is an uncountable set, then the counting measure on (X, P(X)) is not σ-finite.

The following deﬁnition is motivated by considering the collection consisting of the empty set and all bounded or unbounded intervals in R.

Deﬁnition: Let X be a nonempty set. A collection A ⊂ P(X) is called a semi-algebra on X if the following are true: (i) ∅ ∈ A. (ii) A, B ∈ A ⇒ A ∩ B ∈ A. (iii) A ∈ A ⇒ X \ A is a ﬁnite union of pairwise disjoint members of A.

Remark: Note that every algebra is a semi-algebra.

Deﬁnition: Let A be a semi-algebra on a nonempty set X. A function µ : A → [0, ∞] is a pre-measure if (i) µ[∅] = 0, ∪ ∪ ∈ A ∞ ∈ A ∞ (ii) [Countable additivity] If A1,A2,... are pairwise disjoint and n=1 An , then µ[ n=1 An] = ∑∞ n=1 µ[An].

Remark: The concepts of probability pre-measure, finite pre-measure, and σ-finite pre-measure can be defined in the natural manner. In particular, if A is a (semi-)algebra on X, then a pre-measure ∪ A ∈ A ∞ ∞ µ on is said to be σ-finite if there are X1,X2,... such that X = n=1 Xn and µ[Xn] < for every n ∈ N.

Exercise-31: Let A be a semi-algebra on a nonempty set X and let A′ be the collection of all ﬁnite unions of pairwise disjoint members of A. Show that A′ is the algebra generated by A (i.e., the smallest algebra containing A). [Hint: If A1,...,An ∈ A are disjoint and B1,...,Bm ∈ A MEASURE THEORY 31 ∪ ∪ ∪ ∪ n m n m ′ ′ are disjoint, then ( Ai) ∩ ( Bj) = (Ai ∩ Bj) ∈ A . So A is closed under ﬁnite i=1∪ j=1 i=1 j=1 ∪ ∩ c ni c ∈ A′ n c n c ∈ A′ intersections. If Ai = j=1 Aij is a disjoint union, then Ai and so ( i=1 Ai) = i=1 Ai .]

We will show that a pre-measure on a semi-algebra can be extended ﬁrst to a pre-measure on an algebra, and next to an outer measure on P(X). Then we can obtain a measure by [121].

[129] [Outer measure from pre-measure] (i) Let A be a semi-algebra on a nonempty set X, let µ be a pre-measure on A and let A′ be the algebra generated by A. Then A′ is an algebra on X and µ extends uniquely to a pre-measure on A′. (ii) Let A′ be an algebra on a nonempty set X, and let µ : A′ → [0, ∞] be a pre-measure. Define ∑ ∪ ∗ P → ∞ ∗ { ∞ ∈ A′ ⊂ ∞ } ∗ µ : (X) [0, ] as µ [Y ] = inf n=1 µ[An]: An and Y n=1 An . Then µ is an outer measure on X, and µ∗[A] = µ[A] for every A ∈ A′. ∪ ′ k Proof. (i) If A ∈ A and if A = An is any disjoint union with An ∈ A, define µ[A] = ∑ ∪ n=1 ∪ ∪ k m k m µ[A]. If A = Bj is another disjoint union with Bj ∈ A, then A = (An ∩ Bj) n=1 ∪ j=1 ∑ ∑n=i j∑=1 k k k m and Bj = (An∩Bj) are also a disjoint unions. Hence µ[An] = µ[A] = µ[An∩ ∑ n∑=1 ∑ n=1 n=1 j=1 m k ∩ m A′ Bj] = j=1[ n=1 µ[An Bj]] = j=1 µ[Bj], and thus µ is well-defined on . Clearly, µ is finitely additive on A′. ∪ ′ ∞ ′ Now consider pairwise disjoint sets A1,A2,... ∈ A with A := An ∈ A . Write An = ∪ n=1 kn ∈ A ∈ A i=1 An,i, a disjoint union with An,i . Also choose disjoint members B1,...,Bm with ∪ ∪∞ ∪ m kn ∩ A A = j=1 Bj. Then Bj = n=1 i=1(An,i Bj) is a disjoint union of members of . Hence µ[A] = ∑ ∑ ∑∞ ∑ ∑∞ ∑ ∑ ∑∞ m m kn ∩ m kn ∩ j=1 µ[Bj] = j=1 n=1 i=1 µ[An,i Bj] = n=1( j=1 i=1 µ[An,i Bj]) = n=1 µ[An]. The extension is unique since a pre-measure on an algebra has to be finitely additive. (ii) Clearly, µ∗ satisfies monotonicity and µ∗[∅] = 0. To establish countable subadditivity, imitate ′ ∗ ′ the proof of [102](iv). Now consider A ∈ A . By definition, µ [A] ≤ µ[A]. If An ∈ A are such that ∪ ∪ ∞ n ′ A ⊂ An, let B1 = A1 and Bn+1 = An+1 \[ Ai]; then A∩Bn ∈ A are pairwise disjoint and ∪n=1 ∑ i=1 ∑ ∞ ∩ ∞ ∩ ≤ ∞ ≤ ∗ A = n=1(A Bn) so that µ[A] = n=1 µ[A Bn] n=1 µ[An]. This implies µ[A] µ [A].

The following result is useful in establishing the uniqueness of a measure. The proof below is essentially that of the π-λ theorem or monotone class theorem from the literature.

[130] [Dynkin’s theorem] Let A be a semi-algebra on a nonempty set X, and let µ be a σ-ﬁnite e e pre-measure on A. Let A be the σ-algebra generated by A and let µ1, µ2 be two measures on (X, A) such that µ1[A] = µ[A] = µ2[A] for every A ∈ A. Then µ1 = µ2.

Proof. If A′ is the algebra generated by A and if µ is the unique extended pre-measure on A′ given ′ by [129], then it is easily seen that µ1[A] = µ[A] = µ2[A] for every A ∈ A . Therefore, we may as well assume that A itself is an algebra. 32 T.K.SUBRAHMONIAN MOOTHATHU e e Case-1: µ[X] < ∞. Let Aeq = {A ∈ A : µ1[A] = µ2[A]}. We have to show Aeq = A. Let Γ be the collection of all C ⊂ P(X) such that (i) A ⊂ C ⊂ Ae. (ii) If B,C ∈ C and B ⊂ C, then C \ B ∈ C. ∪ ∈ C ∞ ∈ C (iii) If C1,C2,... are pairwise disjoint, then n=1 Cn .

Observe that that Aeq ∈ Γ(∵ for (ii), note that µ1[C \ B] = µ1[C] − µ1[B] = µ2[C] − µ2[B] =

µ2[C \ B]; here we use the assumption µ[X] < ∞). Let C0 be the intersection of all C ∈ Γ. Clearly e C0 ∈ Γ and thus A ⊂ C0 ⊂ Aeq ⊂ A. We will show below that C0 is a σ-algebra. If this is done, e e then we will get C0 = A, and hence we will have Aeq = A. Since X ∈ A ⊂ C0, we see from (ii) that

C0 is closed under complementation. In view of (i) and (iii), to show that C0 is a σ-algebra, now it suffices to show that C0 is closed under finite intersections. We do this in two steps. e Let C1 = {C ∈ C0 : A ∩ C ∈ C0 for every A ∈ A}. We wish to check that C1 ∈ Γ. Clearly C1 ⊂ A, and we have A ⊂ C1 since A is an algebra. The conditions (ii) and (iii) are verified as follows. c c If B,C ∈ C1 are with B ⊂ C and A ∈ A, then A ∩ (C \ B) = A ∩ (C ∩ B ) = (A ∩ C) ∩ B = c c c (A ∩ C) ∩ (A ∪ B ) = (A ∩ C) ∩ (A ∩ B) = (A ∩ C) \ (A ∩ B) ∈ C0 since A ∩ C,A ∩ B ∈ C0 and A ∩ B ⊂ A ∩ C. Hence C \ B ∈ C1. If C1,C2,... ∈ C1 are pairwise disjoint and A ∈ A, then ∪∞ ∪∞ A ∩ ( Cn) = (A ∩ Cn) ∈ C0 since A ∩ C1,A ∩ C2,... are pairwise disjoint members of C0. ∪n=1 n=1 ∞ ∈ C C ∈ C C C Hence n=1 Cn 1. Thus 1 Γ and therefore 1 = 0 by the minimality of 0.

Let C2 = {C ∈ C0 : D ∩ C ∈ C0 for every D ∈ C0}. We may verify in a similar manner that

C2 ∈ Γ (we have A ⊂ C2 by what is proved above). Then C2 = C0. So C0 is closed under ﬁnite intersections. ∪ ∈ A ∞ ∞ ⊂ Case-2: There are En with µ[En] < such that X = n=1 En. We may assume En En+1. f f Let An = {A ∩ En : A ∈ A} and An = {A ∩ En : A ∈ An}. Then An is an algebra on En, and f An is the σ-algebra on En generated by A. Applying the ﬁrst part of the proof, we have that ∪ Af ∈ Ae ∞ ∩ µ1 = µ2 on each n. Now if A , then A is the increasing union A = n=1(A En) so that

µ1[A] = limn→∞ µ1[A ∩ En] = limn→∞ µ2[A ∩ En] = µ2[A].

Example: Let X = (0, 1] and A = {∅} ∪ {(a, b] : 0 ≤ a < b ≤ 1}. Then A is a semi-algebra on X and B(X) is the σ-algebra generated by A. If µ1 is the counting measure on (X, B(X)) and µ2 is the measure deﬁned as µ2[A] = ∞ if A ≠ ∅ and µ[∅] = 0, then µ1 ≠ µ2. But the two measures agree on A. This example shows that the assumption of σ-ﬁniteness is necessary in [130].

[131] [Caratheodory’s extension theorem] Let A be a semi-algebra on a nonempty set X, let µ : A → [0, ∞] be a pre-measure, and let Ae be the σ-algebra on X generated by A. Then there is e a measure µe : A → [0, ∞] such that µe|A = µ. If µ is σ-ﬁnite, then µe is unique. MEASURE THEORY 33

Proof. The uniqueness part follows from [130]. We only have to establish the existence part. By [129](i), we may assume that A is an algebra on X, and let µ∗ be the outer measure on X produced from µ by [129](ii). Then by [121], there is a σ-algebra A∗ on X such that the restriction of µ∗ to A∗ is a (complete) measure. For our purpose, it suffices to show that the algebra A is contained in A∗ (then Ae ⊂ A∗ also). Let A ∈ A. To show A ∈ A∗, we need to establish that µ∗[E] ≥ µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ X. ∗ We may assume µ [E] < ∞ (otherwise the inequality is trivial). Given ε > 0, choose A1,A2,... ∈ ∪ ∑ ∪ ∞ ∞ ∗ ∞ A such that E ⊂ An and µ[An] < µ [E] + ε. Note that A ∩ E ⊂ (A ∩ An), ∪ n=1 n=1 ∑n=1 c ∞ c c ∗ ∞ A ∩E ⊂ (A ∩An), and A∩An,A ∩An ∈ A since A is an algebra. µ [E]+ε > µ[An] = ∑ n=1 ∑ n=1 ∞ ∩ ∞ c ∩ ≥ ∗ ∩ ∗ c ∩ n=1 µ[A An]+ n=1 µ[A An] µ [A E]+µ [A E]. As ε > 0 is arbitrary, we are done. ∏ ⊂ Rd d ⊂ R Example: We say A is a generalized d-box if A = j=1 Ij, where each Ij is a bounded or unbounded interval. For example, (−∞, 3] × (1, 5] × (0, ∞) is a generalized 3-box. Note that ∏ d | | A every generalized d-box A has a well-defined d-dimensional volume, V old(A) = j=1 Ij . If = {∅} ∪ {A ⊂ Rd : A is a generalized d-box}, then A is a semi-algebra on Rd. Define µ : A → [0, ∞] as µ[A] = V old(A). To check countable additivity, it is enough to work with d-boxes. Let A be a ∪ ∑ ∞ k d-box and suppose A = An is a pairwise disjoint union of d-boxes. Since µ[A] ≥ µ[An] n=1 ∑ n=1 ∈ N ≥ ∞ for every k by finite additivity, we get µ[A] n=1 µ[An]. For the other inequality, use the proof of [101]. Thus µ is a pre-measure on A. Since the σ-algebra generated by A is B(Rd), by d d [131] we obtain a measure µe on (R , B(R )) such that µe[A] = V old(A) for every generalized d-box

A. It follows by [130] that µe = µL,d. We could have constructed the Lebesgue measure on the Borel σ-algebra in this way also (and then we could have done a completion process, as in Exercise-25).

To give the next example, we describe the space first. Let m ≥ 2 be an integer and let X = {1, . . . , m}N. Give the finite set {1, . . . , m} the discrete topology, and give X the product topology. −n If x, y ∈ X, define ρ(x, y) = 0 if x = y, and ρ(x, y) = 2 if n ∈ N is the smallest such that xn ≠ yn. n It may be checked that ρ is a metric on the set X. If w = w1 ··· wn ∈ {1, . . . , m} , then w is | | ∈ { }n { ∞ ∈ called a word of length n, and we write w = n. If w 1, . . . , m , let Cw = x = (xn)n=1 X : x1 ··· xn = w}, and we call Cw a cylinder set.

Exercise-32: Let X = {1, . . . , m}N be the space given above. Then, (i) The metric ρ induces the product topology on X.

(ii) Each cylinder set Cw is both open and closed in X. ∪ ⊂ k (iii) If A X is a ﬁnite union of cylinder sets, then A can also be written as A = j=1 Cw(j), where the cylinder sets Cw(j)’s are pairwise disjoint and |w(1)| = ··· = |w(k)|. n (iv) {Cw : n ∈ N and w ∈ {1, . . . , m} } is a base for the topology on X. (v) (X, ρ) is compact, totally disconnected, and is without isolated points. 34 T.K.SUBRAHMONIAN MOOTHATHU

Remark: With more topological arguments, it can be shown that {1, . . . , m}N is homeomorphic to the middle-third Cantor set. Hence {1, . . . , m}N is called a Cantor space.

[132] [Uniform Bernoulli measure on the Cantor space] Let m ≥ 2 be an integer and let X = N −|w| {1, . . . , m} . Then there is a unique probability measure µ on B(X) such that µ[Cw] = m for every cylinder set Cw ⊂ X.

n Proof. Let A = {∅} ∪ {Cw : n ∈ N and w ∈ {1, . . . , m} }. Check that A is a semi-algebra on X. Note that the σ-algebra on X generated by A is B(X) because of Exercise-32(iv). Define −|w| µ : A → [0, ∞] as µ[∅] = 0 and µ[Cw] = m . Clearly µ is finitely additive on A. If A ∈ A is ∪ ∪ ∞ ∈ A ∈ N k A = n=1 An with An , then there is k such that A = n=1 An since the members of A are both compact and open. So countable additivity of µ is automatically satisfied. Thus µ is ∑ m a pre-measure, and by [131], µ extends to a measure µe on B(X). We have µe[X] = µe[Cj] = ∑ ∑ j=1 m m −1 e j=1 µ[Cj] = j=1 m = 1, and hence µ is unique by [130].

Similar argument gives many other probability measures on the Cantor space as stated below.

[132′] [Bernoulli measures on the Cantor space] Let m ≥ 2 be an integer, X = {1, . . . , m}N, and let ∑ m p1, . . . , pm be nonnegative reals such that j=1 pj = 1. Then there is a unique probability measure n µ on B(X) with the property that for every n ∈ N and w = w1 ··· wn ∈ {1, . . . , m} , we have ··· µ[Cw] = pw1 pwn .

Deﬁnition: A measure µ on (R, B(R)) such that µ[I] < ∞ for any bounded interval (i.e., for any 1-box) is called a Lebesgue-Stieltjes measure. We characterize them below by Exercise-33 and [133].

Exercise-33: Suppose µ is a Lebesgue-Stieltjes measure on (R, B(R)), and deﬁne a function fµ :

R → R as fµ(x) = µ[(0, x]] if x > 0, fµ(0) = 0, and fµ(x) = −µ[(x, 0]] if x < 0. Then, fµ is increasing, continuous from the right, and fµ(b) − fµ(a) = µ[(a, b]]. [Hint: for example, if ∩∞ x > 0 and if (xn) is a decreasing real sequence converging to x, then (0, x] = n=1(0, xn] so that fµ(x) = µ[(0, x]] = limn→∞ µ[(0, xn]] = limn→∞ fµ(xn) by Exercise-27].

[133] [Lebesgue-Stieltjes measures] Let f : R → R be increasing and right-continuous. Then there is a unique σ-ﬁnite measure µ on (R, B(R)) such that µ[(a, b]] = f(b) − f(a) for reals a < b.

Proof. Let A = {∅}∪{(a, b]: −∞ ≤ a < b < ∞}∪{(a, ∞): −∞ ≤ a < ∞}. It may be veriﬁed that

A is a semi-algebra on R and the σ-algebra generated by A is B(R). Let f(−∞) = limx↘−∞ f(x) and f(∞) = limx↗∞ f(x). Deﬁne µ : A → [0, ∞] as µ[∅] = 0, µ[(a, b]] = f(b) − f(a), and µ[(a, ∞)] = f(∞) − f(a). Since f is increasing, we observe the following for any A ∈ A: (i) If ∪ ∑ k k A1,...,Ak ∈ A are such that A ⊂ An, then µ[A] ≤ µ[An]. (ii) If A1,...,Ak ∈ A are ∪ n=1 ∑ n=1 k ⊂ k ≤ pairwise disjoint and n=1 Ak A, then n=1 µ[An] µ[A]. MEASURE THEORY 35

From (i) and (ii), µ is ﬁnitely additive on A. Our aim is to show µ is a pre-measure on A. ∪∞ Let A1,A2,... ∈ A be pairwise disjoint members such that A := An ∈ A. We have µ[A] ≥ ∪ ∑ n=1∑ k k ∈ N ≥ ∞ µ[ n=1 An] = n=1 µ[An] for every k by (ii), and therefore µ[A] n=1 µ[An]. Now we need to prove the other inequality.

Case-1: A = (a, b] where a ∈ R. Then necessarily An has the form (an, bn] for every n ∈ N.

Consider ε > 0. Using the right-continuity of f, we may ﬁnd x ∈ (a, b) and cn > bn such that ∪ n+1 ∞ f(x) − f(a) < ε/2 and f(cn) − f(bn) < ε/2 . Since [x, b] ⊂ (an, cn), by the compactness ∪ n=1 ∪ k k of [x, b] there is k ∈ N such that [x, b] ⊂ (an, cn) and hence (x, b] ⊂ (an, cn]. Then by ∑ ∑ n=1 ∑ n=1 ∑ k k k k (i), µ[(x, b]] ≤ µ[(an, cn]] = (f(cn) − f(an)) = (f(cn) − f(bn)) + (f(bn) − ∑ n=1 ∑ n=1 ∑ n=1 n=1 k n+1 k − ≤ ∞ f(an)) < n=1 ε/2 + n=1(f(bn) f(an)) ε/2 + n=1 µ[An]. Therefore, µ[A] = µ[(a, b]] = ∑∞ µ[(a, x]] + µ[(x, b]] = (f(x) − f(a)) + µ[(x, b]] < ε/2 + ε/2 + µ[An]. Since ε > 0 is arbitrary, ∑ n=1 ≤ ∞ we get µ[A] n=1 µ[An].

Case-2: A = (−∞, b]. Let (xk) be a decreasing sequence in A converging to −∞ and let Xk = (xk, b].

We have f(−∞) = limk→∞ f(xk), and therefore µ[A] = f(b) − f(−∞) = limk→∞(f(b) − f(xk)) = ∪∞ limk→∞ µ[Xk]. Since Xk = (Xk ∩ An) is a disjoint union of members of A, by Case-1 we get ∑ ∑n=1 ∑ ≤ ∞ ∩ ≤ ∞ ∈ N ≤ ∞ µ[Xk] n=1 µ[Xk An] n=1 µ[An] for each k . Hence µ[A] n=1 µ[An].

Case-3: Let A = (a, ∞). Let (yk) be an increasing sequence in A converging to ∞ and let Yk = ∑ ∑ ≤ ∞ ∩ ≤ ∞ (a, yk]. As in Case-2, we have µ[A] = limk→∞ µ[Yk], and µ[Yk] n=1 µ[Yk An] n=1 µ[An] ∑∞ for each k ∈ N. Hence µ[A] ≤ µ[An]. n=1 ∪ A R − ∞ Thus µ is a pre-measure on . Since = n∈Z(n, n+1] and µ[(n, n+1]] = f(n+1) f(n) < , µ is a σ-ﬁnite pre-measure. Hence by [131], µ extends uniquely to a σ-ﬁnite measure on (R, B(R)).

Remark: In the above, µ[{x}] = 0 iﬀ f is continuous at x.

Examples: In [133], we have the following: (i) If f = IR, then µ = µL,1 on B(R). (ii) If y ∈ R and f = 1[y,∞), then µ is the dirac probability measure concentrated at the point y. That is, µ[A] = 1 if y ∈ A, and µ[A] = 0 if y∈ / A. (iii) If f is the integer-part function, that is, f(x) = n if x ∈ [n, n+1) for n ∈ Z, then µ[A] = |A∩Z| for any A ∈ B(R). Here note that µ[(0, 1)] < µ[(0, 1]] < µ[[0, 1]], and the completion of B(R) with respect to µ is P(R) and not L(R)(∵ if A ⊂ R, then

A = (A ∩ Z) ∪ (A \ Z) ∈ B(R) ∪ Nµ).

Remark: If µ is a Lebesgue-Stieltjes measure, then clearly µ[K] < ∞ for every compact K ⊂ R, and therefore µ is regular by an earlier remark.

[134] [Approximation theorem for sets of ﬁnite measure] Let (X, A, µ) be a σ-ﬁnite measure space and let A′ be a semi-algebra generating A. Then for any A ∈ A with µ[A] < ∞ and ε > 0, there ∪ ∈ A′ k exist A1,...,Ak such that µ[A∆( n=1 An)] < ε. 36 T.K.SUBRAHMONIAN MOOTHATHU

∗ Proof. Deﬁne an outer measure µ using the pre-measure µ|A′ as in [129], and then get a measure µe on A by restricting µ∗ to A as in the the proof of [131]. By σ-ﬁniteness, we must have µe = µ because ∪ ∗ A ∈ A′ ⊂ ∞ of Dynkin’s theorem. Thus µ = µ on . So there are A1,A2,... such that A n=1 An and ∑∞ ∪∞ µ[An] < µ[A] + ε/2, and therefore µ[( An) \ A] < ε/2. Then for all large k ∈ N, we have n=1 ∪ ∪ ∑ n=1 k ≤ ∞ \ ∞ µ[A∆( n=1 An)] µ[( n=1 An) A] + n=k+1 µ[An] < ε/2 + ε/2 = ε.

Remark: Using the above result, we may deduce the following. (i) If µ is a Lebesgue-Stieltjes measure and A ∈ B(R) is with µ[A] < ∞, then for every ε > 0 there are finitely many intervals ∪ k d I1,...,Ik ∈ B(R) such that µ[A∆( In)] < ε. (ii) If A ∈ L(R ) is with µL,d[A] < ∞ and ε > 0, n=1 ∪ ⊂ Rd k then there are finitely many d-boxes A1,...,Ak such that µL,d[A∆( n=1 An)] < ε. (iii) If X is the Cantor space X = {1, . . . , m}N (m ≥ 2) and µ is a Bernoulli measure on (X, B(X)), then for any A ∈ B(X) and ε > 0, there are finitely many cylinder sets A1,...,Ak ⊂ X such that ∪ k µ[A∆( n=1 An)] < ε.

Example: We give an example to say σ-finiteness is necessary in [134]. Let A′ = {∅} ∪ {(a, b] : 0 ≤ a < b ≤ 1}. Then the semi-algebra A′ generates B((0, 1]). If µ is the counting measure on B((0, 1]) and if A ∈ B((0, 1]) is any nonempty finite set, then it is not possible to approximate A with a ∪ k A′ ∞ ∈ A′ finite union n=1 An of members of as in [134] since µ[An] = 0 or for An .

8. Measurable functions

Definition: Let (X, A), (Y, C) be measurable spaces. A function f : X → Y is said to be A- C measurable (or just measurable) if f −1(C) ∈ A for every C ∈ C. Note the similarity of the definition of a measurable function with the definition of a continuous function between topological spaces.

Remark: Let (Xi, Ai) be measurable spaces for 1 ≤ i ≤ 3, let f : X1 → X2 be A1-A2 measurable and g : X2 → X3 be A2-A3 measurable. Then g ◦ f : X1 → X3 is A1-A3 measurable.

Exercise-34: Let (X, A), (Y, C) be measurable spaces, f : X → Y be a function and let D ⊂ P(Y ) be a collection generating C. If f −1(D) ∈ A for every D ∈ D, then f is A-C measurable. [Hint: Check that {C ∈ C : f −1(C) ∈ A} is a σ-algebra containing D.]

Exercise-35: If X,Y are metric spaces, then every continuous function f : X → Y is B(X)-B(Y ) measurable. [Hint: Use Exercise-34 with D = {open subsets of Y }.]

Exercise-36: Let (X, A), (Y, C) be measurable spaces, f : X → Y be a function, and let X = ∪ ∞ ∈ A A { ∩ ∈ A} | A C j=1 Xj with Xj . Let j = A Xj : A and fj = f Xj . Then f is - measurable iﬀ fj is Aj-C measurable for every j ∈ N. MEASURE THEORY 37

Example: [Fat Cantor sets] We imitate the construction of the middle third Cantor set. Fix m ≥ 3. First remove an open interval of length 1/m from the middle of [0, 1], then remove an open interval of length 1/m2 from the middle of each of the remaining two closed intervals, then remove an open interval of length 1/m3 from the middle of each of the remaining four closed intervals, and so on.

Finally, we are left with a nowhere dense closed subset of [0, 1]; call it Fm. Note that F3 is the middle-third Cantor set. The total length of all open intervals removed in order to construct Fm 2 3 4 is 1/m + 2/m + 4/m + 8/m + ··· = 1/(m − 2) and therefore µ[Fm] = (m − 3)/(m − 2), where

µ = µL,1. Hence µ[Fm] > 0 for m ≥ 4, and we call Fm a Fat Cantor set for m ≥ 4 (this is another way to obtain nowhere dense compact subsets of R having positive Lebesgue measure). It may be proved with some effort that for any m ≥ 4, there is a homeomorphism hm : R → R such that hm(Fm) = F3 (first define hm : R \ Fm → R \ F3 in the natural way, check that this is uniformly continuous, and then extend, etc.). This has some interesting consequences. (i) hm is not L(R)-

L(R) measurable for m ≥ 4. Reason: P(F3) ⊂ N (R) ⊂ L(R), but Fm has non-measurable subsets by [126] since µ[Fm] > 0 for m ≥ 4. (ii) Homeomorphisms of R need not take µ-null sets to µ-null −1 sets; the homeomorphism h4 takes F3 to F4. On the other hand, it may be checked that Lipschitz maps take µ-null sets to µ-null sets. (iii) We get another argument to say B(R) ≠ L(R) as follows.

If B(R) = L(R), then P(F3) ⊂ L(R) = B(R) and hence P(F4) ⊂ B(R) since homeomorphisms must take Borel sets to Borel sets. But we know by [126] that P(F4) \L(R) ≠ ∅.

Discussion: When we deal with real-valued functions on a measurable space (X, A), we have a choice between the following two notions: A-B(R) measurability and A-L(R) measurability. Now, we want the collection of measurable functions to be as large as possible, and we also want the collection to include all continuous functions from X to R when X is a metric space. So we prefer the notion of A-B(R) measurability. Moreover, we may have to deal with extended real- valued functions sometimes (i.e., functions taking values in [−∞, ∞]); for instance, while taking the supremum of a sequence of real-valued functions. Next, when X = Rd, again we have a choice: whether to take A = B(Rd) or A = L(Rd). When we deal with Lebesgue measure, it is convenient to use L(Rd) because of completeness. These considerations motivate the following deﬁnitions.

Deﬁnition: Note that we can topologize [−∞, ∞] by having the usual topology on R and taking (a, ∞] and [−∞, a) as neighborhoods of ∞, −∞ respectively for a ∈ R. So the topological structure on [−∞, ∞] is similar to that on [0, 1]. The Borel σ-algebra B([−∞, ∞]) is the σ-algebra on [−∞, ∞] generated by open subsets of [−∞, ∞]. It may be seen that B([−∞, ∞]) is generated by B(R) ∪ {{−∞}, {∞}} also.

Deﬁnition: Let (X, A) be a measurable space. We say f : X → [−∞, ∞] is measurable if f is A- B([−∞, ∞]) measurable. If f(X) ⊂ R, then measurability of f just means A-B(R) measurability. 38 T.K.SUBRAHMONIAN MOOTHATHU

Deﬁnition: Consider X ∈ L(Rd). Let L(X) := {X ∩A : A ∈ L(Rd)}, and note that it is a σ-algebra on X. We say f : X → [−∞, ∞] is measurable if f is L(X)-B([−∞, ∞]) measurable. For example, if Y ∈ L(X), then 1Y is a (Lebesgue-Borel) measurable function.

Remark: Thus when we say f : Rd → [−∞, ∞] is measurable, we actually mean f is Lebesgue-Borel measurable, which is an asymmetric notion. Two disadvantages of this asymmetry are the following: (i) If f : R → R is a measurable bijection, then f −1 need not be measurable. (ii) If f, g : R → R are measurable, then f ◦ g need not be measurable. For instance, let g : R → R be a homeomorphism −1 with g(F4) = F3 (recall fat Cantor sets), let Y ⊂ F3 be such that g (Y ) ∈/ L(R) and let f : R → R be f = 1Y . Then, f, g are Lebesgue-Borel measurable, but f ◦ g is not Lebesgue-Borel measurable since (f ◦ g)−1({1}) ∈/ L(R). However, we will see in the sequel some of the advantages of using the notion of Lebesgue-Borel measurability (especially while discussing ‘almost everywhere pointwise convergence’ of a sequence of functions).

[135] Let (X, A) be a measurable space. Then for a function f : X → [−∞, ∞], the following are equivalent: (i) f is measurable. (ii) f −1(U) ∈ A for every open U ⊂ [−∞, ∞]. (iii) f −1(−∞), f −1(∞) ∈ A and f −1(U) ∈ A for every open U ⊂ R. (iv) f −1((a, ∞]) = {x ∈ X : f(x) > a} ∈ A for every a ∈ R. (v) f −1([a, ∞]) = {x ∈ X : f(x) ≥ a} ∈ A for every a ∈ R. (vi) f −1([−∞, a)) = {x ∈ X : f(x) < a} ∈ A for every a ∈ R. (vii) f −1([−∞, a]) = {x ∈ X : f(x) ≤ a} ∈ A for every a ∈ R.

Proof. Observe that each statement from (ii)-(vii) is of the form “f −1(D) ∈ A for every D ∈ D” for a generating collection D of B([−∞, ∞]), and use Exercise-34.

[136] Let (X, A) be a measurable space, and let F = {f : X → R : f is measurable}. Then, (i) F is a real vector space under pointwise operations. (ii) If f, g ∈ F, then the pointwise product fg ∈ F; and f/g ∈ F provided g(x) ≠ 0 ∀ x ∈ X. (iii) If f ∈ F and g : R → R is continuous, then g ◦ f ∈ F.

Proof. (i) Let f, g ∈ F and c, a ∈ R. Since {x ∈ X :(−f)(x) > a} = {x ∈ X : f(x) < a}, we have that −f is measurable. If c = 0, then trivially cf is measurable. If c > 0, then {x ∈ X : (cf)(x) > a} = {x ∈ X : f(x) > a/c}, and hence cf is measurable. If c < 0, then cf = −(−cf) is ∪ { ∈ } ∩ { ∈ } measurable. Note that x X :(f + g)(x) > a = r∈Q[Ar Br], where Ar = x X : f(x) > r and Br = {x ∈ X : g(x) > a − r}. Since Ar,Br ∈ A, we have {x ∈ X :(f + g)(x) > a} ∈ A, and thus f + g is measurable. MEASURE THEORY 39

(ii) We wish to check f 2 ∈ F. Consider a ∈ R. If a < 0, then {x ∈ X :(f(x))2 > a} = X ∈ A. If √ √ a ≥ 0, then {x ∈ X :(f(x))2 > a} = {x ∈ X : f(x) > a} ∪ {x ∈ X : f(x) < − a} ∈ A. Thus f 2 is measurable. Now, fg = [(f + g)2 − (f − g)2]/4 is measurable. Next suppose g(x) ≠ 0 for every x ∈ X. We wish to show 1/g ∈ F. Note that if a > 0, then {x ∈ X : (1/g)(x) > a} = {x ∈ X : g(x) < 1/a}; if a < 0, then {x ∈ X : (1/g)(x) > a} = {x ∈ X : g(x) > 1/a}; and {x ∈ X : (1/g)(x) > 0} = {x ∈ X : g(x) > 0}. So (1/g) ∈ F, and therefore, f/g = f(1/g) ∈ F. Statement (iii) This is clear.

Remark: In the above result, we considered only real-valued functions. For extended real-valued functions, there is a problem in deﬁning pointwise addition and pointwise multiplication. If f(x) = ∞ and g(x) = −∞, then f + g and fg are not deﬁned at x.

Deﬁnition: For a function f : X → [−∞, ∞], deﬁne the positive part and negative part as f + = max{f, 0}, f − = max{−f, 0}. Note that f +, f − ≥ 0; f = f + − f − and |f| = f + + f −.

Deﬁnition: If (fn) is a sequence of functions from X to [−∞, ∞], deﬁne { ∈ N} (supn∈N fn)(x) = sup fn(x): n ,

(infn∈N fn)(x) = inf{fn(x): n ∈ N},

lim supn→∞ fn = limk→∞(supn≥k fn) = infk→∞(supn≥k fn), and

lim infn→∞ fn = limk→∞(infn≥k fn) = supk→∞(infn≥k fn).

[137] Let (X, A) be a measurable space, and let F = {f : X → [−∞, ∞]: f is measurable}.

(i) If (fn) is a sequence in F, then sup fn, inf fn, lim sup fn, lim inf fn ∈ F.

(ii) If (fn) is a sequence in F converging pointwise to a function f : X → [−∞, ∞], then f ∈ F. (iii) Let f, g ∈ F with f, g ≥ 0, and let c ∈ R. Then cf, f + g ∈ F. (iv) If f, g ∈ F, then fg ∈ F. (v) If f ∈ F, then f +, f −, |f| ∈ F.

∪∞ Proof. (i) Let a ∈ R. Note that {x ∈ X : (sup fn)(x) > a} = {x ∈ X : fn(x) > a} and ∪ n=1 { ∈ } ∞ { ∈ } ∈ F x X : (inf fn)(x) < a = n=1 x X : fn(x) < a . Hence sup fn, inf fn . Applying this again, we get lim sup fn, lim inf fn ∈ F.

(ii) Note that f = lim sup fn = lim inf fn. (iii) This is similar to the proof of [136].

(iv) Here we cannot imitate the proof of [136] since f +g, f −g may not be deﬁned. Let fn(x) = f(x) if |f(x)| ≤ n, fn(x) = n if f(x) ≥ n and fn(x) = −n if f(x) ≤ −n. Similarly deﬁne gm. Check that the real-valued functions fn, gm are measurable. Hence fngm is measurable by [136]. Since fgm = limn→∞ fngm, we get fgm is measurable by (ii). Then fg = limm→∞ fgm is also measurable. (v) By part (i), f + = max{f, 0} ∈ F and f − = max{−f, 0} ∈ F. Then, |f| = f + + f − ∈ F. 40 T.K.SUBRAHMONIAN MOOTHATHU

9. Approximating measurable functions with simple functions

In the previous section, we did not use the concept of a measure. Now we will consider properties of measurable functions on measure spaces. In this and the coming sections, some results are for extended real-valued functions; while some other results are only for real-valued functions - keep an eye on this distinction.

An arbitrary measurable function can be quite complicated. But we can tackle measurable functions by expressing them as limits of functions which are easier to handle.

Deﬁnition: Let (X, A) be a measurable space, and ϕ : X → R be a measurable function. We say ϕ is a simple function if ϕ(X) is a ﬁnite set, i.e., if there are A1,...,An ∈ A and c1, . . . , cn ∈ R such ∑ n ∵ { } −1 that ϕ = i=1 ci 1Ai ( if ϕ(X) = c1, . . . , cn , let Aj = ϕ (cj)).

Deﬁnition: Let (X, A, µ) be a measure space, and ϕ : X → R be a measurable function. We say ϕ is a µ-bounded simple function if ϕ is simple and if there is A ∈ A with µ[A] < ∞ such that ϕ(x) = 0 for every x ∈ X \ A (this is equivalent to saying that there are A1,...,An ∈ A with µ[Ai] < ∞, ∑ ∈ R n R → R and c1, . . . , cn such that and ϕ = i=1 ci 1Ai ). Eg: ϕ : , ϕ = 2 1Q + 3 1[4,5]. ∑ Rd → R n Deﬁnition: A simple function ψ : is called a step function if ψ has the form ψ = i=1 ci 1Ai , d where each Ai ⊂ R is a d-box.

Remark: Simple functions are real-valued and they do not assume the values ±∞ by deﬁnition.

[138] [Approximating measurable functions with simple functions] Let (X, A) be a measurable space and let f : X → [−∞, ∞] be measurable. Then there is a sequence (ϕn) of simple functions on X converging to f pointwise. Moreover, the following additional properties are true:

(i) If f ≥ 0, we may choose ϕn’s in such a way that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ f.

(ii) If f is bounded, we may choose ϕn’s so that (ϕn) → f uniformly.

(iii) If µ is a σ-ﬁnite measure on (X, A), we may choose ϕn’s as µ-bounded simple functions.

(iv) If the assumptions of (i) &(ii) hold, then ϕn’c can be chosen to satisfy the conclusions of both (i) & (ii). Similarly for the combination (i) & (iii); but not for the combination (ii) & (iii).

n n Proof. The idea is to partition the range of f. Let I(n, k) = [k/2 , (k + 1)/2 ) and Γn = {k ∈ Z : n n −n2 ≤ k < n2 }. Note that Pn := {I(n, k): k ∈ Γn} defines a partition of [−n, n) in such a way −1 that Pn+1 restricted to [−n, n) is a refinement of Pn. Let A(n, k) = f (I(n, k)) for n ∈ N and −1 −1 k ∈ Γn. Also, let A(n, −∞) = f ([−∞, −n)) and A(n, ∞) = f ([n, ∞]). Define ϕn : X → R as ∑ −n ϕ = −n 1 −∞ + k2 1 + n 1 ∞ . Check that (ϕ ) → f pointwise. n A(n, ) k∈Γn A(n,k) A(n, ) n Statements (i) and (ii) are clear from the construction. MEASURE THEORY 41 ∪ ∞ ∈ A ∞ (iii) Write X = j=1 Xj, where Xj and µ[Xj] < . We may also assume Xj’s are pairwise ∞ disjoint. Let fj be the restriction of f to Xj. For each j, choose a sequence (ϕj,n)n=1 of simple functions on Xj as described above, converging to fj pointwise. Define ϕn : X → R as ϕn(x) = ∪ ∈ ≤ ≤ ∈ ∞ ∞ ϕj,n(x) if x Xj for 1 j n, and ϕn(x) = 0 if x j=n+1 Xj. Since µ[Xj] < , each ϕj,n is

µ-bounded and hence ϕn is also µ-bounded. And evidently, (ϕn) → f pointwise. (iv) The assertions about the combinations (i) & (ii), and (i) & (iii) can be seen from the above constructions. If f : R → R is f ≡ 1, then the assumptions of (ii) & (iii) hold, but there does not exist any sequence of µL,1-bounded simple functions converging to f uniformly on R.

Exercise-37: Let f :[−2, 2] → [−∞, ∞] be deﬁned as f(x) = |x + 1| if x ∈ [−2, 0], f(x) = 1/x if x ∈ (0, 1), and f(x) = log(x − 1) if x ∈ [1, 2]. Check that f is measurable and draw the graphs of the simple functions ϕn constructed in the proof of [138] for n = 1, 2, 3.

Deﬁnition: Let (X, A, µ) be a measure space. We say that some property holds µ-almost everywhere (µ-a.e. for short) if there is A ∈ A with µ[A] = 0 such that the property holds on X \ A. For instance, for functions fn, f : X → [−∞, ∞], we say the sequence (fn) converges to f pointwise

µ-almost everywhere if there is A ∈ A with µ[A] = 0 such that (fn) → f pointwise on X \ A.

One advantage of working with a complete measure space is the following.

[139] Let (X, A, µ) be a complete measure space and let f, g, fn : X → [−∞, ∞]. (i) If f is measurable and f = g µ-a.e., then g is measurable.

(ii) If fn’s are measurable, and (fn) → f pointwise µ-a.e., then f is measurable.

Proof. (i) Let A ∈ A be so that µ[A] = 0 and f(x) = g(x) for every x ∈ Ac. Note that {x ∈ X : g(x) > a} = [{x ∈ X : f(x) > a} ∩ Ac] ∪ {x ∈ A : g(x) > a}. Now, {x ∈ X : f(x) > a} ∈ A since f is measurable, and {x ∈ A : g(x) > a} ∈ A by completeness. Hence g is measurable. c (ii) Let A ∈ A be so that µ[A] = 0 and (fn) → f pointwise on A . Deﬁne gn : X → [−∞, ∞] as c gn = fn on A , and gn = f on A. Then gn’s are measurable by part (i). Since (gn) → f pointwise on the whole of X, f is measurable by [137].

Example: Let (X, A, µ) be a measure space which is not complete, and Y ⊂ A ∈ A be with µ[A] = 0 and Y/∈ A. Then 1A is measurable and 1A = 1Y µ-a.e., but 1Y is not measurable.

[140] If f :[a, b] → R is Riemann integrable, then there are step functions ϕn :[a, b] → R such that

ϕ1 ≤ ϕ2 ≤ · · · ≤ f, and (ϕn) → f pointwise µL,1-a.e. Hence f is measurable.

Proof. Assume [a, b] = [0, 1] for simplicity. Let I(n, k) = [(k − 1)/2n, k/2n) for 1 ≤ k < 2n, n n n I(n, 2 ) = [1 − 1/2 , 1] and Pn = {I(n, k) : 1 ≤ k ≤ 2 }. If z(n, k) = inf{f(y): y ∈ I(n, k)}, then ∑ ∈ R → R 2n z(n, k) since f is bounded. Deﬁne the step function ϕn : [0, 1] as ϕn = k=1 z(n, k) 1I(n,k). 42 T.K.SUBRAHMONIAN MOOTHATHU

Since Pn+1 reﬁnes Pn, we have ϕn ≤ ϕn+1 ≤ f. If Y = {x ∈ [0, 1] : f is discontinuous at x}, then

µL,1[Y ] = 0 by [115]. Given x ∈ [0, 1] \ Y and ε > 0, choose δ > 0 by continuity. If n, k are such −n that 2 < δ and x ∈ I(n, k), then we get |f(x) − ϕn(x)| = |f(x) − z(n, k)| ≤ ε. Thus (ϕn) → f pointwise on [0, 1] \ Y , and so f is measurable by [139].

10. Limit theorems for measurable functions

If fn ≡ n and f ≡ ∞, then (fn) → f pointwise, but |f(x) − fn(x)| cannot be made small. Now we introduce some other notions of convergence, where the deﬁnition requires that |f(x) − fn(x)| is small in some sense. So we consider only real-valued functions.

Deﬁnition: Let (X, A, µ) be a measure space and let fn, f : X → R be measurable functions. (i) We say (fn) → f in measure if for every α > 0, we have limn→∞ µ[{x ∈ X : |f(x) − fn(x)| ≥ α} = 0; equivalently, if for every α > 0, there is n0 ∈ N such that µ[{x ∈ X : |f(x) − fn(x)| ≥ α}] ≤ α for every n ≥ n0. (ii) We say (fn) → f uniformly with negligible error (or almost uniformly) if for every α > 0, there is A ⊂ A with µ[A] < α such that (fn) → f uniformly on X \ A. For example, n if fn : [0, 1] → R is fn(x) = x , then (fn) → 0 uniformly with negligible error w.r.to µL,1.

[141] Let (X, A, µ) be a measure space, and let f, fn : X → R be measurable functions such that

(fn) → f uniformly with negligible error. Then, (fn) → f pointwise µ-a.e. and (fn) → f in measure.

Proof. Let Ak ∈ A be so that µ[Ak] < 1/k and (fn) → f uniformly on X \ Ak. Now see that ∩ → ∞ (fn) f pointwise outside the µ-null set k=1 Ak.

Next, choose n(k, α) ∈ N so that {x ∈ X : |f(x) − fn(x)| ≥ α} ⊂ Ak for every n ≥ n(k, α) using the uniform convergence on X \ Ak. Then, µ[{x ∈ X : |f(x) − fn(x)| ≥ α}] ≤ µ[Ak] < 1/k for every n ≥ n(k, α), and thus (fn) → f in measure.

Partial converse to [141] is given by the following two theorems.

[142] [Egorov’s theorem] Let (X, A, µ) be a measure space with µ[X] < ∞, and let f, fn : X → R be measurable functions such that (fn) → f pointwise µ-a.e. Then (fn) → f uniformly with negligible error (and hence (fn) → f in measure).

Proof. Let α > 0. We need to ﬁnd A ∈ A with µ[A] < α such that (fn) → f uniformly on X \ A. ∪∞ Let B ∈ A be such that µ[B] = 0 and (fn) → f pointwise on X \ B. Let A(k, m) = {x ∈ ∩ n=m | − | ≥ } ∈ A ⊃ ∞ ⊂ X : f(x) fn(x) 1/k . Then, A(k, m) , A(k, m) A(k, m + 1) and m=1 A(k, m) B for each k ∈ N because of pointwise convergence on X \ B. Since µ[X] < ∞ and µ[B] = 0, we ∩∞ get limm→∞ µ[A(k, m)] = µ[ A(k, m)] = 0 for each k ∈ N. Hence there is a subsequence (mk) m=1 ∪ ∑ k ∞ ∈ A ∞ k such that µ[A(k, mk)] < α/2 . Let A = k=1 A(k, mk) . Then µ[A] < k=1 α/2 = α. It may MEASURE THEORY 43 be seen that for every x ∈ X \ A and every n ≥ mk, we have |f(x) − fn(x)| < 1/k; from this it follows that (fn) → f uniformly on X \ A.

[143] [Riesz theorem] Let (X, A, µ) be a measure space and let fn, f : X → R be measurable → → functions such that (fn) f in measure. Then there is a subsequence (fnk ) such that (fnk ) f → uniformly with negligible error (and hence (fnk ) f pointwise µ-a.e.).

{ ∈ | − | ≥ −k} Proof. Choose an increasing sequence (nk) such that if Ak = x X : f(x) fnk (x) 2 , then ∑∞ ∪∞ µ[A ] < 2−k. Given α > 0, choose k ∈ N with 2−k < α and put A = A . Then k 0 k=k0 k=k0 k | − | −k ∈ \ ≥ µ[A] < α and moreover f(x) fnk (x) < 2 for every x X A and k k0, from which it follows → \ that (fnk ) f uniformly on X A.

Example: If fn : R → R is fn = 1[n,∞), then (fn) → 0 pointwise, but the convergence is not uniform on R \ A for any A ∈ L(R) with µL,1[A] < ∞. So the assumption µ[X] < ∞ is necessary in Egorov’s theorem. This example also shows that pointwise convergence does not imply convergence in measure.

Example: Let fn : [0, 1] → R be fn = 1(1/(n+1),1/n). Then (fn) → 0 pointwise on [0, 1], and uniformly on [ε, 1] for any ε > 0. But if A ⊂ [0, 1] is a Lebesgue null set, then (1/(n + 1), 1/n) is not contained in A for any n ∈ N and it follows that the convergence (fn) → 0 cannot be uniform on [0, 1] \ A. Thus the conclusion of Egorov’s theorem cannot be improved to uniform convergence outside a null set.

Example: Let (Jn) be a sequence of subintervals of [0, 1] with the following properties: (i) |Jn| → 0 as n → ∞, (ii) for each x ∈ [0, 1], there are inﬁnitely many n with x ∈ Jn, and (iii) for each x ∈ [0, 1], there are inﬁnitely many n with x ∈ [0, 1] \ Jn. For instance, consider the sequence → R [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], [2/3, 1], [0, 1/4],.... Let fn : [0, 1] be fn = 1Jn . Then

(fn) → 0 in measure, but (fn(x)) does not converge for any x ∈ [0, 1]. Hence convergence in measure does not even imply pointwise convergence µ-a.e. In particular, we cannot replace (fnk ) with (fn) in the conclusion of Riesz theorem.

2 2 Example: Let f : R → R be f(x, y) = x. Then f is measurable, and µL,2[{(x, y) ∈ R : |f(x, y)| ≥ n}] = ∞ for every n ∈ N. As a consequence, a sequence of simple functions can never converge to f in measure (∵ simple functions are bounded). Compare with [138]. Also, construct a similar example f : R → R.

Exercise-39: [Linearity of convergence] Let (X, A, µ) be a measure space and let f, g, fn, gn : X → R be measurable functions such that (fn) → f and (gn) → g in measure. Then (afn) → af and

(fn + gn) → f + g in measure, where a ∈ R. Prove similar results for pointwise convergence µ-a.e. 44 T.K.SUBRAHMONIAN MOOTHATHU and uniform convergence with negligible error. [Hint: {x ∈ X : |(f + g)(x) − (fn + gn)(x)| ≥ α} ⊂

{x ∈ X : |f(x) − fn(x)| ≥ α/2} ∪ {x ∈ X : |g(x) − gn(x)| ≥ α/2}.]

Remark: Let X be a metric space and let A, B ⊂ X be nonempty closed subsets with A ∩ B = ∅. Then note there is a continuous function f : X → [0, 1] such that f ≡ 0 on A, and f ≡ 1 on B. In fact, we may take f(x) = dist(x, A)/[dist(x, A) + dist(x, B)].

d d Exercise-40: (i) If A ∈ L(R ), then there are continuous functions fn : R → R such that (fn) → 1A d d in measure. (ii) If A ∈ L(R ) is such that µ[A] < ∞, then there are step functions ψn : R → R d such that (ψn) → 1A in measure. [Hint: (i) Let Fn ⊂ A ⊂ Un ⊂ R be such that Fn is closed, Un d is open and µL,d[Un \ Fn] < 1/n. Let fn : R → [0, 1] be a continuous function such that fn ≡ 1 d d on Fn and fn ≡ 0 on R \ Un. Then, {x ∈ R : 1A(x) ≠ fn(x)} ⊂ Un \ Fn. (ii) Fix n ∈ N.

By the remark given immediately after [134], there are ﬁnitely many d-boxes A1,...,Ak such that ∪ k µL,d[A∆ Aj] < 1/n. We may assume Aj’s are pairwise disjoint. Then for the step function ∑ j=1 ∪ k { ∈ Rd ̸ } k ψn := j=1 1Aj , we have that x : 1A(x) = ψn(x) = A∆ j=1 Aj.]

Exercise-41: Let (X, A, µ) be a measure space and let f, fn, fn,m : X → R be measurable functions.

(i) If (fn) → f in measure, and (fn,m) → fn in measure for each n ∈ N, then there are increasing → ∞ sequences (nk), (mk) such that (fnk,mk ) f in measure. (ii) Assuming µ[X] < if necessary, prove similar results for pointwise convergence µ-a.e., and uniform convergence with negligible ∈ N { ∈ | − | ≥ } error. [Hint: (i) Fix k . First choose nk such that µ[ x X : f(x) fnk (x) 1/k ] < 1/k. { ∈ | − | ≥ } { ∈ Then choose mk such that µ[ x X : fnk (x) fnk,mk (x) 1/k ] < 1/k. Hence µ[ x X : | − | ≥ } → f(x) fnk,mk (x) 2/k < 2/k. Now it is not diﬃcult to see (fnk,mk ) f in measure. (ii) Use part (i) and results [141], [142], [143]. See also the proof below.]

[144] Let B ⊂ Rd be a d-box and f : B → R be a measurable function. Then,

(i) There are step functions ψn : B → R such that (ψn) → f pointwise µL,d-a.e. on B.

(ii) There are continuous functions fn : B → R such that (fn) → f pointwise µL,d-a.e. on B. If B is closed, then we may choose fn’s so that fn|∂B ≡ 0 also (where ∂B is the boundary of B).

Proof. By [138], there is a sequence (ϕn) of simple functions converging to f pointwise on B. Since

µ[B] < ∞,(ϕn) → f in measure by [142] and [141].

(i) By Exercise-39 and Exercise-40, we may ﬁnd step functions ψn,m : B → R such that (ψn,m) → ϕn ∈ N → in measure for each n . Then by Exercise-41, there are (nk), (mk) so that (ψnk,mk ) f in measure. By Riesz theorem, a subsequence of (ψnk,mk ) converges to f pointwise µL,d-a.e. on B.

(ii) By Exercise-39 and Exercise-40, for each n ∈ N we may ﬁnd continuous functions fn,m : B → R → → such that (fn,m) ϕn in measure. Then by Exercise-41, there are (nk), (mk) so that (fnk,mk ) f in measure. By Riesz theorem, a subsequence of (fnk,mk ) converges to f pointwise µL,d-a.e. on B. MEASURE THEORY 45

Now suppose that B is closed and (fn) is a sequence of continuous functions converging to f pointwise on B \ Y , where Y is a Lebesgue null set. Let (An) be a sequence of closed d-boxes such ∪ ⊂ ⊂ · · · ⊂ ∞ → that A1 A2 int[B] and n=1 An = int[B]. Let hn : B [0, 1] be a continuous function so that hn ≡ 1 on An and hn ≡ 0 on ∂B. Then fn(x)hn(x) = 0 for every x ∈ ∂B, and (fnhn) → f pointwise outside the Lebesgue null set Y ∪ ∂B.

[145] [Approximating measurable functions with step functions and with continuous functions of compact support] Let f : Rd → R be measurable. Then, d (i) There are step functions ψn : R → R such that (ψn) → f pointwise µL,d-a.e. d (ii) There are continuous functions fn : R → R such that (fn) → f pointwise µL,d-a.e. We may also arrange fn’s to have compact support. ∪ Rd ∞ | Proof. Write = j=1 Bj, a disjoint union of d-boxes, and let gj = f Bj . By [144], there are step functions ψj,n : Bj → R so that (ψj,n) → gj pointwise µ-a.e. on Bj for each j ∈ N. Deﬁne d ψn : R → R as ψn(x) = ϕj,n(x) if x ∈ Bj and 1 ≤ j ≤ n; and ψn(x) = 0 otherwise. ∪ Rd ∞ | (ii) Write = j=1 Bj, a union of closed d-boxes with pairwise disjoint interiors. Let gj = f Bj .

For each j, let fj,n : Bj → R be continuous functions such that (fj,n) → gj pointwise µL,d-a.e. on d Bj, and fj,n ≡ 0 on ∂Bj (by [144]). Deﬁne fn : R → R as fn(x) = fj,n(x) if x ∈ Bj and 1 ≤ j ≤ n; and fn(x) = 0 otherwise.

[146] [Luzin’s theorem] Let f : Rd → R be a measurable function and ε > 0. Then, ∈ L Rd | Rd \ → R (i) There is Y ( ) such that µL,d[Y ] < ε, and f Rd\Y : Y is continuous. We may also choose Y to be open. d d (ii) There is a continuous function g : R → R such that µL,d[{x ∈ R : f(x) ≠ g(x)}] < ε.

d Proof. (i) Write µ = µL,d. Choose continuous functions fn : R → R so that (fn) → f pointwise

µ-a.e. By Egorov’s theorem, (fn) → f uniformly with negligible error on B(0, j) for each j ∈ N. d j Let Yj ∈ L(R ) be such that Yj ⊂ B(0, j), µ[Yj] < ε/2 and (fn) → f uniformly on B(0, j) \ Yj. If ∪ ∞ \ Rd \ → Y = j=1 Yj, then µ[Y ] < ε. Observe that B(0, j) Y is open in Y , and (fn) f uniformly on B(0, j) \ Y . Hence f|B(0,j)\Y is continuous. Since {B(0, j) \ Y : j ∈ N} is a relatively open cover Rd \ | for Y , we conclude that f Rd\Y is continuous. Now, replacing Y with a slightly bigger open superset (by the outer regularity of Lebesgue measure), we may assume Y is open also. d d (ii) By part (i), there is a closed set A ⊂ R such that µ[R \ A] < ε and f|A is continuous (∵ take d d A = R \ Y ). By Tietze extension theorem, there is a continuous g : R → R with g|A = f.

Remarks: (i) For a set A, saying that f|A is continuous is strictly weaker than saying that f is continuous at every point of A. For instance, if f = 1Q : R → R, then f|Q is continuous, but f is not continuous at any point of Q (or of R). (ii) If F ⊂ R is a fat Cantor set, f : R → R is 46 T.K.SUBRAHMONIAN MOOTHATHU f = 1F , and Y ⊂ R is any Lebesgue null set, then f restricted to R \ Y is not continuous at any point of the nonempty set F \ Y (∵ F \ Y ≠ ∅ since F has positive measure; and for any x ∈ F \ Y , there is a sequence in R \ (F ∪ Y ) converging to x since F is nowhere dense and Y is a null set). This shows that we cannot improve Luzin’s theorem by choosing Y to be a null set in [146]. (iii) d d If f : R → [−∞, ∞] is a measurable function with µL,d[{x ∈ R : f(x) = ±∞}] > 0, then clearly the conclusion of Luzin’s theorem fails for f.

Exercise-42: [Converse of Luzin] Let f : Rd → R be a function having the property that for each ∈ L Rd | Rd \ → R ε > 0, there is Y ( ) such that µL,d[Y ] < ε, and f Rd\Y : Y is continuous. Then f is −n d d measurable. [Hint: Choose Yn for ε = 2 , and let fn : R be fn = 0 on Yn and fn = f on R \ Yn. ∩ ∪ → ∞ ∞ Check that fn’s are measurable, and (fn) f pointwise outside the null set k=1 n=k Yn. Then use [139].]

The following result can be thought of as a functional analogue of the fact that for any A ∈ L(Rd) there is B ∈ B(Rd) such that A∆B is a Lebesgue null set.

[147] If f : Rd → R is (Lebesgue-Borel) measurable, then there is a B(Rd)-B(R) measurable d function g : R → R such that f = g µL,d-a.e.

d d d Proof. By [145], there are step functions ψn : R → R and D ∈ L(R ) with µL,d[R \ D] = 0 so that (ψn) → f pointwise on D. We may assume D is an Fσ set (hence Borel) by [124](v). d Note that any step function is Borel-Borel measurable. Deﬁne gn = ψn 1D. Then gn’s are B(R )- d B(R) measurable. If g = f 1D, then (gn) → g pointwise everywhere on R . If K ⊂ R is a nonempty compact set, let Um = {y ∈ R : dist(y, K) < 1/m} (which is an open set) and verify ∩ ∪ −1 ∞ ∞ −1 −1 ∈ B Rd B R that g (K) = m=1 n=m gn (Um). Hence g (K) ( ). Since ( ) is generated by compact d subsets of R, we get that g is B(R )-B(R) measurable. And by deﬁnition, f = g µL,d-a.e.

11. Integration with respect to a measure

Let (X, A, µ) be a measure space. We describe below an integration theory for measurable functions f : X → [−∞, ∞]. Strictly speaking, the measurability of f is not needed to deﬁne the integral, but measurability is needed to derive many properties of the integral. The integral of f with respect to µ will be deﬁned in three steps: (i) when f is simple and ≥ 0, (ii) when f ≥ 0, (iii) for arbitrary f.

Deﬁnition: Let (X, A) be a measurable space and let ϕ : X → R be a simple function. Suppose y1, . . . , ym ∈ R are the distinct elements of ϕ(X), and let Xj = {x ∈ X : ϕ(x) = yj} ∈ A for ∑ ∑ m m 1 ≤ j ≤ m. Then yj 1X is called the canonical representation of ϕ. If yj 1X is the j=1 j ∪ j=1 j m canonical representation of ϕ, then note that X = j=1 Xj is a disjoint union. MEASURE THEORY 47

Integration, step-1: Let (X, A, µ) be a measure space. If ϕ : X → [0, ∞) is a simple function with ∑ ∫ m canonical representation ϕ = yj 1X , deﬁne the integral of ϕ with respect to µ as ϕdµ = ∑ j=1 j X m j=1 yjµ[Xj].

Remark: The reason for the assumption ϕ ≥ 0 is that ∞ − ∞ is undeﬁned. For instance consider the simple function ϕ : R → R given by ϕ = 2 1(∞,0) + (−3) 1[0,∞), and try to imitate the deﬁnition given above to see what goes wrong.

[148] Let (X, A, µ) be a measure space, let ϕ ≥ 0, ψ ≥ 0 be simple functions on X, and let c ∈ [0, ∞). Then, ∫ ∫ (i) (cϕ)dµ = c ϕdµ. ∫X X∫ ∫ (ii) (ϕ + ψ)dµ = ϕdµ + ψdµ. X∑ X X ∫ n (iii) If ai 1A is any representation (not necessarily canonical) of ϕ with ai ≥ 0, then ϕdµ = ∑ i=1 i X n aiµ[Ai]. i=1∫ (iv) ϕdµ = 0 iﬀ ϕ = 0 µ-a.e. X ∫ ∫ (v) If ϕ ≤ ψ µ-a.e., then ϕdµ ≤ ψdµ. ∫X ∫X (vi) If ϕ = ψ µ-a.e., then X ϕdµ = X ψdµ. ∑ m Proof. (i) The case c = 0 is clear. If c > 0 and if yj 1X is the canonical representation of ϕ, ∑ j=1 j m then j=1 cyj 1Xj is the canonical representation of cϕ. ∑ ∑ m n (ii) Let ϕ = aj 1A , ψ = bk 1B be the canonical representations. Then, we see that ∪ ∪ j=1 j k=1 k m n ∩ ∈ ∩ X = j=1 k=1(Aj Bk) is a disjoint union and (ϕ + ψ)(x) = aj + bk if x Aj Bk. Let c1, . . . , ct ∈ R be the distinct elements of (ϕ + ψ)(X). Let Γi = {(j, k): aj + bk = ci} and let Ci = ∪ ∑ t ∈ (Aj ∩Bk) for 1 ≤ i ≤ t. Then ci 1C is the canonical representation of ϕ+ψ. Therefore, ∫ (j,k) Γi ∑ ∑ i=1∑ i ∑ ∑ t t t (ϕ + ψ)dµ = ciµ[Ci] = ci( ∈ µ[Aj ∩ Bk]) = ( ∈ ((aj + bk)µ[Aj ∩ X ∑ ∑ i=1 i=1 ∑(j,k) Γi ∑ i=1∑ (j,k)∑Γi m n m n n m Bk]) = ((aj +bk)µ[Aj ∩Bk]) = aj( µ[Aj ∩Bk])+ bk( µ[Aj ∩Bk]) = ∑ j=1 k∑=1 ∫ ∫j=1 k=1 k=1 j=1 m n j=1 ajµ[Aj] + k=1 bkµ[Bk] = X ϕdµ + X ψdµ. ∫ ∫ ∑ ∑ ∫ ∑ n n n (iii) We have X ϕdµ = X ( i=1 ai 1Ai )dµ = i=1 ai X 1Ai dµ = i=1 aiµ[Ai] by (i) and (ii). ∫ (iv) This is direct from the deﬁnition of X ϕdµ, considering the canonical representation of ϕ. ∫ (v) In view of (iv), we may assume ϕ ≤ ψ everywhere. Also we may suppose ψdµ < ∞. If ∫ ∫ X ∫ ϕ1 = ψ − ϕ, then note that ϕ1 ≥ 0 is a simple function. Hence by (ii), ψdµ = ϕdµ + ϕ1dµ. ∫ ∫ X X X ≥ ∞ ≥ Since all the terms are 0 and < , we get X ψdµ X ϕdµ. (vi) This follows from (v) since ϕ ≤ ψ and ψ ≤ ϕ µ-a.e.

Integration, step-2: Let (X, A, µ) be a measure space. If f : X → [0, ∞] is a measurable function, ∫ ∫ { ≤ ≤ } we define the integral of f with respect to µ as X fdµ = sup X ϕdµ : ϕ is simple and 0 ϕ f . If f ≥ 0 is a simple function, then this definition agrees with the previous one by [148](iii). Also 48 T.K.SUBRAHMONIAN MOOTHATHU ∫ ∫ → ∞ ≤ ≤ note that if f, g : X [0, ] are measurable functions with f g, then X fdµ X gdµ by definition.

Exercise-43: [Markov’s inequality] Let (X, A, µ) be a measure space, f : X → [0, ∞] be a measurable ∫ function, and let At = {x ∈ X : f(x) ≥ t}. Then tµ[At] ≤ fdµ for 0 < t < ∞.[Hint: ∫ ∫ X ≤ ≤ ≤ 0 t 1At f and hence X t 1At dµ X fdµ.] Exercise-44: Let (X, A, µ) be a measure space, f : X → [0, ∞] be a measurable function, let ∫ A = {x ∈ X : f(x) = ∞} and B = {x ∈ X : f(x) > 0}. (i) If fdµ < ∞, then µ[A] = 0. ∫ X ∈ ∞ (ii) If X fdµ = 0, then µ[B] = 0. [Hint: (i) If µ[A] > 0, choose t (0, ) large enough ∫ ∪∞ so that tµ[A] > X fdµ; this contradicts Markov’s inequality. (ii) Write B = n=1 Bn, where

Bn = {x ∈ X : f(x) > 1/n} and use Markov’s inequality.]

Integration, step-3: Let (X, A, µ) be a measure space. If f : X → [−∞, ∞] is a measurable ∫ ∫ function, and if min{ f +dµ, f −dµ} < ∞, we define the integral of f with respect to µ as ∫ ∫ ∫ X X ∫ ∫ ∫ + − − + − ∞ X fdµ = X f dµ X f dµ. If both X f dµ and X f dµ are equal to , the integral X fdµ is undefined (since ∞ − ∞ is undefined).

Definition Let (X, A, µ) be a measure space. A measurable function f : X → [−∞, ∞] is µ- ∫ ∫ ∫ ∫ | | ∞ { + − } ∞ integrable if X fdµ is defined and X fdµ < ; equivalently if max X f dµ, X f dµ < . Definition: Let (X, A, µ) be a measure space, let A ∈ A and let f : A → [−∞, ∞] be a measurable ∫ ∫ function. We define the integral of f on A with respect to µ as fdµ = f 1 dµ, if it exists. ∫ ∫ A X A | | ∞ We say f is µ-integrable on A if A fdµ is defined and A fdµ < .

Remark: Let (X, A, µ) be a measure space and let f = f1 + if2 : X → C be a function. If both ∫ ∫ ∫ f1, f2 are µ-integrable, we say f is µ-integrable and deﬁne X fdµ = X f1dµ + i X f2dµ. Thus we can carry over our integration theory to complex valued functions in a natural way, but we will not discuss this here.

Deﬁnition: Let X ∈ L(Rd) and let f : X → [−∞, ∞] be a measurable function. We say f is Lebesgue integrable if f is µL,d-integrable on X. Note that the theory of Lebesgue integrable functions is a special case of the general integration theory; however, the general integration theory is motivated by Lebesgue integration theory.

12. Properties of the integral, and limit theorems

As seen in the theory of measurable functions, often a complicated function can be approximated by a sequence of functions which are more elementary in nature and easier to handle. Therefore, ∫ ∫ we are primarily interested in knowing whether the equality X fdµ = limn→∞ X fndµ holds if

(fn) → f in some sense. That is, we want to know whether ‘limit’ and ‘integration’ can be MEASURE THEORY 49 ∫ ∫ interchanged (i.e., whether X (limn→∞ fn)dµ = limn→∞ X fndµ). We give some examples to say that this need not happen always.

Example-1: Let X = [0, 2], µ = µ , and fn = n 1 . Then (fn) → 0 pointwise on X, but ∫ L,1 ∫ (1/n, 2/n) ̸ limn→∞ X fndµ = limn→∞ 1 = 1 = 0 = X 0dµ.

Example-2: Let µ = µ and fn = (1/n) 1 . Then fn’s are uniformly bounded by the constant L,1 (0,n) ∫ ∫ function 1, and (fn) → 0 uniformly on R. But limn→∞ R fndµ = limn→∞ 1 = 1 ≠ 0 = R 0dµ.

Example-3: Let µ = µ and fn = (1/n) 1 ∞ . Then f1 > f2 > ··· > 0 and (fn) → 0 uniformly ∫ L,1 (n, ) ∫ on R. But limn→∞ R fndµ = limn→∞ ∞ = ∞= ̸ 0 = R 0dµ.

However, we can interchange limit and integration in certain situations.

[149] [Monotone (increasing) convergence theorem] Let (X, A, µ) be a measure space and let fn :

X → [0, ∞] be measurable functions on X with 0 ≤ f1 ≤ f2 ≤ · · · . If (fn) converges pointwise to ∫ ∫ → ∞ a function f : X [0, ], then X fdµ = limn→∞ X fndµ. ∫ ∫ Proof. By [137], f is measurable. Since fn ≤ f, we have fndµ ≤ fdµ and therefore ∫ ∫ X X limn→∞ fndµ ≤ fdµ. To prove the other inequality it suﬃces to show the following: if X X ∫ ∫ ϕ is a simple function on X with 0 ≤ ϕ ≤ f, then limn→∞ fndµ ≥ ϕdµ. ∑ X X m { ∈ Suppose ϕ = j=1 yj 1Xj is the canonical representation. Let 0 < λ < 1 and Xj,n = x Xj : fn(x) ≥ λyj}. Then Xj,n ⊂ Xj,n+1 since fn ≤ fn+1. Consider x ∈ Xj. Since f(x) ≥ ϕ(x) = yj, we ∪ ≥ ∈ N ∞ have that fn(x) λyj for all large n . This says that n=1 Xj,n = Xj for each j. As this is an increasing union, we obtain (µ[Xj,n]) → µ[Xj] as n → ∞ for each j. ∑ ∫ m If ϕn = λ yj 1X , then ϕn is a simple function with 0 ≤ ϕn ≤ fn. Therefore fndµ ≥ ∫ ∑j=1 j,n ∫ ∑ X m m ϕndµ = λ yjµ[Xj,n]. Taking the limit n → ∞, we get limn→∞ fndµ ≥ λ yjµ[Xj] = X∫ j=1 X j=1 ∈ λ X ϕdµ. Since λ (0, 1) can be arbitrarily close to 1, we are through.

Exercise-45: Let (X, A, µ) be a measure space and let f : X → [0, ∞] be a measurable function.

If (ϕn) is any sequence of simple functions on X such that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ f and (ϕn) → f ∫ ∫ pointwise, then X fdµ = limn→∞ X ϕndµ.[Hint: Direct application of [149].]

Remark: Also note that for any measurable f : X → [0, ∞], a sequence (ϕn) as mentioned in

Exercise-45 exists by [138]; moreover observe from the proof of [138] that ϕn’s are constructed by partitioning the range of f (where as in Riemann integration theory we partition the domain of f).

Before continuing with the limit theorems and their corollaries, we deduce below some basic properties of integration.

[150] [Basic properties of integration] Let (X, A, µ) be a measure space and let f, g : X → [−∞, ∞] be µ-integrable functions. Then, 50 T.K.SUBRAHMONIAN MOOTHATHU ∫ ∫ ∫ (i) [Linearity] af + bg is µ-integrable and (af + bg)dµ = a fdµ + b gdµ for every a, b ∈ R. ∫ X X X (ii) fdµ = 0 iﬀ f = 0 µ-a.e. X ∫ ∫ (iii) [Monotonicity] If f ≤ g µ-a.e., then fdµ ≤ gdµ. ∫ ∫ X (iv) If f = g µ-a.e., then X fdµ = X gdµ.

Proof. The proofs of (ii), (iii), and (iv) are left as exercises (use [148] and the deﬁnition of integration). We prove (i) below. After modifying on a set of measure zero, we may assume f, g are real-valued (see Exercise-44). We will consider f + g and cf separately, c ∈ R.

First suppose f ≥ 0, g ≥ 0. Let (ϕn), (ψn) be increasing sequences of nonnegative simple functions converging pointwise to f, g respectively. Then (ϕn + ψn) is an increasing sequence of nonnegative simple functions converging pointwise to f + g. Hence by Exercise-45 we have ∫ ∫ ∫ ∫ ∫ (f + g)dµ = limn→∞ (ϕn + ψn)dµ = limn→∞[ ϕndµ + ψndµ] = limn→∞ ϕndµ + X ∫ ∫ X ∫ X X X limn→∞ X ψndµ = X fdµ + X gdµ. Now consider the general case, and let h = f + g. Note that h+ + h− = |h| = |f + g| ≤ |f| + |g| = ∫ ∫ + − + − + − ∞ f +f +g +g . Hence by the additivity proved above, we get X h dµ+ X h dµ < , and this implies h is µ-integrable. One common mistake now is to think that (f +g)+ = f + +g+ which is not true. However, we have h+−h− = h = f +g = (f +−f −)+(g+−g−) or h++f −+g− = h−+f ++g+. ∫ ∫ ∫ ∫ ∫ ∫ Using additivity, we have h+dµ+ f −dµ+ g−dµ = h−dµ+ f +dµ+ g+dµ. Since all X X X X X∫ ∫X ∫ the functions involved are integrable, we may group them back to get hdµ = fdµ + gdµ. ∫ ∫ X X X Next we would like to show (cf)dµ = c fdµ. First suppose f ≥ 0 and c ∈ [0, ∞). From X X ∫ ∫ [148] and the deﬁnition of integration, we easily get (cf)dµ = c fdµ. Since (−f)+ = 0 and ∫ ∫ ∫ X ∫ X ∫ (−f)− = f, we have (−f)dµ = 0dµ − fdµ = − fdµ. It follows that (−cf)dµ = ∫ X X X X X − c X fdµ also. Use these and the additivity proved above to establish the general case. ∫ ∫ ∫ Remark: The proof of [150](i) also gives the identities (f + g)dµ = fdµ + gdµ and ∫ ∫ X X X → ∞ ∈ ∞ X (cf)dµ = c X fdµ for measurable f, g : X [0, ] and c [0, ). Similar remark about properties (ii), (iii), (iv) for nonnegative measurable functions.

Now we discuss some consequences of the monotone convergence theorem. The following result says that Lebesgue integration theory is indeed a generalization of Riemann integration theory. ∫ [151] Let f :[a, b] → R be Riemann integrable. Then, f is Lebesgue integrable and fdµ = ∫ [a,b] L,1 b a f(x)dx.

Proof. Write µ = µL,1 and assume [a, b] = [0, 1]. By [140] we know that f is measurable. Since f is Riemann integrable, f is bounded. Therefore, replacing f with f + c for some constant c > 0, we may assume f ≥ 0. Let I(n, k) = [(k − 1)/2n, k/2n) for 1 ≤ k < 2n and I(n, 2n) = [1 − 1/2n, 1].

Let y(n, k) = inf{f(x): x ∈ I(n, k)}, and deﬁne the step (simple) function ϕn : [0, 1] → R as ϕn = MEASURE THEORY 51 ∑ 2n y(n, k) 1 . From the proof of [140] we know that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ f and (ϕn) → f k=1 I(n,k) ∫ ∫ pointwise µ-a.e. Hence by [149] and [150](iii), we conclude that fdµ = limn→∞ ϕndµ. ∫ ∑ ∑ [0,1] [0,1] 2n | | 2n n − − n Now [0,1] ϕndµ = k=1 y(n, k) I(n, k) = k=1 y(n, k)(k/2 (k 1)/2 ), which is the lower n Riemann sum L(f, Pn) corresponding to the partition Pn = {I(n, k) : 1 ≤ k ≤ 2 }. Since the ∫ → ∞ → 1 maximal length of an interval in Pn goes to 0 as n , we have (L(f, Pn)) 0 f(x)dx by Riemann’s theory.

Exercise-46: Let (X, A, µ) be a measure space and let f : X → [−∞, ∞] be measurable. Then, (i) ∫ ∫ | | | | ≤ | | f is µ-integrable iﬀ f is µ-integrable, (ii) if f is µ-integrable, then X fdµ X f dµ.[Hint: (i) If f is µ-integrable, then use |f| = f + +f − and additivity from [150](i) to conclude |f| is integrable. For the converse, note that max{f +, f −} ≤ |f|. (ii) Use similar ideas.]

Remark: Thus in our integration theory, f is integrable iﬀ |f| is integrable. Now, inﬁnite summation is a discrete analogue of integration, and for a real series we know that convergence does not imply absolute convergence. There is an integration theory known as Henstock-Kurzweil theory (which is in a sense a further generalization of Lebesgue integration theory) where the integrability of f does not necessarily imply the integrability of |f|.

Exercise-47: Let (X, A, µ) be a measure space and let fn : X → [0, ∞] be measurable for n ∈ N. ∫ ∑ ∑ ∫ ∞ ∞ ··· Then X ( n=1 fn)dµ = n=1 X fndµ.[Hint: Apply [149] and additivity to gn = f1 + + fn.]

−3 Example: Let µ = µL,1. Let (In) be a sequence of pairwise disjoint intervals in R with |In| = n and ∑ ∫ ∑ ∫ ∑ ∞ ∞ −2 ∞ 2 ∞ −1 ∞ let f = n=1 n 1In . Then by Exercise-47, R fdµ = n=1 n < and R f dµ = n=1 n = . This shows that the product of two µ-integrable functions need not be µ-integrable.

Exercise-48: [Another way of constructing measures] Let (X, A, µ) be a measure space, let f : X → ∫ ∞ A → ∞ [0, ] be a measurable function, and deﬁne β : [0, ] as β[A] = A fdµ. Then (i) β is a measure on (X, A), and (ii) µ[A] = 0 implies β[A] = 0 for every A ∈ A (we say β is absolutely continuous with respect to µ)[Hint: (i) For countable additivity, use Exercise-47 with fn = f 1An .]

Remark: The converse of Exercise-48 is also true if µ, β are σ-ﬁnite measure. That is, β satisfying ∫ → ∞ property (ii) can be realized in the form β[A] = A fdµ for some measurable function f : X [0, ]. This non-trivial converse is known as Radon-Nikodym theorem.

Exercise-49: [Borel-Cantelli lemma] Let (X, A, µ) be a measure space, let A1,A2,... ∈ A be such ∑ ∩ ∪ ∞ ∞ ∞ ∞ { ∈ ∈ ∈ N} that n=1 µ[An] < and let A = k=1 n=k An = x X : x An for inﬁnitely many n . ∑∞ ∫ Then µ[A] = 0. [Hint: If f = 1A , then fdµ < ∞ by Exercise-47; hence µ[A] = µ[{x ∈ n=1 n ∪ X ∑ ∞} ≤ ∞ ≤ ∞ → → ∞ X : f(x) = ] = 0. Or note that µ[A] µ[ n=k An] n=k µ[An] 0 as k .] 52 T.K.SUBRAHMONIAN MOOTHATHU

Remark: Taking (An) to be the sequence [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], [2/3, 1], [0, 1/4],..., ∑ ∞ ∞ → we see that the hypothesis n=1 µ[An] < cannot be weakened to (µ[An]) 0 in Borel-Cantelli.

Exercise-50: Let (X, A, µ) be a measure space and f : X → [−∞, ∞] be µ-integrable. Then for ∫ every ε > 0 there is δ > 0 such that |f|dµ < ε for every A ∈ A with µ[A] < δ.[Hint: We may A ∪ ≥ { ∈ ≤ } \ ∞ { assume f 0. Let An = x X : f(x) n and fn = f 1An . Then µ[X n=1 An] = µ[ x : ∪∞ f(x) = ∞}] = 0, and (fn) increasingly converges to f pointwise on An. By [149], there is n ∫ n=1 ∫ ∫ ∫ − ∈ − with X (f fn)dµ < ε/2. Choose any δ (0, ε/2n) and use the splitting f = (f fn) + fn.] ∫ For a sequence (fn) of nonnegative measurable functions, even when we cannot say lim fn = ∫ (lim fn), we can say which side is bigger.

[152] [Fatou’s lemma] Let (X, A, µ) be a measure space and let fn : X → [0, ∞] be measurable ∫ ∫ ∈ N ≤ functions for n . Then X (lim infn→∞ fn)dµ lim infn→∞ X fndµ.

Proof. Let g = lim infn→∞ fn and g = inf ≥ fn. Then g and g ’s are measurable, 0 ≤ g1 ≤ g2 ≤ k n k ∫ k ∫ · · · ≤ g and (g ) → g pointwise. Hence by [149], we have gdµ = lim →∞ g dµ. If n ≥ k, then k ∫ ∫ X k X k ≤ ≤ gk fn and therefore X gkdµ infn≥k X fndµ. Use this above.

Remark: One way to remember the direction of the inequality in Fatou’s lemma is to keep the example fn = 1(n,∞) in mind.

In monotone convergence theorem and Fatou’s lemma, we considered only nonnegative functions. But often we may have to deal with a sequence of functions taking both positive and negative values. A basic convergence theorem for such cases is the following.

[153] [Lebesgue’s dominated convergence theorem] Let (X, A, µ) be a mesure space and let fn :

X → [−∞, ∞] be µ-integrable functions for n ∈ N. Suppose (fn) converges pointwise µ-a.e. to a measurable function f : X → [−∞, ∞]. Also suppose there is a µ-integrable function g : X → [0, ∞] such that |fn| ≤ g µ-a.e. for every n ∈ N. Then, (i) f is µ-integrable. ∫ ∫ (ii) fdµ = limn→∞ fndµ. X ∫ X | − | (iii) limn→∞ X f fn dµ = 0.

Proof. After modifying on a set of measure zero, we may assume that the µ-integrable functions fn and g are real valued (see Exercise-44(i)). By a similar modiﬁcation, we may also assume that

(fn) → f pointwise everywhere and |fn| ≤ g everywhere. ∫ ∫ | | ≤ → | | ≤ | | ≤ ∞ (i) We have f g since (fn) f pointwise and fn g. Hence X f dµ X gdµ < by Fatou’s lemma, and thus f is µ-integrable by Exercise-46. MEASURE THEORY 53

(ii) Note that g + fn ≥ 0 and g − fn ≥ 0, so we may apply Faou’s lemma to these functions. ∫ ∫ ∫ ∫ ∫ We have gdµ + fdµ = (g + f)dµ ≤ lim infn→∞ (g + fn)dµ = lim infn→∞[ gdµ + ∫ X∫ X ∫X ∫ X ∫ ∫X fndµ] = gdµ+lim infn→∞ fndµ. Hence fdµ ≤ lim infn→∞ fndµ. Similarly, gdµ− ∫X ∫ X X ∫ X ∫ X ∫ ∫ X fdµ = (g − f)dµ ≤ lim infn→∞ (g − fn)dµ = lim infn→∞[ gdµ − fndµ] = gdµ − X X∫ ∫ X ∫ X X X ≥ lim supn→∞ X fndµ, and hence X fdµ lim supn→∞ X fndµ.

(iii) Let hn = |f − fn|. Then hn’s are µ-integrable, (hn) → 0 pointwise and |hn|’s are bounded ∫ by the µ-integrable function |f| + |g|. So by applying part (ii) to hn’s we obtain |f − fn|dµ → ∫ X X 0dµ = 0.

Remark: In [153], the measurability of f is part of the hypothesis. But it can also be obtained as a conclusion by [139] provided the measure space (X, A, µ) is complete.

Exercise-51: Let (X, A, µ) be a measure space and f : X → [−∞, ∞] be µ-integrable. (i) Let ∪∞ ∫ ∫ A, A ∈ A be such that A ⊂ A ⊂ · · · and A = A . Then fdµ = lim →∞ fdµ. (ii) n 1 2 n=1 n A n An ∪∞ ∫ ∑∞ ∫ Let B ∈ A be pairwise disjoint and let B = B . Then fdµ = fdµ.[Hint: For n n=1 n B n=1 Bn | | ∪ · · · ∪ (i), apply [153] with fn = f 1An and g = f 1A . For (ii), use (i) with An = B1 Bn.] ∪∞ Remark: Let µ = µ and f : R → R be µ-integrable. If R = (an, bn) is an increasing union, ∫ L,1 ∫ ∫ n=1 ∫ then fdµ = lim →∞ fdµ = lim →∞ fdµ = lim →∞ fdµ by Exercise-51. R n (an,∞) n (−∞,bn) n (an,bn)

13. Lp spaces

Deﬁnition: [Lebesgue spaces Lp(µ)] Let (X, A, µ) be a complete measure space and 1 ≤ p < ∞. ∫ p { → R | |p ∞} We deﬁne L (µ) = f : X : f is measurable and X f dµ < , with the understanding that we identify f and g if f = g µ-a.e. (so the members of Lp(µ) are equivalence classes, strictly speaking). Note that L1(µ) is just the space of µ-integrable real-valued functions. Let L∞(µ) = {f : X → R : f is measurable and there is M ∈ [0, ∞) such that |f| ≤ M µ-a.e.}, again with the understanding that we identify f and g if f = g µ-a.e.

Seminar topic: For 1 ≤ p ≤ ∞, the space Lp(µ) is a real vector space under pointwise operations. ∫ ∥ ∥ | |p 1/p p ≤ ∞ ∥ ∥ { ∈ ∞ | | ≤ If we deﬁne f p = ( X f dµ) on L (µ) for 1 p < , and f ∞ = inf M [0, ): f ∞ p M µ-a.e.} on L (µ), then it can be proved with some eﬀort that ∥·∥p is a complete norm on L (µ) for 1 ≤ p ≤ ∞ (Minkowski inequality, Riesz-Fischer theorem). Also if f ∈ Lp(µ), g ∈ Lq(µ) and 1 1/p + 1/q = 1, then fg ∈ L (µ) and ∥fg∥1 ≤ ∥f∥p · ∥g∥q (Holder inequality).

Example: In general, there is no inclusion relation between Lp(µ) and Lq(µ) for 1 ≤ p < q ≤ ∞. ∑ R ∈ ∞ R → R ∞ −(1/α) ∈ p Let µ = µL,1 on and let α (1, ). (i) If f : is f = n=1 n 1(n,n+1), then f / L (µ) p for 1 ≤ p ≤ α and f ∈ L (µ) for α < p ≤ ∞. (ii) Let (In) be a sequence of pairwise disjoint intervals 54 T.K.SUBRAHMONIAN MOOTHATHU ∑ | | −(1+α) R → R ∞ ∈ p ≤ ∈ p with In = n . If f : is f = n=1 n 1In , then f L (µ) for 1 p < α but f / L (µ) for α ≤ p ≤ ∞.

Exercise-52: Let (X, A, µ) be a complete measure space with µ[X] < ∞. Then Lp(µ) ⊃ Lq(µ) for 1 ≤ p ≤ q ≤ ∞.[Hint: |f|p ≤ |f|q on A := {x ∈ X : |f(x)| > 1} and |f|p ≤ 1 on X \ A.]

Example: [Series summation is discrete analogue of integration] If µ is the counting measure on ∑ ∑ N P N p p { ∈ RN ∞ | |p ∞} ∥ ∥ ∞ | |p 1/p ∈ p ( , ( )), then L (µ) = l = x : n=1 xn < and x p = ( n=1 xn ) for x L (µ) ≤ ∞ ∞ ∞ { ∈ RN | | ∞} ∥ ∥ | | and 1 p < . Similarly, L (µ) = l = x : supn∈N xn < and x ∞ = supn∈N xn for x ∈ L∞(µ). Also, lp ⊂ lq if 1 ≤ p ≤ q ≤ ∞ (direction of inclusion opposite to that in Exercise-52).

We will discuss two properties of L1(µ) below.

1 [154] [Riesz-Fischer theorem] If (X, A, µ) is a complete measure space, then (L (µ), ∥ · ∥1) is a complete normed space (corresponding result is true for 1 < p ≤ ∞ also).

1 Proof. The verification that (L (µ), ∥ · ∥1) is a normed space is left to the reader. We prove only 1 completeness. Consider a Cauchy sequence (fn) from (L (µ), ∥ · ∥1). Since a Cauchy sequence −n converges iff a subsequence converges, we may assume ∥fn+1 − fn∥1 ≤ 2 for every n ∈ N, after ∑∞ passing onto a subsequence. Let g : X → [0, ∞] be defined as g(x) = |f1(x)|+ |fn+1(x)−fn(x)| ∫ n=1 { ∈ ∞} | | ∞ and let A = x X : g(x) = . By Exercise-47, we have X g dµ < , and hence µ[A] = 0 by ∑∞ Exercise-44. Define f : X → R as f(x) = 0 if x ∈ A and f(x) = f1(x) + (fn+1(x) − fn(x)) if ∫ ∫ n=1 x ∈ X \ A. Then f is measurable, and |f| ≤ g so that |f|dµ ≤ gdµ < ∞. Thus f ∈ L1(µ). ∑ X∑ X ∞ − ∞ − ∈ N Observe that f1(x) + k=1(fk+1(x) fk(x)) = fn(x) + k=n(fk+1(x) fk(x)) for any n .

Hence (fn) → f pointwise on X \ A, and |f − fn| ≤ g for every n ∈ N. Applying [153] to the ∫ − | − | → ∥ − ∥ → → ∞ sequence (f fn), we have X f fn dµ 0, that is, f fn 1 0 as n .

∑ ≤ ∞ n Remark: If 1 p < and if ϕ = j=1 aj 1Aj is the canonical representation of a simple function, p then ϕ ∈ L (µ) iﬀ µ[Aj] < ∞ for aj ≠ 0.

1 1 [155] [Approximation in L ] Let µ = µL,d. Then the following collections are dense in (L (µ), ∥·∥1): 1 (i) F1 = {ϕ ∈ L (µ): ϕ is a simple function}. 1 (ii) F2 = {ψ ∈ L (µ): ψ is a step function}. 1 (iii) F3 = {f ∈ L (µ): f is a continuous function with compact support}.

1 Proof. (i) Consider g ∈ L (µ). First suppose g ≥ 0. By [138](i), there are simple functions ϕn such 1 1 that 0 ≤ ϕ1 ≤ ϕ2 ≤ · · · ≤ g and (ϕn) → g pointwise. We have ϕn ∈ L (µ) since g ∈ L (µ). Hence ∫ | − | → ∥ − ∥ → → ∞ by [153], we have Rd g ϕn dµ 0, i.e., g ϕn 1 0 as n . In the general case write + − + − 1 g = g − g , and consider g , g separately to see F1 is dense in (L (µ), ∥ · ∥1). MEASURE THEORY 55 ∑ ∈ F ∈ F ∥ − ∥ ∵ n (ii) If ϕ 1 and ε > 0, then we can ﬁnd ψ 2 with ϕ ψ 1 < ε ( write ϕ = j=1 cj 1Aj with

µ[Aj] < ∞ and approximate each Aj with a ﬁnite union of d-boxes by [134]). Hence F2 is dense in 1 (L (µ), ∥ · ∥1). (iii) If A is a d-box and ε > 0, choose an open d-box B such that A ⊂ B and µ[B \ A] < ε. Let f : Rd → [0, 1] be a continuous function such that f ≡ 1 on A and f ≡ 0 on Rd \ B (see Exercise-40 ∫ ∫ ∈ F ∥ − ∥ ≤ | | ≤ \ also). Then f 3 and f 1A 1 B\A f dµ B\A 1dµ = µ[B A] < ε. By linearity, members 1 of F3 can approximate any step function ψ ∈ F2. Hence F3 is also dense in (L (µ), ∥ · ∥1).

Remark: Result [155] remains valid for 1 ≤ p < ∞; however the case of L∞(µ) is a little diﬀerent. Simple functions are dense in L∞(µ) because of [138](ii), but step functions and continuous functions ∞ ∞ are not dense in L (µ). If g : R → R is g = 1[0,∞) ∈ L (µ), then ∥g −ψ∥ ≥ 1 for any step function ∞ ∞ ψ ∈ L (µ) and ∥g − f∥∞ ≥ 1/3 for any continuous function f ∈ L (µ). ∑ F ′ 1 n Remark: Let µ = µL,d and 2 be the collection of all step functions in L (µ) of the form j=1 aj 1Aj , d where aj ∈ Q and Aj’s are d-boxes with vertices in Q . Then, by the help of [155](ii), we may see F ′ 1 p ≤ ∞ that 2 is a countable dense subset of L (µ) (more generally, of L (µ) for 1 p < ).

Exercise-53: [Translation invariance of Lebesgue integral] Let µ = µ and f ∈ L1(µ). Then, ∫ ∫ L,d ∫ f(x+y)dµ(x) = f(x)dµ(x) for every y ∈ Rd.[Hint: If A ∈ L(Rd), then 1 (x+y)dµ(x) = ∫Rd Rd ∫ Rd A 1 − (x)dµ(x) = µ[A − y] = µ[A] = 1 (x)dµ(x). By the linearity given by [150], we have ∫Rd A y ∫ Rd A Rd ϕ(x + y)dµ(x) = Rd ϕ(x)dµ(x) for every simple function ϕ. Now use [155].] ∫ Exercise-54: [Decay of Fourier coeﬃcients] Let µ = µ . Then we have limn→∞ f(x) sin(nx)dµ = ∫ L,1 R 1 0 = limn→∞ R f(x) cos(nx)dµ = 0 for every f ∈ L (µ). [Hint: Let [a, b] ⊂ R and ε > 0. Let n0 ∈ N ′ be such that 2π/n0 < ε/2. If n ≥ n0, ﬁnd integers k, m such that a = 2kπ/n ∈ (a − ε/2, a) and ∫ ∫ ∫ ′ b = 2mπ/n ∈ (b, b+ε/2). We have | sin(nx)dµ| ≤ | ′ ′ sin(nx)dµ|+ ′ ′ | sin(nx)|dµ ≤ ∫ [a,b] [a ,b ] [a , b ]\[a,b] 0 + ε. This shows limn→∞ R 1[a,b](x) sin(nx)dµ = 0. Now use linearity and [155](ii).]

Now we mention a few points about convergence in Lp.

∞ Remark: Suppose (fn) → f in L (µ) and let nk be such that ∥f − fn∥∞ ≤ 1/k for every n ≥ nk.

Then there are µ-null sets Ak ⊂ X such that |f(x) − fn(x)| ≤ 1/k for every x ∈ X \ Ak and n ≥ nk. ∪ → ∞ So (fn) f uniformly outside the µ-null set k=1 Ak. In particular, we see that convergence in L∞(µ) implies uniform convergence with negligible error, convergence in measure and pointwise convergence µ-a.e. Behavior in Lp(µ) for 1 ≤ p < ∞ is diﬀerent, as shown below.

Example: Let µ = µL,1 and let (Jn) be the sequence of intervals [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], → p ≤ ∞ [2/3, 1], [0, 1/4],.... If fn = 1Jn , then (fn) 0 in L (µ) for 1 p < . Here, this means only p that the area under the graph of fn goes to 0, so we cannot expect even pointwise convergence. In 56 T.K.SUBRAHMONIAN MOOTHATHU fact, we know already that (fn(x)) does not converge for any x ∈ [0, 1] as the sequence contains ∞ inﬁnitely many 0’s and inﬁnitely many 1’s. Also (fn) does not converge in L (µ).

p p [156] [L convergence - necessity] Let 1 ≤ p < ∞ and suppose (fn) → f in L (µ). Then,

(i) (fn) → f in measure. → (ii) There is a subsequence (fnk ) such that (fnk ) f uniformly with negligible error. → (iii) There is a subsequence (fnk ) such that (fnk ) f pointwise µ-a.e. ∫ ∫ Proof. (i) Let α > 0. If A = {x ∈ X : |f(x) − f (x)| ≥ α}, then µ[A ] = 1dµ ≤ α−p|f − n n n An An p −p p fn| dµ ≤ α ∥f − fn∥p → 0 as n → ∞ for 1 ≤ p < ∞. Now (ii) and (iii) follow from [143].

Examples: Let µ = µL,1 on R. (i) Let fn = (1/n) 1(0,en). Then (fn) → 0 uniformly and hence in ∞ p p n p L (µ). But (fn) does not converge in L (µ) for 1 ≤ p < ∞ since ∥fn∥p = e /n → ∞ as n → ∞. p (ii) Fix α ∈ [1, ∞) and let fn = (1/n) 1(0,nα). Then (fn) → 0 uniformly. Also (fn) → 0 in L (µ) iﬀ p > α. (iii) Fix α ∈ [1, ∞), and let fn = n 1(n−α,2n−α). Then (fn) → 0 pointwise; and (fn) p p converges in L (µ) iﬀ p < α. (iv) If fn = 1(n,n+1), then (fn) → 0 pointwise, but not in L (µ) for

1 ≤ p ≤ ∞. (v) If fn = n 1(1/n,2/n), then (fn) → 0 pointwise as well as in measure, but not in p n L (µ) for 1 ≤ p ≤ ∞. (vi) Let fn(x) = x 1(0,1)(x). Then |fn| ≤ 1; (fn) → 0 pointwise but not p ∞ uniformly; (fn) → 0 in L for 1 ≤ p < ∞, but (fn) does not converge in L (µ).

p p [157] [L convergence - suﬃciency] Let 1 ≤ p < ∞ and f, fn ∈ L (µ) be such that (fn) → f in one of the following sense - (i) uniformly with negligible error, (ii) in measure, or (iii) pointwise µ-a.e. p p If there is g ∈ L (µ) such that |fn| ≤ g µ-a.e., then (fn) → f in L (µ).

Proof. Since (i) implies both (ii) and (iii), it suﬃces to consider only (ii) and (iii). First suppose p p p p p 1 (fn) → f pointwise µ-a.e. If 1 ≤ p < ∞, then |f − fn| ≤ 2 g µ-a.e. and 2 g ∈ L (µ). Hence ∫ ∥ − ∥p | − |p → by Lebesgue’s dominated convergence theorem, f fn p = X f fn dµ 0. Now suppose p (fn) → f in measure. If the convergence does not happen in L (µ), there exist δ > 0 and a ∥ − ∥ ≥ subsequence (fnk ) such that f fnk p δ. By passing onto a further subsequence, we may → → p assume (fnk ) f pointwise µ-a.e. by [143]. Then (fnk ) f in L (µ) by what we proved above, a contradiction.

14. Product spaces and product measures ∏ A ∈ Deﬁnition: Let (Xα, α) be measurable spaces for α J, and let X = α∈J Xα. The product

σ-algebra A on X is deﬁned as the smallest σ-algebra on X such that each projection πα : X → Xα A A A { −1 ∈ ∈ A } is - α measurable. That is, is the σ-algebra generated by πα (Aα): α J and Aα α .

We are mainly interested in ﬁnite products, especially the product of two measurable spaces.

Let (Xi, Ai) be measurable spaces for i = 1, 2. Then by deﬁnition, the product σ-algebra A on MEASURE THEORY 57

X = X1 × X2 is generated by {A1 × X2 : A1 ∈ A1} ∪ {X1 × A2 : A2 ∈ A2}. It is then easy to see that A is also generated by A1 × A2 = {A1 × A2 : A1 ∈ A1 and A2 ∈ A2}.

Remarks: (i) Let (X, A1), (Y, A2) be measurable spaces and A be the product σ-algebra on X × Y .

If A ∈ A, then its projections πX (A), πY (A) may not belong to A1, A2 respectively. This is essentially because projections do not commute with complementation and countable intersection. ′ ′ To understand it better, consider A = {A ∈ A : πX (A) ∈ A1}, make an attempt to prove A is a σ-algebra, and ﬁnd out what goes wrong. To see what can be said positively, see Exercise-55 below. (ii) Lebesgue assumed wrongly that projections would commute with countable intersections, and this led him to the wrong belief that projections of Borel sets are Borel. Luzin corrected this mistake. It is a non-trivial fact that a projection of a Borel set need not be Borel.

Example: The projection of a Lebesgue measurable set need not be Lebesgue measurable also. If ∈ P R \L R ∗ × { } × { } ∈ L R2 × { } A ( ) ( ), then µL,2[A 0 ] = 0 and hence A 0 ( ). And A 0 projects to A.

y Deﬁnition: For A ⊂ X × Y , let Ax = {y ∈ Y :(x, y) ∈ A} and A = {x ∈ X :(x, y) ∈ A}.

Exercise-55: Let (X, A1), (Y, A2) be measurable spaces, and let A be the product σ-algebra on y X × Y . If A ∈ A, then Ax ∈ A2 for every x ∈ X and A ∈ A1 for every y ∈ Y .[Hint: Let ′ ′ A = {A ∈ A : A has the given property}. Check that A is a σ-algebra containing A1 × A2.]

Exercise-56: Let (X, A1), (Y, A2) be measurable spaces, and let A be the product σ-algebra on

X × Y . If C1 ⊂ A1, C2 ⊂ A2 are generating collections, then C1 × C2 generates A.[Hint: Let ′ ′ A be the σ-algebra generated by C1 × C2. Clearly, A ⊂ A. Now ﬁx C1 ∈ C1 and note that A′ { ∈ A × ∈ A′} C A′ A 2 := A2 2 : C1 A2 is a σ-algebra containing 2. Hence 2 = 2. Use this to show A′ { ∈ A { } × A ⊂ A′} A′ A A × A ⊂ A′ similarly that if 1 = A1 1 : A1 2 , then 1 = 1. Thus 1 2 also.]

[158] [Product measure] Let (Xi, Ai, µi) be measure spaces for i = 1, 2. Let X = X1 × X2 and

A be the product σ-algebra on X. Then there is a measure µ on (X, A) such that µ[A1 × A2] =

µ1[A1]µ2[A2] for every A1 × A2 ∈ A1 × A2. This µ is unique if µ1 and µ2 are σ-ﬁnite measures.

Proof. Verify that A1 × A2 is a semi-algebra on X. Define µ : A1 × A2 → [0, ∞] as µ = µ1 × ∪ × ∈ A × A × ∞ × µ2. Suppose A B 1 2 is such that A B = n=1(An Bn), a disjoint union with ∑∞ An × Bn ∈ A1 × A2. Then 1 (x) 1 (y) = 1 (x) 1 (y) for (x, y) ∈ X1 × X2. Fixing A B n=1 An Bn ∫ x ∈ X and integrating with respect to µ , we obtain by Exercise-47 that 1 (x) 1 (y)dµ (y) = 1 2 A X2 B 2 ∫ ∑∞ ∑∞ ∫ ∑∞ 1 (x) 1 (y)dµ (y) = 1 (x) 1 (y)dµ (y) = 1 (x)µ [B ]. Thus X2 n=1 An Bn 2 n=1 An X2 Bn 2 n=1 An 2 n ∑∞ 1A(x)µ2[B] = 1A (x)µ2[Bn]. Now integrating this similarly with respect to µ1, we have ∑ n=1 n ∑ ∞ × ∞ × µ1[A]µ2[B] = n=1 µ1[An]µ2[Bn]. That is, µ[A B] = n=1 µ[An Bn]. Thus µ is a pre-measure on the semi-algebra A1 × A2. Hence µ extends to a measure on A by Caratheodory’s extension theorem [131]. Now suppose µ1 and µ2 are σ-finite. So there are An ∈ A1, Bm ∈ A2 such that 58 T.K.SUBRAHMONIAN MOOTHATHU ∪ ∪ ∪ ∪ ∞ ∞ ∞ ∞ ∞ ∞ × X1 = n=1 An, X2 = m=1 Bm, µ1[An] < and µ2[Bm] < . Then X = n=1 m=1(An Bm) and µ[An × Bm] < ∞. Thus µ is σ-finite, and hence the extension is unique by [131].

[159] (i) If B is the product σ-algebra on Rd+k generated by B(Rd) × B(Rk), then B = B(Rd+k). (ii) L(Rd) × L(Rk) ⊂ L(Rd+k) (iii) If A is the product σ-algebra on Rd+k generated by L(Rd) × L(Rk), then A ̸= L(Rd+k), but d+k the completion of A w.r.to µL,d+k is L(R ) and µL,d+k is the unique product measure.

Proof. (i) This follows from Exercise-56 (take C1 = {d-boxes} and C2 = {k-boxes}) by noting that A ⊂ Rd+k is a (d + k)-box iﬀ A is the product of a d-box with a k-box.

d k d (ii) Any Y ∈ L(R ) × L(R ) can be written as Y = (B1 ∪ C1) × (B2 ∪ C2), where B1 ∈ B(R ), C1 ∈ d k k N (R ), B2 ∈ B(R ), and C2 ∈ N (R ). Then we see that Y can be written as Y = (B1 × B2) ∪ Z, d+k d+k d+k where Z ∈ N (R ). Also B1 × B2 ∈ B(R ) by part (i). Hence Y ∈ L(R ). A ⊂ L Rd+k ∈ P Rd \L Rd ∈ Rk ∗ ×{ } (iii) We have ( ) by part (ii). If A ( ) ( ), and y , then µL,d+k[A y ] = 0 and hence A × {y} ∈ L(Rd+k). But A × {y} ∈/ A by Exercise-55 and thus A ̸= L(Rd+k). Since B(Rd)×B(Rk) ⊂ L(Rd)×L(Rk), we have B(Rd+k) ⊂ A by part (i). Thus B(Rd+k) ⊂ A ⊂ L(Rd+k). d+k Now take completions to see L(R ) is the completion of A. Uniqueness of µL,d+k is clear since it is determined by its value on (d + k)-boxes.

d+k k y d d+k Remark: If A ∈ B(R ), then Ax ∈ B(R ) and A ∈ B(R ) for every (x, y) ∈ R by Exercise- 55 and [159](i). We cannot replace Borel σ-algebra with Lebesgue σ-algebra here because of the problem explained in the proof of [159](iii). However, it can be proved (with some eﬀort) that if d+k k d y d k A ∈ L(R ), then Ax ∈ L(R ) for almost every x ∈ R and A ∈ L(R ) for almost every y ∈ R .

[160] [Fubini-Tonelli Theorem] (Rough statement) Let (X, A1, µ1), (Y, A2, µ2) be σ-ﬁnite measure spaces, let µ be the unique product measure on X×Y and let f : X×Y → [−∞, ∞] be a measurable ∫ ∫ ∫ ∫ ∫ ≥ function. If f is either µ-integrable or 0, then X×Y fdµ = X ( Y fdµ2)dµ1 = Y ( X fdµ1)dµ2. Remark: The proof of [160] is left as a reading assignment. We just remark that the main step in the proof is to establish the result when f = 1A for some measurable A ⊂ X × Y . Once this is done, then the result extends to simple functions by linearity and to arbitrary measurable functions by approximation. The necessity of the σ-ﬁnite assumption can be seen from the following.

Example: Let X = Y = [0, 1], A1 = A2 = B([0, 1]), let µ1 = µ and let µ2 be the counting ∫ L,1 2 measure. If A = {(x, y) ∈ [0, 1] : x = y} and f = 1 , then f(x, y)dµ2 = µ2[Ax] = µ2[{x}] = 1 ∫ ∫ A∫ Y ∫ ∫ y { } ̸ and X f(x, y)dµ1 = µ1[A ] = µ1[ y ] = 0. Hence X ( Y fdµ2)dµ1 = 1 = 0 = Y ( X fdµ1)dµ2. MEASURE THEORY 59

15. Fundamental Theorem of Calculus (FTC) ∫ ∫ Let µ = µ for this section. We will write b fdµ for fdµ. Let f :[a, b] → R be Lebesgue L,1 ∫ a [a,b] → R x ′ integrable and let g :[a, b] be g(x) = a fdµ. Can we say that g is diﬀerentiable and g = f?

The answer is no; consider for example f = 1[0,1/2] on [0, 1]. An obstruction to getting a positive answer is the fact that if g is diﬀerentiable, then g′ should have intermediate value property (by Darboux theorem), but f may not have intermediate value property. However, if we ignore what happens on a small set, we can say something positive.

Deﬁnition: (Recall) We say f :[a, b] → R is absolutely continuous if for every ε > 0 there is δ > 0 such that for any ﬁnite collection of points x1 ≤ y1 ≤ x2 ≤ y2 ≤ · · · ≤ xn ≤ yn in [a, b] with ∑ ∑ n | − | n | − | j=1 yj xj < δ, we have j=1 f(yj) f(xj) < ε. ∫ Exercise-57: Let µ = µ and let f :[a, b] → R be µ-integrable. (i) If g :[a, b] → R is g(x) = x fdµ, L,1 ∫ a then g is absolutely continuous. (ii) If x fdµ = 0 for µ-a.e. x ∈ [a, b], then f = 0 µ-a.e. [Hint: (i) a ∫ ∫ x ∈ y Use Exercise-50. (ii) By the continuity of g, a fdµ = 0 for every x [a, b]. Hence x fdµ = 0 for a ≤ x ≤ y ≤ b. If U ⊂ [a, b] is open, then U is a disjoint union of countably many intervals, and ∫ ∫ ∫ ∫ ⊂ b − hence U fdµ = 0 by Exercise-51(ii). If K [a, b] is closed, then K fdµ = a fdµ [a,b]\K fdµ = 0.

Now An = {x : f(x) > 1/n} and Bn = {x : f(x) < −(1/n)} are µ-null by inner regularity.] ∫ → R b ′ ≤ − Exercise-58: Let g :[a, b] be increasing. Then a g dµ g(b) g(a). [Hint: By [112], g is diﬀerentiable µ-a.e. Also g′ ≥ 0 since g is increasing. Extend g to [a, b + 1] by setting g(x) = g(b) for x ≥ b. If hn :[a, b] → R is hn(x) = n(g(x + 1/n) − g(x)), then hn’s are measurable and ′ ′ (hn) → g pointwise µ-a.e. in [a, b]; so g is measurable. And, hn ≥ 0. So by Fatou’s lemma, ∫ ∫ ∫ ∫ ∫ b ′ b b b b+1/n g dµ ≤ lim infn→∞ hndµ. Now, hndµ = n (g(x + 1/n) − g(x))dµ = n[ g(x)dµ − ∫a ∫ a ∫ a ∫ a ∫ a+1/n b b+1/n − a+1/n ≤ b+1/n − a+1/n − a g(x)dµ] = n[ b gdµ a gdµ] n[ b g(b)dµ a g(a)dµ] = g(b) g(a).] [161] [FTC - ﬁrst part] Let µ = µ . Let f :[a, b] → R be µ-integrable and g :[a, b] → R be ∫ L,1 x ′ ∈ g(x) = a fdµ. Then g is absolutely continuous and g (x) = f(x) for µ-a.e. x [a, b].

Proof. By Exercise-57, g is absolutely continuous. Now we may assume f ≥ 0 by considering f + and f − separately. Then g is increasing, and hence differentiable µ-a.e. by [112]. From the hint of ∫ ′ −1 c+δ Exercise-58, we also see that g is measurable. For later use observe that lim → (δ gdµ) = ∫ δ 0 c ∈ −1 c+δ g(c) for any c (a, b) since g is continuous and g(c) = δ c g(c)dµ. ′ Case-1: f is bounded, say |f| ≤ M. Let hn(x) = (g(x + 1/n) − g(x))/(1/n). Then (hn) → g ∫ ∫ | | | x+1/n | ≤ | x+1/n | | ′| ≤ pointwise µ-a.e. We have hn = n x fdµ n x Mdµ = M and hence g M µ-a.e. also. The constant function M is µ-integrable over [a, b] since [a, b] has finite measure. Hence by ∫ Lebesgue dominated convergence theorem, for any c ∈ (a, b) we have c g′dµ = ∫ ∫ ∫ a ∫ c c+1/n − a+1/n − c limn→∞ a hndµ = limn→∞[n c g(x)dµ n a g(x)dµ] = g(c) g(a) = g(c) = a fdµ. 60 T.K.SUBRAHMONIAN MOOTHATHU ∫ c ′ − ∈ ′ So a (g f)dµ = 0 for all c (a, b) and hence g = f µ-a.e. by Exercise-57. ∫ x ′ Case-2: f is unbounded. Let fn = min{f, n} and gn(x) = fndµ. By case-1, g = fn µ-a.e. If ∫ a n x ′ sn(x) = (f − fn)dµ, then sn is monotone increasing (∵ f − fn ≥ 0), and hence s ≥ 0 µ-a.e. a ∫ n x ′ ′ ′ ≥ ∈ N Since g(x) = a fdµ = gn(x) + sn(x), we conclude that g = gn + sn fn µ-a.e., for each n . By tending n → ∞, we get g′ ≥ f µ-a.e. Now to show that g′ = f µ-a.e., it suffices to show ∫ b(g′ − f)dµ = 0 because of Exercise-44(ii). a ∫ Since g′ ≥ f µ-a.e., we have b(g′ − f)dµ ≥ 0. On the other hand, by Exercise-58 we get ∫ ∫ a ∫ b ′ ≤ − b b ′ − ≤ a g dµ g(b) g(a) = g(b) = a fdµ, or a (g f)dµ 0, so we are through.

Exercise-59: [Devil’s staircase aka Cantor function] There is an increasing (hence of bounded variation), continuous surjection f : [0, 1] → [0, 1] such that f ′ = 0 µ-a.e. Let K be the middle-third Cantor set. First deﬁne f on [0, 1] \ K by setting f ≡ 1/2 on (1/3, 2/3), f ≡ 1/4 on (1/9, 2/9), f ≡ 3/4 on (7/9, 8/9), and so on. Then f on [0, 1] \ K is uniformly continuous and hence it has a continuous extension to [0, 1]. Verify the stated properties. A similar, strictly increasing example g : [0, 1] → [0, 2] with certain other interesting properties is obtained by taking g(x) = f(x) + x. Learn more about this from books/other sources.

Remark: The Cantor function is not absolutely continuous because of the following.

Exercise-60: Let f :[a, b] → R be absolutely continuous. If f ′ = 0 µ-a.e., then f is constant. [Hint: We have to use Vitali’s lemma for this. See p.109 of H.L. Royden, Real Analysis, 3rd Edn.]

[162] [FTC - second part] Let f :[a, b] → R be absolutely continuous. Then f ′ exists µ-a.e., f ′ is ∫ b ′ − µ-integrable, and a f dµ = f(b) f(a).

Proof. We know from [114] that f is of bounded variation, and f ′ exists µ-a.e. We may write f = g − h where g, h are monotone increasing, by (the proof of) [113]. Now f ′ is measurable because g′, h′ are measurable (see the hint of Exercise-58). Also, |f ′| ≤ g′ + h′. Using Exercise-58 ∫ ∫ we obtain b |f ′|dµ ≤ b(g′ + h′)dµ ≤ g(b) + h(b) − g(a) − h(a) < ∞, and hence f ′ is µ-integrable. ∫ a a x ′ ′ ′ If s(x) = a f dµ, then s is absolutely continuous by Exercise-57, and s = f µ-a.e. by [161]. So the function f − s is absolutely continuous and (f − s)′ = f ′ − s′ = 0 µ-a.e. Therefore there is c ∈ R such that f(x) = s(x) + c for every x ∈ [a, b] by Exercise-60. Since s(a) = 0, we have c = f(a), and ∫ − x ′ ∈ thus f(x) f(a) = a f dµ for every x [a, b].

Remarks: (i) The example of Cantor function shows that we cannot weaken the hypothesis of absolute continuity in [162] by either the bounded variation or the monotone increasing property. (ii) From [161] and [162] we can conclude the following: a function f :[a, b] → R is absolutely ∫ ′ ′ x ′ ∈ continuous iﬀ f exists µ-a.e., f is µ-integrable and f(x) = f(a) + a f dµ for every x [a, b]. MEASURE THEORY 61

16. Measures on metric spaces

The following deﬁnitions are motivated by considering the construction and properties of Lebesgue (outer) measure on Rd.

Definition: Let X be a metric space and let B(X) denote the Borel σ-algebra on X, i.e., the σ-algebra generated by open subsets of X (equivalently, by closed subsets of X). (i) Let µ∗ be an outer measure on X. We say A ⊂ X is µ∗-measurable if A satisfies the Caratheodory condition: µ∗[E] = µ∗[A ∩ E] + µ∗[Ac ∩ E] for every E ⊂ X. Recall from [121] that µ∗ restricted to the σ-algebra of all µ∗-measurable subsets of X is a complete measure. (ii) An outer measure µ∗ on X is a Borel outer measure if every Borel subset of X is µ∗-measurable. Similarly, a measure µ on some σ-algebra A on X is called a Borel measure if B(X) ⊂ A. (iii) A Borel outer measure µ∗ on X is Borel regular if for any A ⊂ X, there is B ∈ B(X) such that A ⊂ B and µ∗[A] = µ∗[B]. (iv) A Borel measure µ on X is locally finite if X can be covered by open subsets of finite measure. When X is locally compact, µ is locally finite ⇔ µ[K] < ∞ for every compact K ⊂ X. (v) A Borel measure µ on X is inner regular if µ[B] = sup{µ[K]: K ⊂ X compact and K ⊂ B} for every B ∈ B(X). (vi) A Borel measure µ on X is outer regular if µ[B] = inf{µ[U]: U ⊂ X open and B ⊂ U} for every B ∈ B(X). Note that inner regularity implies outer regularity when µ[X] < ∞ (apply inner regularity to Bc), and outer regularity implies inner regularity when X is compact and µ[X] < ∞.

Examples: (i) The Lebesgue outer measure on Rd is Borel regular, and the Lebesgue measure on Rd is locally ﬁnite, inner regular, and outer regular.

(ii) Fix y ∈ R and let µy be the Dirac measure at y on (R, P(R)) defined as µy[A] = 1 if y ∈ A and µy[A] = 0 if y∈ / A. As an outer measure µy is Borel regular. As a measure, µy is locally finite, inner regular, and outer regular. If we modify the definition and put µ[A] = ∞ if y ∈ A and µ[A] = 0 if y∈ / A, then µ is not locally finite, though µ remains inner regular and outer regular.

(iii) Write Q = {y1, y2,...} and let µ be the finite Borel measure on (R, P(R)) given by µ = ∑ ∞ −j ⊂ R j=1 2 µyj , where µyj is the Dirac measure at yj. For A , the cocountable (hence Borel) set B := A ∪ (R \ Q) satisfies A ⊂ B and µ[A] = µ[B]. Thus µ is Borel regular. Given B ∈ B(R), c put Kj = B ∩ {y1, . . . , yj} and Uj = B ∪ {y1, . . . , yj} . Then Kj is compact, Uj is open since it is cofinite, and µ[B] = sup{µ[Kj]: j ∈ N} = inf{µ[Uj]: j ∈ N}. Thus µ is inner regular and outer ∑ e ∞ e regular. If we modify µ and define µ = j=1 µyj , then µ is neither locally finite nor outer regular since µe[[0, 1]] = ∞ but µe[{0}] = 1. However, µe is inner regular: argue as in the case of µ. 62 T.K.SUBRAHMONIAN MOOTHATHU

(iv) Let X be a metric space and µ be the counting measure on (X, P(X)), i.e., µ[A] is the cardinality of A if A ⊂ X is a finite set, and µ[A] = ∞ otherwise. As an outer measure µ is Borel regular since any non-Borel subset of X must be infinite. As a measure µ is inner regular since any infinite subset of X contains finite sets of arbitrarily large cardinality. Except when X is a discrete metric space, µ is neither locally finite nor outer regular.

(v) Let µ be a Borel measure on (R, B(R)) defined by the condition that µ[B] = 0 or = ∞ according to whether B ∈ B(R) is of first category or second category in R. Then µ is neither locally finite nor outer regular since every nonempty open subset of R is of second category in R. Any compact subset of R contained in R \ Q is nowhere dense in R, and hence µ is not inner regular.

(vi) Fix a non-Borel set Y ⊂ R and deﬁne a measure µ on (R, P(R)) as µ[A] = ∞ if A ∩ Y ≠ ∅ and µ[A] = 0 if A ∩ Y = ∅ for A ⊂ R. It may be seen that µ, considered as an outer measure, is not Borel regular since for any Borel set B ⊂ R containing R \ Y we have µ[R \ Y ] = 0 ≠ ∞ = µ[B]. By considering singletons in Y , we see that µ is not locally ﬁnite. However, µ is inner regular. Now √ √ suppose Y = Z ∪ (Q + 2), where Z ⊂ R is a non-Borel subset disjoint with Q ∪ (Q + 2). Then Y is (non-Borel and) dense in R. Hence µ[U] = ∞ for any nonempty open set U ⊂ R containing Q. On the other hand, µ[Q] = 0, and thus µ is not outer regular.

Deﬁnition: Let X be a metric space. An outer measure µ∗ on X is called a metric outer measure if µ∗[A ∪ B] = µ∗[A] + µ∗[B] whenever A, B ⊂ X are subsets with dist(A, B) > 0. For example, the Lebesgue outer measure on Rd is a metric outer measure by [102](ix).

∗ Exercise-61: Let µ be a metric outer measure on a metric space X, and Aj ⊂ X for j ∈ N. ∪ ∑ ∗ ∞ ∞ ∗ (i) If dist(Ai,Aj) > 0 for i ≠ j, then µ [ Aj] = µ [Aj]. j=1 j=1 ∪ ∪ ∗ n ∗ ∞ (ii) If dist(Aj,Aj+2k) > 0 for every j, k ∈ N, then limn→∞ µ [ Aj] = µ [ Aj]. ∪ ∪ j=1 ∑ j=1 ∗ ∞ ∗ n n ∗ [Hint: Enough to show ‘≥’ in both. (i) µ [ Aj] ≥ µ [ Aj] = µ [Aj] for every n ∈ N. ∪ ∪ j=1 j=1 j=1 ∗ ∗ (ii) Let B1 = ∈ N− Aj and B2 = ∈ N Aj. We may assume µ [B1] < ∞ and µ [B2] < ∞, for j 2 1 j 2 ∑ ∗ ∗ otherwise the result can be deduced using part (i). Now ∈ N− µ [Aj] = µ [B1] < ∞ by (i), and ∑ j 2 1 ∑ ∗ ∗ ∗ similarly, ∈ N µ [Aj] = µ [B2] < ∞. Therefore, given ε > 0 we have µ [Aj] < ε for all j 2 ∪ ∪ ∪ ∪ j>n ∗ n ≥ ∗ n ∗ ≥ ∗ ∞ large n so that µ [ j=1 Aj] + ε µ [ j=1 Aj] + µ [ j>n Aj] µ [ j=1 Aj].]

[163] On a separable metric space X, any metric outer measure µ∗ is a Borel outer measure.

Proof. Observe that B(X) is generated by open sets, every open set is a countable union of open balls by separability, and every open ball is a countable intersection of closed balls. Therefore it is enough to show every closed ball A := {x ∈ X : d(a, x) ≤ r} is µ∗-measurable, where a ∈ X ∪ c ∞ and r > 0. For E ⊂ X, we have A ∩ E = Ej, where E1 = {x ∈ E : d(a, x) > r + 1} and j=1 ∪ { ∈ 1 ≤ 1 } ≥ n ∗ c ∩ Ej = x E : r + j < d(a, x) r + j−1 for j 2. Letting Yn = j=1 Ej, we see µ (A E) = MEASURE THEORY 63

∗ ∗ ∗ ∗ ∗ limn→∞ µ [Yn] by Exercise-61(ii). Moreover µ [E] ≥ µ [(A ∩ E) ∪ Yn] = µ [A ∩ E] + µ [Yn] since ∗ ∗ ∗ c dist(A ∩ E,Yn) > 0. Thus µ [E] ≥ µ [A ∩ E] + µ [A ∩ E]. And ‘≤’ is true always.

Exercise-62: Let µ∗ be a Borel outer measure on a metric space X, and let Y ⊂ X be an arbitrary ∗ ∗ ∗ ∩ subset. Then we may deﬁne another outer measure µY on X as µY [A] = µ [A Y ]. Then, ∗ (i) µY is a Borel outer measure (even if Y is not Borel). ∗ ∗ ⊂ (ii) If µ is Borel regular and Y is Borel, then µY is Borel regular. [Hint: (ii) Given A X, choose B ∈ B(X) with A ∩ Y ⊂ B and µ∗[A ∩ Y ] = µ∗[B]. Then A ∩ Y ⊂ B ∩ Y ⊂ B so that ∩ ∪ c ∗ ∗ ∩ ∗ ∩ ∗ B1 := [B Y ] Y is a Borel superset of A with µY [B1] = µ [B Y ] = µ [A Y ] = µY [A].]

[164] Let µ be a ﬁnite Borel measure on a metric space X. Then for any B ∈ B(X) and ε > 0, there exist a closed set F ⊂ X and an open set U ⊂ X such that F ⊂ B ⊂ U and µ[U \ F ] < ε (in particular, µ[U \ B] < ε and µ[B \ F ] < ε.)

Proof. Let B0 be the collection of all B ∈ B(X) such that for each ε > 0 there exist a closed set

F ⊂ X and an open set U ⊂ X with F ⊂ B ⊂ U and µ[U \ F ] < ε. To show B0 = B(X), it suﬃces to prove B0 is a σ-algebra containing all closed subsets of X. Clearly ϕ, X ∈ B0. If B ∈ B0, then c c c B ∈ B0 since F \ U = U \ F . Suppose Bj ∈ B0 for j ∈ N and ε > 0. Choose Fj ⊂ Bj ⊂ Uj with ∪ −j−1 ∞ Fj closed, Uj open, and µ[Uj \ Fj] < 2 ε. Let U = Uj, which is open. Choose k ∈ N large ∪ ∪ ∪ j=1 ∪ ∞ k k ∞ with µ[ Fj \ Fj] < ε/2 and put F = Fj, which is closed. Then F ⊂ Bj ⊂ U j=1 j=1∪ ∪ j=1 ∪ j=1 \ ≤ \ ∞ ∞ \ ∞ ∈ B ⊂ and µ[U F ] µ[U j=1 Fj] + µ[ j=1 Fj F ] < ε/2 + ε/2 = ε. Thus j=1 Bj 0. If B X is closed, then for Un := {x ∈ X : dist(x, B) < 1/n} we have µ[B] = limn→∞ µ[Un] since µ[X] < ∞.

Given ε > 0, taking F = B and U = Un for large n, we see B ∈ B0.

We have some corollaries:

[165] Let µ be a Borel measure on a metric space X. (i) If B ∈ B(X) is with µ[B] < ∞, then for any ε > 0 there is a closed set F ⊂ X such that F ⊂ B and µ[B \ F ] < ε; and consequently, there is an Fσ set E ⊂ X such that E ⊂ B and µ[B \ E] = 0. (ii) [Ulam’s theorem] Suppose X is a complete separable metric space and µ[X] < ∞. Then for every B ∈ B(X) and ε > 0, there is a compact set K ⊂ X such that K ⊂ B and µ[B \ K] < ε. Consequently, µ is inner regular and outer regular. (iii) Suppose X is locally compact and second countable and µ is locally ﬁnite. Then µ is inner regular and outer regular. Moreover, for every B ∈ B(X), there exist a σ-compact set E ⊂ X and a Gδ set G ⊂ X such that E ⊂ B ⊂ G and µ[E] = µ[B] = µ[G].

Proof. (i) Apply [164] to µB as deﬁned in Exercise-62. 64 T.K.SUBRAHMONIAN MOOTHATHU

(ii) Let ε > 0 be given. We claim there is compact K0 ⊂ X with µ[X \ K0] ≤ ε/2. Let {xj : j ∈ N} ⊂ X be dense. For each k ∈ N, choose nk ∈ N large enough so that the closed set ∪ ∩ nk k+1 ∞ Ak := B(xj, 1/k) satisﬁes µ[X \ Ak] < ε/2 . Then K0 := Ak is closed and totally j=1 ∑ k=1 \ ≤ ∞ k+1 bounded, and hence compact. Moreover, µ[X K0] k=1 ε/2 = ε/2, proving our claim. Now for B ∈ B(X), there is closed F ⊂ X with F ⊂ B and µ[B\F ] < ε/2 by (i). Then K := F ∩K0 ⊂ B is compact and µ[B \ K] < ε. Hence µ is inner regular, and also outer regular since µ[X] < ∞. ∪ ∞ ⊂ ∞ (iii) Write X = j=1 Kj where Kj’s are compact and Kj int[Kj+1]. Note that µ[Kj] < for every j ∈ N. Consider B ∈ B(X) and ε > 0. By [164], for each j there is a closed set Fj ⊂ X with ⊂ \ ∩ ⊂ Fj B and µKj [B Fj] < 1/j. Replacing Fj with Fj Kj, we may assume Fj Kj, and then Fj is also compact. If µ[B] = ∞, then µ[Fj] → ∞ = µ[B] as j → ∞. If µ[B] < ∞, then for a given \ \ ε > 0 choose j0 large enough with µ[B Kj0 ] < ε/2 and 1/j0 < ε/2. Then we see µ[B Fj0 ] < ε, and therefore, µ[B] = limj→∞ µ[Fj], thereby proving inner regularity. To prove outer regularity, it is enough to consider the case µ[B] < ∞. Let ε > 0 be given. Let Vj = int[Kj+1], and choose open −j supersets Uj of B with µV [Uj \B] < 2 ε. Replacing Uj with Uj ∩Vj, we may assume Uj ⊂ Vj, and ∪ j ∑ ∑ ∞ ⊂ \ ≤ ∞ \ ∩ ∞ −j put U = j=1 Uj. Then U is open, B U, and µ[U B] j=1 µ[Uj B Vj] < j=1 2 ε = ε. This establishes outer regularity. ∪ ∞ ∞ ⊂ Now we prove the last assertion. If µ[B] = , take G = X, and E = j=1 Fj, where Fj X is compact with Fj ⊂ B and µ[Fj] ≥ j. If µ[B] < ∞, let Fj ⊂ Bj ⊂ Uj be such that Fj is compact in ∪ ∩ \ ∞ ∞ X, Uj is open in X, and µ[Uj Fj] < 1/j. Put E = j=1 Fj and G = j=1 Uj.

Deﬁnition: Let X be a locally compact second countable metric space. Let C(X, R) = {f : X →

R : f is continuous} and Cc(X, R) = {f ∈ C(X, R): f has compact support}.

Exercise-63: Let X be locally compact second countable metric space and let µ, β be locally ﬁnite Borel measures on X. (i) If µ[K] = β[K] for every compact K ⊂ X, or if µ[U] = β[U] for every open U ⊂ X, then µ = β. ∫ ∫ ∈ R ≥ (ii) If X fdµ = X fdβ for every f Cc(X, ) with f 0, then µ = β.

[Hint: (i) is a corollary to [165](iii). To prove (ii), consider compact K ⊂ X, let Un = {x ∈ X : dist(x, K) < 1/n}, and fn : X → [0, 1] be continuous with fn ≡ 1 on K and fn ≡ 0 on X \ Un.

Then there is k ∈ N with Un is compact and fn ∈ Cc(X, R) for n ≥ k. Also, (fn) → 1K pointwise. Using 1 ∈ L1(µ) ∩ L1(β) as the dominating function, or replacing f with f ··· f , assuming Uk n 1 n 1 1 0 ≤ fn+1 ≤ fn and using f ∈ L (µ)∩L (β) as the dominating function, apply Lebesgue dominated k ∫ ∫ ∫ convergence theorem to (fn) ≥ . Then, µ[K] = 1 dµ = limn→∞ fndµ = limn→∞ fndβ = ∫ n k X K X X X 1K dβ = β[K]. Now use (i).] MEASURE THEORY 65

Exercise-64: Let X be a locally compact second countable metric space and µ be a locally finite p Borel measure on X. Then for 1 ≤ p < ∞, the complete normed space (L (µ), ∥ · ∥p) is separable ∫ R ∥ ∥ | |p 1/p and Cc(X, ) is dense in it, where f p := ( X f dµ) .[Hint: We argue for the case where p = 1. The collection of all µ-integrable simple Borel functions on X is dense in L1(µ) by the definition of integration. Now let U be a countable base of open balls with compact closures in X and let V be all ∑ U k finite unions of members of . Consider a µ-integrable simple Borel function g = i=1 ci 1Bi . Since

µ is regular by [165], we can approximate each Bi in measure by an open superset W , and then there is V ∈ V such that µ[W ∆V ] is suﬃciently small. Also each ci can be approximated by a rational. ∑ F { k ∈ N ∈ Q ∈ V} 1 Hence the countable collection = i=1 ci 1Vi : k , ci ,Vi is dense in L (µ). Now 1 to show Cc(X, R) is dense in L (µ), it is enough to show that 1V , V ⊂ X open with µ[V ] < ∞, can be approximated by members of Cc(X, R) since Cc(X, R) is a vector subspace and since F is ∪ 1 ⊂ ∞ ∈ R dense in L (µ). Let Kn’s be compact with Kn Kn+1 and V = n=1 Kn. Let fn Cc(X, ) be such that 0 ≤ fn ≤ 1, fn ≡ 1 on Kn, and fn ≡ 0 on X \ V . Then, ∥ 1V − fn∥1 ≤ µ[V \ Kn] → 0.]

17. Measure Theory and Functional Analysis

Deﬁnition: Let µ, β be measures on an abstract measurable space (X, A). We say β is absolutely continuous with respect to µ if µ[A] = 0 ⇒ β[A] = 0 for every A ∈ A (notation: β << µ); and we say µ, β are orthogonal (or singular) if there is A ∈ A with β[A] = 0 = µ[X \ A] (notation: µ ⊥ β).

Examples: (i) Let µ be the Lebesgue measure on [0, 1], β1[A] := β[A ∩ [0, 1/2]], and β2[A] :=

[A ∩ [1/2, 1]] for Lebesgue measurable A ⊂ [0, 1]. Then β1, β2 << µ, β1 ⊥ β2 and µ = β1 + β2.

(ii) Let µ be the Lebesgue measure on [0, 1], and let β1, β2 be Borel measures on [0, 1] deﬁned by: ∫ ∈ ∈ → R β1[A] = 1 if 0 A, β1[A] = 0 if 0 / A, and β2[A] = A fdµ where f : [0, 1] is µ-integrable and

≥ 0 (here, the countable additivity of β2 is checked using the Monotone Convergence Theorem).

Then β1 ⊥ µ and β2 << µ. (iii) Let f : R → R be a Lipschitz continuous homeomorphism, µ be the Lebesgue measure on R and deﬁne β[A] = µ[f(A)]. Then β is a Borel measure on R. Let λ > 1 be a Lipschitz constant for f. If B ⊂ R is Borel with µ[B] = 0 and ε > 0, then there are countably many intervals Jn such that ∪∞ ∑∞ ∑∞ B ⊂ Jn and µ[Jn] < ε/λ. If Kn = f(Jn), then µ[Kn] ≤ λµ[Jn] so that µ[Kn] < ε. n=1 ∪ n=1 n=1 ⊂ ∞ Also f(B) n=1 Kn. Hence β[B] = µ[f(B)] = 0. Thus β << µ.

Exercise-65: [Reason for the name ‘absolutely continuous’] Let µ, β be ﬁnite measures on (X, A). Then β << µ iﬀ for every ε > 0 there is δ > 0 such that β[A] < ε for every A ∈ A with µ[A] < δ.

[Hint: Let ε > 0. If there is no δ > 0 with the required property, then there are An ∈ A such that ∩ ∪ −2 ≥ ∞ ∞ ≥ µ[An] < n and β[An] ε. If B = k=1 n=k An, then µ[B] = 0 and β[B] ε > 0. To obtain ‘⇐’, note that if β[A] > ε > 0 and if δ > 0 is chosen for this ε, then µ[A] ≥ δ > 0.] 66 T.K.SUBRAHMONIAN MOOTHATHU

Exercise-66: Let µ, β be measures on (X, A) with β << µ and β ⊥ µ. Then β ≡ 0.

Recall the following elementary geometric fact: if u, v are two vectors in the plane, then v can be written uniquely as v = v1 + v2 so that the vector v1 is parallel to u and v2 ⊥ u. A result of the same spirit for measures is:

[166] [Lebesgue Decomposition Theorem] (true for σ- finite measures also) Let β, µ be two finite measures on a measurable space (X, A). Then there are unique measures βa, βs on (X, A) such that β = βa + βs, βa << µ and βs ⊥ µ. ∫ Proof. Note that β << β + µ. The idea is to obtain a suitable g such that β[A] = A gd(β + µ) for every A ∈ A. Then βa, βs can be defined with the help of g. We have (β + µ)[X] < ∞ and hence ∫ 2 ⊂ 1 2 → R L (β + µ) L (β + µ). Consider the linear functional ϕ : L (β + µ) given by ϕ(h) = X hdβ. By Cauchy-Schwarz,we have ∫ ∫ 1/2 |ϕ(h)| ≤ |h|dβ ≤ |h|d(β + µ) = ∥h∥1 = ∥h · 1∥1 ≤ ∥h∥2∥1∥2 = ∥h∥2[(β + µ)[X]] , X X and hence ϕ is bounded. By Riesz representation theorem, there is g ∈ L2(β + µ) such that ϕ(h) = ⟨h, g⟩. That is, ∫ ∫ (i) hdβ = hgd(β + µ) for every h ∈ L2(β + µ), and X ∫ X ∈ A (ii) β[A] = A gd(β + µ) for every A (taking h = 1A in (i)). We use (ii) for the rest of the proof; (i) will be used to prove the next theorem. First we claim −1 −1 that 0 ≤ g ≤ 1. Let An = {x ∈ X : g(x) ≤ −n } and Bn = {x ∈ X : g(x) ≥ 1 + n }. Using −1 A = An in (ii), β[An] ≤ −n (β +µ)[An] and hence (β +µ)[An] = 0. Similarly, A = Bn in (ii) gives −1 β[Bn] ≥ (1 + n )(β[Bn] + µ[Bn]) which implies β[Bn] = 0 and then µ[Bn] = 0. Thus 0 ≤ g ≤ 1 (β + µ)-almost everywhere. ∪ { ∈ } \ ∞ { ∈ ≤ − −1} Let S = x X : g(x) = 1 , T = X S. Then, T = n=1 Tn, Tn = x X : g(x) 1 n .

Taking A = S in (ii), we get µ[S] = 0. For Y ∈ A with µ[Y ] = 0, taking A = Y ∩ Tn in (ii), we ∑ ∩ ≤ − −1 ∩ ∩ ≤ ∞ ∩ have β[Y Tn] (1 n )β[Y Tn] implying β[Y T ] n=1 β[Y Tn] = 0. Thus if we deﬁne

βs[Y ] = β[Y ∩ S], βa[Y ] = β[Y ∩ T ], then βs ⊥ µ, βa << µ and β = βa + βs.

Uniqueness: Suppose β = αa + αs is another decomposition with respect to µ. Then αa[S] = 0 since µ[S] = 0. Hence βs[Y ] = β[Y ∩ S] = αa[Y ∩ S] + αs[Y ∩ S] = αs[Y ∩ S] ≤ αs[Y ]. Thus

βs ≤ αs and hence βa ≥ αa. Then the measure ν = αs − βs = βa − αa satisﬁes ν ⊥ µ and ν << µ. Therefore ν ≡ 0.

[167] [Radon-Nikodym Theorem for ﬁnite measures] Let β, µ be ﬁnite measures on a measurable space (X, A) and suppose that β << µ. Then there exists a non-negative µ-integrable function ∫ → R ∈ A f : X such that β[A] = A fdµ for every A . Moreover, f is unique µ-almost everywhere. MEASURE THEORY 67

(Thus β << µ ⇔ β is obtained by integrating against µ.)

Proof. Let ϕ, g, S, T be as in the proof of [166]. Let β = βa +βs be the Lebesgue decomposition of β with respect to µ. Since β << µ, we have βa = β and βs = 0 by the uniqueness of the decomposition. Now, let us make a guess about f. Taking h = f 1 in (i) of the proof of [166] we see that the ∫ ∫ A ∫ required f must satisfy fdβ = fgd(β + µ), or (adding fdµ to both sides and rearranging,) ∫ ∫ A A A ∫ ∫ − A f(1 g)d(β + µ) = A fdµ. On the other hand, we must have A fdµ = β[A] = A gd(β + µ) by (ii) from the proof of [166], so that we anticipate the identity f(1 − g) = g. Deﬁne f : X → R ∫ ∫ as f ≡ 0 on S and f = g/(1 − g) on T . Then f(1 − g)d(β + µ) = f(1 − g)d(β + µ) = ∫ A A∩T gd(β + µ) = (by (ii) from [166] )β[A ∩ T ] = βa[A] = β[A]. But we have already derived above A∩T ∫ ∫ ∫ − ∈ A that A f(1 g)d(β + µ) = A fdµ. Thus β[A] = A fdµ for every A .

Uniqueness µ-almost everywhere: If h is another such function and if An = {x ∈ X : f(x) − h(x) ≥ ∫ ∫ ∫ n−1}, then fdµ = β[A ] = hdµ and hence 0 = (f −h)dµ ≥ n−1µ[A ] implying µ[A ] = 0 An n An An n n and therefore µ[{x ∈ X : f(x) > h(x)}] = 0. Similarly µ[{x ∈ X : h(x) > f(x)}] = 0 also.

[167′] [Radon-Nikodym Theorem] Let β, µ be σ-ﬁnite measures on a measurable space (X, A) and let β << µ. Then there exists a non-negative measurable function f : X → R such that ∫ ∈ A β[A] = A fdµ for every A . Moreover, f is unique µ-almost everywhere. ∪ ∞ ∞ Proof. Write X = i=1 Xi, a disjoint union of measurable sets Xi with (β + µ)[Xi] < . Get → R → R | fi : Xi by [167] and let f : X be f Xi = fi.

We know that locally ﬁnite Borel measures on locally compact second countable metric spaces are σ-ﬁnite. Therefore, we have:

[167′′] [Radon-Nikodym Theorem for locally compact second countable metric spaces] Let β, µ be locally ﬁnite Borel measures on a locally compact second countable metric space X and let β << µ. ∫ → R Then there exists a non-negative Borel function f : X such that β[A] = A fdµ for every Borel A ⊂ X. Moreover, f is unique µ-almost everywhere.

Remark: The function f given by the Radon-Nikodym Theorem is called the Radon-Nikodym derivative of β with respect to µ and is denoted by dβ . This notation is justiﬁed by the following dµ ∫ ∫ observation. If g :(X, A) → R is a non-negative measurable function, then gdβ = g dβ dµ [∵ ∫ ∫ ∫ X X dµ dβ dβ if g = 1A, then X gdβ = β[A] = A dµ µ = X g dµ µ; extend this to nonnegative simple functions by linearity and then use approximation].

Exercise-67: Let µ be the Lebesgue measure on [0, 1], let f : [0, 1] → [0, 1] be a twice differentiable homeomorphism with f ′ > 0 on (0, 1) (example: f(x) = xn) and define β[A] = µ[f(A)] for Borel A ⊂ [0, 1]. We already know that β << µ (since f is Lipschitz by Mean value theorem). Show that 68 T.K.SUBRAHMONIAN MOOTHATHU dβ = f ′.[Hint: Since f ′ > 0, f is increasing on (0, 1) and hence on [0, 1]. If [a, b] ⊂ [0, 1], then ∫dµ ∫ ∫ f ′dµ = a f ′(t)dt = f(b) − f(a) = µ[f([a, b])] = β[[a, b]] = dβ dµ. Since closed intervals [a,b] b ∫ ∫ [a,b] dµ ′ dβ ⊂ generate the Borel σ-algebra on [0, 1], we deduce A f dµ = A dµ dµ for every Borel A [0, 1]. By dβ ′ uniqueness, dµ = f .] Exercise-68: Let Y,X be measurable spaces, g : Y → X be a measurable surjection and let β be a finite measure on Y . Define µ[A] = β[g−1(A)] for measurable A ⊂ X. Then µ is a finite ∫ ∫ measure on X and f ◦ gdβ = fdµ for integrable f : X → R.[Hint: If f = 1 , then ∫ ∫ Y X ∫ A ◦ −1 Y f gdβ = g−1(A) 1dβ = β[g (A)] = µ[A] = X fdµ.] Exercise-69: Let X be a compact metric space and consider C(X, R) with the supremum norm. If ϕ : C(X, R) → R is linear, then the following are equivalent: (i) ϕ is positive (i.e., ϕ(f) ≥ 0 for every f ∈ C(X, R with f ≥ 0). (ii) ϕ(f) ≤ ϕ(g) for f, g ∈ C(X, R) with f ≤ g. (iii) ϕ is bounded and ∥ϕ∥ = ϕ(1). [Hint:(i) ⇔ (ii) is easy. To see (ii) ⇒ (iii), consider f ∈ C(X, R) with |f| ≤ 1. Then f ≤ |f| ≤ 1 and −f ≤ |f| ≤ 1 so that ϕ(f) ≤ ϕ(1) and −ϕ(f) = ϕ(−f) ≤ ϕ(1). Hence |ϕ(f)| ≤ ϕ(1), and equality holds when f = 1. To prove (iii) ⇒ (i), it suffices to show ϕ(f) ≥ 0 for f ∈ C(X, R) with 0 ≤ f ≤ 1. Since 0 ≤ 1 − f ≤ 1, ϕ(1) − ϕ(f) = ϕ(1 − f) ≤ ∥ϕ∥ = ϕ(1) and so ϕ(f) ≥ 0.]

Remarks: (i) The above result is true for linear ϕ : C(X, C) → C, but the proof is more involved. (ii) If X is only a locally compact second countable metric space, a positive linear functional

ϕ :(Cc(X, R), sup norm) → R need not be bounded. For instance, consider ϕ : Cc(X, R) → R ∫ given by ϕ(f) = R f(x)dx.

Next we show how positive linear functionals on certain function spaces can be identified with locally finite Borel measures. If µ is a locally finite Borel measure on a locally compact second ∫ R countable metric space X and if we define ϕ : Cc(X, ) as ϕ(f) = X fdµ, then it is easy to see that ϕ is a positive linear functional. Our aim is to prove the converse that every positive linear functional on Cc(X, R) has this form. We need some preparation.

[168] If X is a compact metric space, then there is a continuous surjection f : {0, 1}N → X.

Proof. Let A(1),...,A(n1) ⊂ X be finitely many nonempty compact sets with diam[A(i)] < 1 and ∪ ∏ n1 ∈ k { } X = i=1 A(i). At the (k+1)th step, for each (i1, . . . , ik) j=1 1, 2, . . . , nj , choose finitely many nonempty compact sets A(i1, . . . , ik, 1),...,A(i1, . . . , ik, n(k+1)) having diameter < 1/(k + 1) such ∏ N ∞ n that their union equals A(i1, . . . , ik). Since {0, 1} = {0, 1} k , it is easy to find a continuous ∏ k=1 { }N → ∞ { } surjection g : 0, 1 k=1 1, 2, . . . , nk . Hence it suffices to find a continuous surjection h : MEASURE THEORY 69

∏∞ ∏∞ {1, 2, . . . , nk} → X (for then we may take f = h◦g). For y = (y1, y2,...) ∈ {1, 2, . . . , nk} k=1 ∩ k=1 ∞ deﬁne h(y) to be the unique point in k=1 A(y1, . . . , yk). Verify that this does the job.

We prove Riesz Representation Theorem for Measures (RRTM) in three steps2: ﬁrst for {0, 1}N, then for compact metric spaces, and then for locally compact second countable metric spaces.

[169] [RRTM for {0, 1}N] Let X = {0, 1}N and let ϕ : C(X, R) → R be a positive linear functional. ∫ ∈ R Then there is a unique ﬁnite Borel measure µ on X such that ϕ(f) = X fdµ for every f C(X, ).

Proof. We know from [132] that A := {∅} ∪ {all cylinder sets Cw} is a semi-algebra on X, where k ′ Cw = {(xn) ∈ X : x1 ··· xk = w} for w ∈ {0, 1} . By Exercise-31, the algebra A generated by A is equal to the collection of all finte unions of pairwise disjoint members of A. Also recall that every ′ member of A, and hence every member of A , is both open and compact in X. Hence, 1A ∈ C(X, R) ′ ′ for every A ∈ A . If µ : A → [0, ∞) is defined as µ[A] = ϕ( 1A), then µ[A ∪ B] = ϕ( 1A∪B) = ′ ϕ( 1A + 1B) = ϕ( 1A) + ϕ( 1B) = µ[A] + µ[B] for disjoint A, B ∈ A , and thus µ is finitely additive ∪ A′ A′ ∞ ∈ A′ ∈ A′ on . Since every member of is both compact and open, n=1 An / if An are pairwise disjoint nonempty sets. Thus trivially, µ is countably additive on A′. Also µ[X] = ϕ(1) ∈ [0, ∞). So µ is a finite pre-measure on A′. Since A′ is also a base for the topology of X, the algebra generated by A′ is the Borel σ-algebra. So µ extends to a finite Borel measure µ on X by [131]. ∑ G ∈ N Let be the collection of all functions of the form g = w∈{0,1}k aw1Cw , where k and aw ∈ R. If f ∈ C(X, R), then for every ε > 0, we may find k ∈ N by the uniform continuity of f such that |f((xn)) − f((yn))| < ε for every (xn), (yn) ∈ X with x1 ··· xk = y1 ··· yk. From this observation, it follows that F is dense in C(X, R) with respect to the supremum norm. Define ∫ R → R ∈ G ψ : C(X, ) as ψ(f) = X fdµ. Then ψ is linear and positive, and ψ(g) = ϕ(g) for every g by the definition of µ and linearity. As both ϕ and ψ are continuous by Exercise-69, and since G is ∫ R ∈ R dense in C(X, ), it follows that ϕ = ψ, which means ϕ(f) = X fdµ for every f C(X, ). ′ Uniqueness: If β is another such measure, then β[A] = ϕ( 1A) = µ[A] for every A ∈ A and then β[U] = µ[U] for every open U ⊂ X by approximation. Then β = µ by Exercise-63.

[169′] [RRTM for compact metric spaces] Let X be a compact metric space and let ϕ : C(X, R) → R be a positive linear functional. Then there is a unique ﬁnite Borel measure µ on X such that ∫ ∈ R ϕ(f) = X fdµ for every f C(X, ).

Proof. Let Y = {0, 1}N and let p : Y → X be a continuous surjection given by [168]. Deﬁne T : C(X, R) → C(Y, R) as T (f) = f ◦ p. Then T is a linear isometry and T (f) ≥ 0 whenever f ≥ 0. So C(X, R) may be identiﬁed with a vector subspace of C(Y, R) and ϕ may be considered as a positive linear functional on this subspace. By Exercise-69, ϕ is bounded and ∥ϕ∥ = ϕ(1). By

2We adapt the proof given by V.S. Sunder, who also mentions an earlier proof by V.S. Varadarajan. 70 T.K.SUBRAHMONIAN MOOTHATHU

Hahn-Banach Theorem, there is a bounded linear functional ϕe : C(Y, R) → R such that ϕe(T (f)) = e e e e ϕ(f) for every f ∈ C(X, R) and ∥ϕ∥ = ∥ϕ∥ = ϕ(1X ) = ϕ(T (1X )) = ϕ(1Y ). So ϕ is positive by Exercise-69. Therefore, by [169] there is a unique finite Borel measure β on Y such that ∫ ϕe(g) = gdβ for every g ∈ C(Y, R). Define µ[A] = β[p−1(A)] for Borel A ⊂ X. Then µ is a Y ∫ ∫ finite Borel measure on X and f ◦ pdβ = fdµ for every f ∈ C(X, R) by Exercise-68. Hence ∫ Y ∫ X e e ◦ ◦ ∈ R ϕ(f) = ϕ(T (f)) = ϕ(f p) = Y f pdβ = X fdµ for every f C(X, ). Uniqueness of µ follows from Exercise-63. ∫ ∥ ∥ Remark: It may be noted that ϕ = ϕ(1) = X 1dµ = µ[X], so that positive linear functionals of norm 1 on C(X, R) correspond to Borel probability measures on X, when X is a compact metric space.

Exercise-70: Let Y be a compact metric space and ψ : C+(Y, R) → R be such that ψ(f + h) =

ψ(f) + ψ(h) and ψ(αf) = αψ(f) for every f, h ∈ C+(Y, R) and α ≥ 0. Then ψ has a unique extension to a linear functional on C(Y, R). [Hint: Let f1, f2, h1, h2 ∈ C+(Y, R) and suppose f1 − f2 = h1 − h2. Then f1 + h2 = h1 + f2 so that by applying ψ and rearranging we obtain

ψ(f1) − ψ(f2) = ψ(h1) − ψ(h2). For f ∈ C(Y, R), deﬁne ψ(f) = ψ(h1) − ψ(h2), where h1, h2 ∈

C+(Y, R) are such that f = h1−h2 (for example, f = f+−f−). Then this extension ψ is well-deﬁned by the above argument. Check other properties.]

[169′′] [Riesz Representation Theorem for Measures for locally compact second countable metric spaces] Let X be a locally compact second countable metric space and let ϕ : Cc(X, R) → R be a positive linear functional. Then there is a unique locally ﬁnite Borel measure µ on X such that ∫ ∈ R ϕ(f) = X fdµ for every f Cc(X, ).

Proof. Step-1: Let K ⊂ X be compact and Un = N1/n(K). Note that Un+1 ⊂ Un for all n and Un is compact for all large n. Let gn ∈ C(X, [0, 1]) be such that gn ≡ 1 on Un+1 and gn ≡ 0 on X \ Un.

Then 0 ≤ gn+1 ≤ gn ≤ 1. e Let C+(K, R) = {f ∈ C(K, R): f ≥ 0}. If f ∈ C+(K, R), we see that there is f ∈ Cc(X, R) with e e e e f ≥ 0 and f|K = f (∵ choose open U with K ⊂ U, let f = f on K, f ≡ 0 on X \ U and apply e e e Tietze). Since 0 ≤ fgn+1 ≤ fgn,(ϕ(fgn)) is a decreasing sequence of nonnegative reals and hence e converges. Deﬁne ϕK : C+(K, R) → R as ϕK (f) = limn→∞ ϕ(fgn). e To verify that ϕK (f) is independent of the choice of the extension f, consider another extension ∗ ∗ f ∈ Cc(X, R) of f with f ≥ 0. Suppose that Un is compact for every n ≥ n0. Given ε > 0 e ∗ let An = {x ∈ Un : |f(x) − f (x)| ≥ ε} for n ≥ n0. Then An’s are compact, An+1 ⊂ An and ∩ ∩ ∩ A ⊂ U = K so that A = ∅ (∵ f = f = f ∗ on K). So there is n(ε) ≥ n n≥n0 n n≥n0 n n≥n0 n 0 e ∗ such that An(ε) = ∅. Then |f(x) − f (x)| < ε for every x ∈ Un(ε). Now gn ≡ 0 on X \ Un(ε) for MEASURE THEORY 71

≥ | e − ∗ | ≤ ≤ ≥ ≥ e − ∗ ≤ n n(ε). Hence fgn f gn εgn εgn0 for every n n(ε) n0. That is, fgn f gn εgn0 ∗ − e ≤ ≥ | e − ∗ | ≤ ≥ and f gn fgn εgn0 for n n(ε). Applying ϕ, ϕ(fgn) ϕ(f gn) εϕ(gn0 ) for every n n(ε) e ∗ which implies limn→∞ ϕ(fgn) = limn→∞ ϕ(f gn).

Next note that ϕK has the following properties: ϕK (f + h) = ϕK (f) + ϕK (h) and ϕK (αf) =

αϕK (f) for every f, h ∈ C+(K, R) and α ≥ 0. By Exercise-70, ϕK extends uniquely to a linear functional ϕK on C(K, R). The extended linear functional ϕK : C(K, R) → R is positive (∵ if e ′ f ≥ 0, then ϕ (f) = limn→∞ ϕ(fgn) ≥ 0 since ϕ is positive). So by [169 ], there is a unique K ∫ | finite Borel measure µK on K with ϕK (f) = K fdµK . Also observe that ϕK (f K ) = ϕ(f) for e nonnegative f ∈ Cc(X, R) with supp(f) ⊂ K (∵ taking f = f, ϕK (f|K ) = limn→∞ ϕ(fgn) = ϕ(f) since fgn = f). ∪ ∞ ⊂ Step-2: Write X = j=1 Kj, where Kj’s are compact and Kj int[Kj+1]. Step-1 gives a positive linear functional ϕj : C(Kj, R) → R defined using ϕ and let µj be the unique finite Borel measure | on Kj corresponding to ϕj. We claim that µj+1 Kj = µj. e Choose gn’s for Kj as in Step-1 and for f ∈ C+(Kj, R), let f ∈ Cc(X, R) be an extension with e e e e e f ≥ 0 and supp(f) ⊂ Kj+1. Then supp(fgn) ⊂ Kj+1 and hence ϕj+1(fgn| ) = ϕ(fgn) for all ∫ Kj+1 e e large n by the observation at the end of Step-1. Hence ϕ(fgn) = fgndµj+1 for all large n ∫ ∫ Kj+1 e e and therefore, fdµj = ϕj(f) = limn→∞ ϕ(fgn) = limn→∞ fgndµj+1, which is equal, by Kj ∫ ∫ Kj+1 Dominated convergence theorem, to f 1 dµj+1 = fd(µj+1| ). Hence the claim holds Kj+1 Kj Kj Kj by the uniqueness part in [169′].

For Borel A ⊂ X, deﬁne µ[A] = limj→∞ µj[A ∩ Kj] ∈ [0, ∞]. It may be checked that µ is a

Borel measure. For countable additivity, note the fact that if (xn,j) is a double sequence of reals increasing in each variable, then limn→∞ limj→∞ xn,j = limj→∞ limn→∞ xn,j, and use this with xn,j = µj[An ∩ Kj]. If K ⊂ X is compact, then K ⊂ Kj for some j and this implies µ[K] ≤

µj[K] < ∞ and so µ is locally ﬁnite. Also, if f ∈ Cc(X, R) is nonnegative, then supp(f) ⊂ Kj for ∫ ∫ ∫ ∫ some j so that ϕ(f) = ϕj(f| ) = fdµj = fdµ = fdµ, and therefore ϕ(f) = fdµ for Kj Kj Kj X K every f ∈ Cc(X, R) since f = f+ − f−. | Uniqueness: if β is another such measure and if βj = β Kj , then βj = µj by the uniqueness in [169′] and hence β = µ.

Remark: RRTM has other formulations when X is a compact metric space: (i) bounded linear functional on C(X, R) correspond to ﬁnite signed real-valued Borel measures, (ii) bounded linear functional on C(X, C) correspond to ‘ﬁnite’ complex-valued Borel measures, etc. 72 T.K.SUBRAHMONIAN MOOTHATHU

Remark: Let X be a compact metric space. Then the dual of the Banach space (C(X, C), sup norm) can be identified with M(X) := {all complex-valued ‘finite’ Borel measures on X} by Riesz Theo- rem. Now recall from Functional Analysis that if Y is a Banach space, then the weak* topology on the dual Y ∗ is the smallest topology such that the evaluation maps ϕ → ϕ(y) from Y ∗ to C are continuous for every y ∈ Y ; the closed unit ball of Y ∗ is weak* compact (Alaoglu’s Theorem) and it is weak* metrizable if Y is separable. Hence the closed unit ball of M(X) = C(X, C)∗ is weak* compact and weak* metrizable. For µ, µn ∈ M(X), (µn) → µ in the weak* topology iff ∫ ∫ ∫ → C ∈ C 7→ ( X fdµn) X fdµ in for every f C(X, ), since ϕ ϕ(f) = X fdµ is the evaluation ∗ map if ϕ ∈ C(X, C) corresponds to µ ∈ M(X). If (µn) is a bounded sequence in M(X), then it is contained in some closed ball (which is compact and metrizable in the weak* topology) and hence there exist µ ∈ M(X) and a subsequence (µn ) such that (µn ) weak* converges to µ, or ∫ ∫ k k → ∈ C equivalently ( X fdµnk ) X fdµ for every f C(X, ). Reading assignment: Theorem 6.19 of W. Rudin, Real and Complex Analysis, 3rd Edn.

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