FUNDAMENTALS of REAL ANALYSIS by Do˘Gan C¸Ömez II

FUNDAMENTALS OF REAL ANALYSIS by

Do˘ganC¸¨omez

II. MEASURES AND MEASURE SPACES

II.1. Prelude

Recall that the Riemann integral of a real-valued function f on an interval [a, b] is defined as n Z b X f(x)dx := lim f(xi)(xi − xi−1), (if the limit exists) kPk→0 a i=1 where P = {a = x1 < x2 < ··· < xn−1 < xn = b} is a partitioning of [a, b], kPk = maxi |xi − xi−1|, and xi ∈ [xi−1, xi] is arbitrary. Then, it follows that if f :[a, b] → R is a continuous function, then it is Riemann-integrable. Also, if f :[a, b] → R has finitely many bounded jump discontinuities, then it is Riemann-integrable. Indeed, there are functions with infinitely many bounded jumps which are still Riemann-integrable. However, there are “very simple” functions with countably many bounded jumps which are not Riemann-integrable. For example, let 1 if x ∈ [0, 1] ∩ Q f(x) = c 0 if x ∈ [0, 1] ∩ Q . Pn Pn Then, lim supkPk→0 i=1 f(xi)(xi − xi−1) = 1 6= 0 = lim infkPk→0 i=1 f(xi)(xi − xi−1)!

Riemann integral also suﬀers from a convergence problem: let {ri} be an enumeration (listing) of rational numbers in [0, 1], and deﬁne a sequence of functions {fn} on [0, 1] by

1 if x = r1, r2, r3, . . . , rn f (x) = n 0 otherwise, and f be the function in the example above. Then each fn is Riemann-integrable, fn → f pointwise on [0, 1], but f is not Riemann-integrable. Verdict: Riemann integral has several deﬁcits!

Next, observe that if f is a positive, continuous function on [a, b] or has finite jump discon- R b tinuities, then a fdx gives the area of the region R = {(x, y): a ≤ x ≤ b, 0 ≤ y ≤ f(x)}. However, if the discontinuities of f is more “complicated” or if f is continuous but has oscilla- tions on a “large” set, then f need not be Riemann-integrable; hence, for such regions “area” is not defined. Question. Can we define area of a region without involving Riemann-integral? If so, how? This was the question asked by mathematicians of late 19th century that led to the concept of “measure” that would measure the size (length, area, volume, . . . etc.) of a set (in Rn). For, one needs a working definition of measure.

Recall that the area of a region A ⊂ R2 bounded by two “nice” (continuous, piecewise linear) functions as A = {(x, y): a ≤ x ≤ b, f(x) ≤ y ≤ g(x)} is given by the Riemann- R b integral a |f(x) − g(x)|dx. This deﬁnition is still valid if f and g have ﬁnitely many bounded jump discontinuities. This is a good start; however, as seen in the example above, there are functions with countably many bounded jump discontinuities which are not Riemann-integrable. 1 2

Furthermore, there are continuous curves (so called “space-filling curves”) which do not fit to this scheme. Therefore, defining area via Riemann-integral has serious inadequacies. Here comes Peano. In order to address this deficiency, Peano proposed the following geometric approach (inspired from Archimedes’ idea on calculating areas enclosed by conic sections). For S ⊂ R2, define

ai(S) = inner area of S = supremum of areas of all polygons contained in S,

ao(S) = outer area of S = infimum of areas of all polygons containing S, and define A(S) = ai(S) if both the inner and outer area exist and the equality ao(S) = ai(S) holds. Although this is a significant improvement over the definition of area via Riemann integral, it still has some drawbacks. For instance, if S = {(x, y): x, y ∈ [0, 1] ∩ Q}, then ao(S) = 1 6= 0 = ai(S). Thus, some rather simple regions do not have area! Jordan’s try. Inspired by Peano, Jordan proposed using the concept of “content” to address the problem. For S ⊂ [a, b], let P = {(xi, xi+1)} be any finite partition of [a, b] by open intervals. Let X Ji(S) = { l(Ii): S ⊃ Ii ∈ P contains interior points of S}, X Jo(S) = { l(Ii): Ii ∈ P contains points of S ∪ ∂S}, and let ci(S) = supP Ji(S) and ci(S) = infP Jo(S). Then, if ci(S) and co(S) exist and are equal, define c(S) = content of S = ci(S) and call S Jordan-measurable. It turns out that form many sets in R their content is well defined (which includes regions defined by space filling curves). Jordan also defined integral of a real-valued function f on [a, b] as follows. Let Z b X f = sup inf f(x).c(Ei) a P Ei Z b X f = inf sup f(x).c(Ei), P a Ei R b R b where P = {Ei} is a partitioning of [a, b] into Jordan measurable sets. If a f = a f, then the common value is called the integral of f. Jordan integral improves Riemann integral; furthermore, the collection of Jordan-measurable sets is richer than those having area in the sense of Peano. On the other hand, the content c concept also has some problems; for example, ci([0, 1]∩Q) = 0 = ci([0, 1]∩Q ) and co([0, 1]∩Q) = c 1 = co([0, 1] ∩ Q ); thus, both of these sets are not Jordan-measurable. Furthermore, there are even some open (or closed) bounded sets which are not Jordan-measurable, which is a serious drawback. Despite it’s shortcomings, Jordan measure (content) and Jordan measurable sets have the following “nice” properties:

1) If A and B are Jordan measurable, then so are A ∩ B,A ∪ B,A \ B,B \ A (closure under set operations). 2) c(E) ≥ 0 (non-negativity). 3) If E and F are disjoint Jordan measurable sets, then c(E ∪ F ) = c(E) + c(F ) (ﬁnite additivity); if E and F are not necessarily disjoint, c(E∪F ) ≤ c(E)+c(F ) (subadditivity). 4) If A ⊂ B are Jordan measurable sets, then c(A) ≤ c(B) (monotonicity). 5) If A is Jordan measurable and α is a real number, then α + A is also Jordan measurable and c(A) = c(α + A) (translation invariance).

Exercise Prove the properties 2)-5) stated above. 3

Borel’s approach. The drawback of Jordan-measurability led Borel to deﬁne the properties the measure of sets should satisfy explicitly. He stated that:

1) measure of a set should be non-negative 2) measure of a (finite or infinite) union of disjoint sets should be the sum of measures of the individual sets 3) if A ⊂ B, then measure of B \ A should be measure of B minus measure of A 4) every set whose measure is non-zero should be uncountable, and 5) the location of the set should not affect the measure of the set (i.e. measure of A and measure of α + A should be the same).

Notice that, these are essentially the features of Jordan content. Having described properties that a measure must carry, Borel also restricted it to those sets which are constructible, where he deﬁned a set constructible if it is

a) a union (ﬁnite or countably inﬁnite) of disjoint intervals, b) the complement of a constructible set A wrt any other constructible set B covering A.

He prescribed that the measure of an interval would be its length. With these, he hoped that: (i) any subset of R would be constructible, and (ii) a measure satisfying the properties listed above could be constructed. Bad news! (Or, may be good!?) It turns out that it is impossible to construct a measure on subsets of R having all five properties above (see pp: 19-20 in the text). However, when n = 1, 2, it is possible to construct a set function on subsets of Rn if only finite additivity is required (Banach), and when n ≥ 3 even this is impossible (Hausdorff). On the other hand, constructing a set function on subsets of Rn satisfying first four properties is possible (though very delicate). The property (5) above is, for obvious reasons (it makes sense!), very essential; hence, having a set function satisfying only first four properties does not address the problem. Furthermore, it turns out that not every subset of R is constructible either (we’ll see an example of such a set later). Thus, since we would like to preserve as much of the properties (1)-(5) and since many sets constructed over intervals seem to satisfy them, we have one option to follow: weaken one of the properties/requirements of Borel. Weakening (1), (3) and (4) does not make sense. As seen above, although weakening (4) by requiring only finitely additivity does not solve the problem, one can weaken (2) in an appropriate fashion while keeping measure well-defined. These considerations provide compelling rationale to give up defining a measure on all of P(R), and, instead, let a measure be defined on certain (proper) subclasses of P(R). First of all, we must be careful in applying “additivity” of a measure m. If additivity also includes adding over not-necessarily countable index sets, then we would encounter paradoxical situations. For instance, (0, 1] = ∪x∈(0,1]{x}; hence, X X 1 = m((0, 1]) = m({x}) = 0 = 0! x∈(0,1] x∈(0,1] Therefore, it’d be safer to restrict it to “countable additivity”. This clears way to Program. Develop: (1) a good understanding of constructible sets, and (2) a good concept of measure on constructible sets. Let’s start with (1). Following Borel’s prescription of constructibility, we see that constructible sets must satisfy the following axioms: 4

(i) if A and B are constructible, then A ∪ B is also constructible 0 (i ) if {Ai} is a countable collection of constructible sets, then ∪iAi is also constructible (ii) if A is constructible, then so is Ac (iii) R and ∅ are constructible. Notice that, since intervals are a priori constructible, all open (closed) sets are constructible (why?), and so are arbitrary union (intersection) of closed (open) sets.

Deﬁnition. The smallest collection B of all subsets of R which contains all open sets and satisfying (i0), (ii) and (iii) is called Borel sets (or Borel σ-algebra). Remark. From the deﬁnition, the collection B includes:

(a) all closed sets (b) all arbitrary union of closed sets (Fσ sets) (c) all arbitrary intersection of open sets (Gδ sets) (d) all arbitrary intersection of sets in (b) (Fσδ sets) (e) all arbitrary union of sets in (c) (Gδσ sets) and so on. Notice that this process goes on indeﬁnitely; hence the deﬁnition of B is stated above in a “peculiar” manner.

Exercises. 1. Show that Z, Q, Qc, C are Borel sets, where C is the Cantor set. Which one of the classes Fσ, Gδ, Fσδ,... each of these sets belong?

2. Prove that translate of any Fσ-set ( Gδ-set) is a Fσ-set ( Gδ-set) Question. Does the collection B actually exist? The answer is a resounding Yes as the following statement proves.

Fact.1 Given any collection O of subsets of R, there exists a smallest collection A of subsets of R satisfying (i0), (ii), (iii) and containing O. Note. Such a family A is called a σ-algebra (of subsets of R). Proof. (of Fact.1) Let ∆ be the family of all σ-algebras of subsets of R that contain O. Then ∆ is nonempty (it contains P (R)). Let A = ∩{D : D ∈ ∆}. Then, O ⊂ A. Furthermore, (i) if A, B ∈ A, then A, B ∈ D for all D ∈ ∆. So, since such a D is a σ-algebra, A ∪ B ∈ D for all D ∈ ∆. Hence, A ∪ B ∈ A. (ii) if A ∈ A, then A ∈ D for all D ∈ ∆; and it follows that Ac ∈ D for all D ∈ ∆. Hence, Ac ∈ A.

(iii) By definition, ∅, R ∈ D for all D ∈ ∆; and hence, ∅, R ∈ A. Therefore, A is a σ-algebra. From the definition of A, if D is a σ-algebra containing O, then A ⊂ D; hence, A ⊂ D for all D ∈ ∆. Hence, A is the smallest σ-algebra containing O. Remark. |B| = c. (See Real and Abstract Analysis, by Hewitt & Stromberg, Theorem 10.23, pp:133-134.) It is well known that |P(R)| = 2c > c; hence, B is a proper subset of P(R)! Thus, it appears that the smallest collection of sets that are constructible over all intervals in R does not include all subsets of R. This remark suggests that in order to complete the program identified above, we might keep (1), (3)-(5) intact and restrict (2) to countable additivity and have measures defined on largest possible subcollection (σ-algebra) A smaller than P(R). 5

Now, let’s see how can we approach the problem of constructing a measure m on a σ-algebra. Following Borel, the starting point is defining measure on finite sets and intervals. Define measure of an interval as its length. So m([a, b]) = l([a, b]) = l((a, b)) = l((a, b]) = l([a, b)) = b − a = m((a, b)). If a = −∞ or b = ∞, then m([a, b]) = l([a, b] = ∞. Since [a, b] = [a, b) ∪ {b}, we must have m({b}) = 0 for any singleton {b} ⊂ R; consequently, measure of any finite set is 0. Also, m([a, b]) = l([a, b]) < l([a, 2b]) = m([a, 2b]); hence m(A) ≤ m(B) should hold if A ⊂ B. This discussion, with a little luck, tells us that if we are to construct a measure on subsets of R, on that collection we’d hope it to satisfy the following properties: (i) 0 ≤ m(A) ≤ ∞ (ii) m(A) ≤ m(B) if A ⊂ B (monotonicity) (iii) m(∅) = 0 = m({a}), ∀a ∈ R (iv) m(I) = l(I) for any interval I ⊂ R ∞ P∞ (v) m(∪k=1Ak) = k=1 m(Ak) for all disjoint collection of sets {Ak} (countable additivity) (vi) m(a + A) = m(A), ∀a ∈ R (translation invariance).

Exercises

1. Recall that for S ⊂ [a, b] and P = {(xi, xi+1)} any finite partition of [a, b] by open intervals, we define X Ji(S) = { l(Ii): Ii ⊂ S,Ii ∈ P contains interior points of S}, X Jo(S) = { l(Ii): Ii ∈ P contains points of S ∪ ∂S}, and let ci(S) = supP Ji(S) and co(S) = infP Jo(S). Then, if ci(S) and co(S) exist and are equal, we define c(S) = content of S = ci(S) and call S Jordan-measurable. Prove that (i) c(E) ≥ 0 (non-negativity). (ii) If A ⊂ B are Jordan-measurable sets, then c(A) ≤ c(B) (monotonicity). (iii) If E and F are disjoint Jordan-measurable sets, c(E∪F ) = c(E)+c(F ) (finite additivity); if E and F are not necessarily disjoint, c(E ∪ F ) ≤ c(E) + c(F ) (subadditivity). (iv) If A is Jordan measurable and α is a real number, then α + A is also Jordan measurable and c(A) = c(α + A) (translation invariance).

2. Provide simple justiﬁcations that any Fσ, Gδ, Fσδ and Gδσ-set is a Borel set.

3. (i) Show that Z, Q, Qc, C are Borel sets, where C is the Cantor set. Which one of the classes Fσ, Gδ, Fσδ,... each of these sets belong?

(ii) Prove that translate of any Fσ-set ( Gδ-set) is a Fσ-set ( Gδ-set).

II.2. Lebesgue Measure Having Jordan’s improvement on Peano’s original ideas and Borel’s speciﬁcations of a measure and constructible sets, Lebesgue developed a theory of measure that addressed the limitations of Jordan measure and applicable to a large subclass of P(R) that avoids pathological sets mentioned above. In this section, we will construct this measure known as Lebesgue measure. As expected, we’ll begin with measure of intervals. Clearly, we have the following for bounded intervals: 6

a) 0 ≤ l(I) < ∞ b) l(I) ≤ l(J) if I ⊂ J (monotonicity) c) l(I ∪ J) = l(I) + l(J) if I,J are disjoint (ﬁnite additivity) d) l(a + I) = l(I), ∀a ∈ R (translation invariance).

Exercise. Let I be a bounded open interval and I1,I2,...,In be a ﬁnite collection of bounded n Pn open intervals. If I ⊂ ∪i=1Ii, then `(I) ≤ i=1 `(Ii).

Fact.2 Let I be a bounded open interval and I1,I2,...,In,... be a countable collection of ∞ P∞ bounded open intervals such that I ⊂ ∪i=1Ii. Then `(I) ≤ i=1 `(Ii). Proof. Let I = (a, b) and let > 0 be arbitrary. Consider I0 = (a − , a + ) and I00 = (b − , b + ). 4 4 4 4 ∞ 0 00 Then the collection {Ii}i=1 ∪ {I ,I } is an open cover for [a, b]. Hence by Heine-Borel theorem, ∞ 0 00 there exists a ﬁnite sub-cover, say {Ii}i=1 ∪ {I ,I }. Now, apply the Exercise above.

Remarks. 1. If the collection I1,I2,...,In,... in Fact.2 above were a disjoint, countable P∞ collection of bounded open intervals covering I, then `(I) ≤ i=1 `(Ii) would follow (almost) trivially from (c) above. (Exercise)

2. If, furthermore, {Ii} is a disjoint, countable collection of bounded open intervals with ∞ P∞ I = ∪i=1Ii, then `(I) = i=1 `(Ii). (Exercise)

Now, we would like to extend these (nice) properties of the measure m(= `) on intervals to more general sets of real numbers.

∗ # Deﬁnition. The set function m : P(R) → R+ deﬁned by ( ∞ ∞ ) ∗ X [ m (A) = inf `(Ik): A ⊆ Ik, each Ik is a bounded open interval k=1 k=1 is called the (Lebesgue) outer measure.

Remarks. It is easy to show that:

(i) m∗(∅) = 0 and, m∗({a}) = 0 for all a ∈ R. (ii) m∗(A) ≤ m∗(B) if A ⊂ B.

Exercises. 1. Calculate m∗(A) if (i) A is a singleton or a ﬁnite set, (ii) A = Q ∩ [0, 1]. 2) Show that m∗(A) = m∗(α + A) for all α ∈ R.

Fact.3 For any interval I ⊂ R, m∗(I) = `(I). Proof. Assume I = (a, b), then , given any > 0, we have I ⊂ (a − 2 , b + 2 ). Then, by Remark (ii) above, m∗(I) ≤ `(I) < b − a + which implies that m∗(I) ≤ b − a = `(I). On the other hand, b − a is a lower bound for the set of numbers

( ∞ ∞ ) X [ U = `(Ii): I ⊂ Ii, {Ii} open . i=1 i=1 Since m∗(I) = inf(U), which must be greater than or equal to any other lower bound, we have m∗(I) ≥ b − a = `(I). Hence m∗(E) = `(I). 7

Next assume that I = [a, b]. Then, for all > 0 (a, b) ⊂ I ⊂ (a − , b + ). Thus, m∗(a, b) ≤ m∗(I) ≤ m∗(a − , b + ) ⇒ `(I) ≤ m∗(I) ≤ `(E) + 2ε ⇒ `(I) = m∗(I). The other two bounded cases are proved similarly. If I is an unbounded interval (i.e., if a = −∞ or b = ∞), then there exists an interval I0 ⊂ I with `(I0) = n. Then m∗(I) ≥ m∗(I0) = n for all n ≥ 1; hence, m∗(I) = ∞ = `(I).

∗ Fact.4 The outer measure m is countably sub-additive, namely, if A1,A2,... is a sequence of subsets of R, then ∞ ! ∞ ∗ [ X ∗ m Ai ≤ m (Ai). i=1 i=1

Proof. Fix > 0. Given any Ai, pick a cover of Ai consisting of open intervals {Ii,n} such that ∞ X ε m∗(A ) ≤ `(I ) < m∗(A ) + . i i,n i 2i n=1 ∞ S∞ Next consider the collection {Ii,n}i,n=1, which covers i=1 Ai. Hence ∞ ! ( ∞ ∞ ∞ ) ∗ [ X [ [ m Ai = inf `(Jk): Jk ⊃ Ai . i=1 k=1 k=1 i=1 Therefore, ∞ ! ∞ ∞ ∞ ∞ ∞ [ X X X X ε X m∗ A ≤ `(I ) = `(I ) < m∗(A ) + = m∗(A ) + ε. i i,n i,n i 2i i i=1 i,n=1 i=1 n=1 i=1 i=1

Remark. It was shown by Vitali (1905) that there are A, B ⊂ R disjoint such that m∗(A) > 0, m∗(B) > 0, and m∗(A ∪ B) 6= m∗(A) + m∗(B).

Fact.5 m∗ is ﬁnitely additive implies m∗ is countably additive.

∞ ∗ Proof. If {Ai}i=1 is a disjoint family of subsets of R and m is finitely additive, then by Fact.4 n n ! ∞ ! ∞ X ∗ ∗ [ ∗ [ X ∗ m (Ai) = m Ai ≤ m Ai ≤ m (Ai). i=1 i=1 −=1 i=1 ∗ S∞ P∞ ∗ Now by letting n → ∞ we have m ( i=1 Ai) = i=1 m (Ai). Consequently, in order to have m∗ be countably additive, we need to seek a smaller class of subsets of R (than P(R)) on which m∗ is finitely additive. A simplification: From Fact.4 and Fact.5, in order to show that m∗ is finitely additive on a class of sets, all we need to show is that m∗(A ∪ B) ≥ m∗(A) + m∗(B) for any pair A, B in that class.

Deﬁnition. A set E ⊂ R is called measurable if given any A ⊂ R, then m∗(A) = m∗(A ∩ E) + m∗(A ∩ Ec), i.e., E splits A m∗-additively.

Remarks. 1. ∅ and R are measurable. 2. E measurable implies that Ec is measurable. 8

Question. What is the structure and the size of the the class of measurable sets? In order to answer this question we need to make an observation. Fact.6 If m∗(E) = 0, then E is measurable.

Proof. Let A ∈ R. Then (A ∩ E) ⊂ E and (A ∩ Ec) ⊂ A. Hence by monotonicity of m∗ m∗(A ∩ E) + m∗(A ∩ Ec) ≤ m∗(E) + m∗(A) = m∗(A). Hence m∗(A ∩ E) + m∗(A ∩ Ec) ≤ m∗(A). Conversely we notice that m∗(A) ≥ m∗(A ∩ Ec) = 0 + m∗(A ∩ Ec) = m∗(A ∩ E) + m∗(A ∩ Ec). Therefore, E is measurable.

Deﬁnition. A nonempty collection G of subsets of R is called an algebra if

i) R ∈ G, n n ii) if {Ai}i=1 ⊂ G, then ∪i=1Ai ∈ G, and iii) A ∈ G implies Ac ∈ G.

Remark. It follows from (i) and (iii) that ∅ ∈ G. Also, it follows from (ii) and (iii) that if n n {Ai}i=1 ⊂ G, then ∩i=1Ai ∈ G.

Deﬁnition. A nonempty collection G of subsets of R is called a σ-algebra if it is an algebra 0 ∞ ∞ and (ii ) if {Ai}i=1 ⊂ G, then ∪i=1Ai ∈ G.

Theorem.1 F = {E ⊂ R : E is measurable} is a σ−algebra. Proof. We first note that ∅, R ∈ F. Hence (i) is satisfied. Also E ∈ F which implies that c ∞ S∞ E ∈ F. Thus (iii) is also satisfied. Therefore, we must show {Ei}i=1 ⊂ F ⇒ i=1 Ei ∈ F.

First we will consider the ﬁnite case. Let E1,E2 ∈ F. In order to prove that E1 ∪ E2 ∈ F we need to show for all A ⊂ R that ∗ ∗ ∗ c m (A) ≥ m (A ∩ (E1 ∪ E2)) + m (A ∩ (E1 ∪ E2) ). Observe that c A ∩ (E1 ∪ E2) = (A ∩ E1) ∪ (A ∩ E1 ∩ E2). Now ∗ ∗ c ∗ ∗ c c m (A ∩ (E1 ∪ E2)) + m (A ∩ (E1 ∪ E2) ) = m (A ∩ (E1 ∪ E2)) + m (A ∩ E1 ∩ E2)

∗ ∗ c ∗ c c ≤ m (A ∩ E1) + m (A ∩ E2 ∩ E1) + m (A ∩ E1 ∩ E2) ∗ ∗ c ∗ = m (A ∩ E1) + m (A ∩ E1) = m (A) since E2 ∈ F. Hence (E1 ∪ E2) ∈ F. Therefore F is an algebra.

Claim. If {E1,E2,...,En} is a disjoint family of sets in F, then for all A ⊂ R we have n !! n ∗ [ X ∗ m A ∩ Ei = m (A ∩ Ei). i=1 i=1 (Exercise: Prove this claim. Hint: Use induction.)

Next let {Ei} ⊂ F be an arbitrary (countable) family of sets. Without loss of generality, we S∞ can assume that it is a disjoint family. Let E = i=1 Ei. We now need to show for all A ⊂ R m∗(A) ≥ m∗(A ∩ E) + m∗(A ∩ Ec). 9

First observe that ∞ n n !c c \ c \ c [ E = Ei ⊂ Ei = Ei i=1 i=1 i=1 for any n. Then, n !! n !c! ∗ ∗ [ ∗ [ m (A) = m A ∩ Ei + m A ∩ Ei since F is an algebra i=1 i=1 n !! ∗ [ ∗ c ≥ m A ∩ Ei + m (A ∩ E ) by the observation above. i=1 By the Claim above, we have n !! n ∗ [ ∗ c X ∗ ∗ c m A ∩ Ei + m (A ∩ E ) = m (A ∩ Ei) + m (A ∩ E ). i=1 i=1

Now if we let n → ∞, we have ∞ ∗ X ∗ ∗ c m (A) ≥ m (A ∩ Ei) + m (A ∩ E ) i=1 ∞ !! ∗ [ ∗ c ≥ m A ∩ Ei + m (A ∩ E ) i=1 = m∗ (A ∩ E) + m∗(A ∩ Ec).

∗ ∗ ∗ c S∞ Hence m (A) ≥ m (A ∩ E) + m (A ∩ E ) and i=1 Ei ∈ F. This proves that F is a σ−algebra.

Proposition. All intervals are in F (Hence B ⊂ F).

Proof. We will show that (a, ∞) ∈ F for a ∈ R. Let A ⊂ R, which without loss of generality can be assumed with m∗(A) < ∞. Now, from the deﬁnition of m∗, we can ﬁnd a countable family of open intervals {Ek} covering A such that for ε > 0 ∞ ∗ X ∗ m (A) ≤ `(Ik) < m (A) + ε. k=1 0 00 0 00 For any k, let Ik = Ik ∪ Ik where Ik = Ik ∩ (a, ∞) and Ik = Ik ∩ (−∞, a]. Then ∗ 0 00 ∗ 0 ∗ 00 m (Ik) = `(Ik) = `(Ik) + `(Ik ) = m (Ik) + m (Ik ). Hence ∞ ! ! ∞ ! ∞ ∗ ∗ [ ∗ [ 0 X ∗ 0 m (A ∩ (a, ∞)) ≤ m Ik ∩ (a, ∞) = m Ik ≤ m (Ik). k=1 k=1 k=1 Similarly we have have ∞ ∗ X ∗ 00 m (A ∩ (−∞, a]) ≤ m (Ik ). k=1 Then ∞ ∞ ∗ ∗ X ∗ 0 ∗ 00 X ∗ ∗ m (A ∩ (a, ∞)) + m (A ∩ (−∞, a]) ≤ [m (Ik) + m (Ik )] = m (Ik) ≤ m (A) + ε. k=1 k=1 Hence (a, ∞) ∈ F. Proof for other types of intervals are similar and left as an exercise. 10

Remark. It follows from this proposition that the σ-algebra F is rather large (at least as large as B). We will see below that B ⊂ F ⊂ P(R), an each of these inclusions is proper.

Exercise. Let E ⊂ R with m∗(E) > 0. Show that there exists a bounded subset of E that also has positive outer measure.

∗ Deﬁnition. The set function m = m |F : F → [0, ∞] is called the Lebesgue measure and the triple (R, F, m) is sometimes called the Lebesgue measure space. Fact.7 The Lebesgue measure m inherits all of its properties from m∗, namely:

i) m(∅) = 0, m(R) = ∞ ii) If A ⊂ B, then m(A) ≤ m(B) for all A, B ∈ F iii) m(A ∪ B) = m(A) + m(B) for any A, B ∈ F with A ∩ B = ∅ (ﬁnite additivity) iv) m(A) = `(A) if A is an interval v) m is translation invariant

Furthermore, m is countably additive, namely,

∞ P∞ vi) m(∪i=1Ai) = i=1 m(Ai) for any countable disjoint collection {Ai} ⊂ F. Proof. All these properties, except (iii), (v) and (vi), easily follow from the properties of the Lebesgue outer measure. SinceA ∩ B = ∅, it follows that Ac ∩ B = B. Hence, since A is measurable, m∗(A ∪ B) = m∗(A ∩ (A ∪ B)) + m∗(Ac ∩ (A ∪ B)) = m∗(A) + m∗(B), which proves (iii). Proof of (v) is straightforward (exercise). For (vi), ﬁrst observe that from the claim in Theorem.1 it follows that m∗ is ﬁnitely additive on F, which together with Fact.5 implies (vi).

∞ Another proof goes as follows: We need to show that if {Ek}k=1 ⊂ F is a disjoint collection, then ∞ ! ∞ [ X m Ek = m(Ek). k=1 k=1 We already know (from ﬁnitely additivity) that, for any n ≥ 1 n ! n [ X m Ek = m(Ek). k=1 k=1 So by monotonicity ∞ ! n ! n [ [ X m Ek ≥ m Ek = m(Ek). k=1 k=1 k=1 Now if we let n → ∞ we get the desired result. Now, we will exhibit some important features of Lebesgue measurable sets and Lebesgue measure. Let’s begin with the approximation property of the Lebesgue measurable sets.

Theorem.2 Let E ⊂ R. The following are equivalent: (i) E ∈ F (ii) For all ε > 0, there exists O open such that E ⊂ O and m∗(O \ E) < ε. (iii) For all ε > 0, there exists C closed such that C ⊂ E and m∗(E \ C) < ε. 11

∗ (iv) There exists G ∈ Gδ with E ⊂ G such that m (G \ E) = 0. ∗ (v) There exists F ⊂ Fσ with F ⊂ E such that m (E \ F ) = 0. Furthermore, if E ∈ F with m(E) < ∞, then (ii)-(v) are equivalent to

∗ (vi) For all ε > 0, there exists a ﬁnite union U of open intervals s.t. m (U M E) < ε where A M B = (A \ B) ∪ (B \ A). Proof. First we make an important observation: if m∗(A) < ∞ is measurable and A ⊂ B, then m∗(B) = m∗(A ∪ (B \ A)) = m∗(A) + m∗(B \ A) ⇒ m∗(B \ A) = m∗(B) − m∗(A).

We will prove (i) ⇒ (ii) ⇒ (iv) ⇒ (i) and (ii) ⇒ (iv) ⇒ (i). The other implications will be left as an exercise. (i) ⇒ (ii): Assume first that m∗(E) < ∞. From definition, given ε > 0 there exists an open cover {Ik} of E such that, ∞ X ∗ l(Ik) < m (E) + ε. k=1 ∞ ∗ Let O = ∪k=1Ik, then, since m is an outer measure, ∞ ! ∗ ∗ [ ∗ m (O) = m Ik < m (E) + ε. k=1 Hence m∗(O) − m∗(E) < ε, and since E has finite outer measure, we have m∗(O \ E) = m∗(O) − m∗(E) < ε.

∗ ∗ Next let m (E) = ∞ and let Ek = E ∩ [−k, k] for any k ≥ 1. So m (Ek) < ∞. Hence by above, there exists Ok open containing Ek such that ε m∗(O \ E ) < . k k 2k S∞ S∞ Since E = k=1 Ek ⊂ k=1 Ok = O, where O is open. Then ∞ ! ∞ !! ∗ ∗ [ [ m (O \ E) == m Ok \ Ek . k=1 k=1

S∞ S∞ S∞ Claim. ( k=1 Ek) \ ( k=1 Ek) ⊂ k=1(Ok \ Ek). (Exercise: Prove the claim.) Hence ∞ ! ∞ ∞ [ X X ε m∗(O \ E) ≤ m∗ (O \ E ) ≤ m∗(O \ E ) < = ε. k k k k 2k k=1 k=1 k=1

∗ 1 (ii) ⇒ iv): Given E, by (ii), pick an open set Ok such that m (Ok \ E) = k . Now let T∞ G = k=1 Ok, then G ∈ Gδ. Since E ⊂ Ok for all k, E ⊂ G and we have ∞ ! ∞ \ \ G \ E = Ok \ E = (Ok \ E) ⊂ Ok \ E k=1 k=1 ∗ ∗ 1 ∗ for any k. Then m (G \ E) ≤ m (Ok \ E) < k for any k ≥ 1.. This implies that m (G \ E) = 0. (iv) ⇒ i): For any set E ⊂ G, we have G = E ∪ (G \ E) and E = G ∩ (G \ E)c. If G is a set satisfying (iv), then from the fact that Gδ ⊂ F we have G ∈ F. Also, m∗(G \ E) = 0 ⇒ G \ E ∈ F ⇒ (G \ E)c ∈ F. 12

Hence E = G ∩ (G \ E)c ∈ F. (ii) ⇒ (vi): Since E ∈ F, there exists G open with G ⊃ E such that ∗ S m (G \ E) < ε/2. Since G is open, G = (k disjoint) Ik. Then ! ! [ [ [ ε I = E ∪ I \ E ⇒ m∗ I < m∗(E) + . k k k 2 k disjoint k disjoint k disjoint Note that ! ∗ [ X ∗ X m Ik = m (Ik) = `(Ik). k disjoint k k P So k `(Ik) is convergent. Hence there exists N such that ∞ X ε `(I ) < . k 2 k=N+1

SN ∗ ε Now let U = k=1 Ik. Then U \ E ⊂ G \ E ⇒ m (U \ E) < 2 . Thus N !c N ! c [ \ c E \ U = E ∩ U = E ∩ Ik = E ∩ Ik . k=1 k=1 SN S∞ Hence E \ U = k=1(E \ Ik) ⊂ k=N+1 Ik and ∞ ! ∞ [ X ε m∗(E \ U) < m∗ I ≤ `(I ) < . k k 2 k=N+1 k=N+1 ∗ Therefore m (U M E) < . (vi) ⇒ (i): We begin by making two observations. First, for any sets U, V ⊂ R, we have c c c c c c ∗ U \ V = V \ U (and V \ U = U \ V ); hence, U M V = U M V . Consequently, m (U M V ) = ∗ c c m (U M V ). ∗ ∗ ∗ ∗ Secondly, if m (U) ≥ m (V ), then U ⊂ V ∪ (U M V ), which implies that m (U) ≤ m (V ) + ∗ ∗ ∗ ∗ m (U M V ). Similarly, if m (V ) ≥ m (U), then V ⊂ U ∪ (V M U), and therefore, m (V ) ≤ ∗ ∗ ∗ ∗ ∗ m (U) + m (U M V ). Thus, |m (U) − m (V )| ≤ m (U M V ). Now, assume (vi); so, given > 0, let B be an open set (ﬁnite union of open intervals) such ∗ ∗ ∗ ∗ c ∗ c that m (A M B) < 2 . Hence, we have |m (A) − m (B|) < 2 and |m (A ) − m (B )| < 2 . Since B is measurable, we have, for any X ⊂ R, m∗(X) = m∗(B ∩ X) + m∗(Bc ∩ X). Thus, |m∗(A ∩ X) + m∗(Ac ∩ X) − m∗(X)| = |m∗(A ∩ X) + m∗(Ac ∩ X) − m∗(B ∩ X) + m∗(Bc ∩ X)| ≤ |m∗(A ∩ X) − m∗(B ∩ X)| + |m∗(Ac ∩ X) − m∗(Bc ∩ X)| ∗ ∗ c c ∗ ∗ c c ≤ m ((A M B) ∩ X) + m ((A M B ) ∩ X) ≤ m (A M B) + m (A M B ) < . Therefore, m∗(A ∩ X) + m∗(Ac ∩ X) = m∗(X), implying that A ∈ F.

Corollary. For all E ∈ F, E = B ∪ N where B ∈ B and m∗(N) = 0.

∗ Proof. By the previous theorem (v), there exists F ∈ Fσ such that m (E \ F ) = 0 and F ⊂ E. Then E = F ∪ (E \ F ). Take B = F and N = E \ F . This corollary says that the ”completion” of B is F. 13

Remark. The analysis above also proves that on (all of ) P(R) Lebesgue outer measure cannot be additive (hence, cannot be countably additive); however, it’s additive on a (proper) subset of it, i.e., on F. Exercises. 1. Prove that A ∈ F if and only if for any > 0 there is a closed set C and an open set O with C ⊂ A ⊂ O such that m∗(O \ C) < .

2. Let E ⊂ R have ﬁnite outer measure. Prove that E ∈ F if and only if for each bounded interval (a, b), b − a = m∗((a, b) ∩ E) + m∗((a, b) \ E).

Theorem.3 (Continuity of m, or Monotone convergence theorem for Lebesgue measure) Let {Ek} ⊂ F. S∞ i) If Ek ⊂ Ek+1 for all k ≥ 1, then m( k=1 Ek) = limn→∞ m(En). T∞ ii) If Ek ⊃ Ek+1 for all k ≥ 1 and m(E1) < ∞, then m( k=1 Ek) = limn→∞ m(En).

Proof. (i) Assume m(Ek) < ∞ for all k ≥ 1. Let

A1 = E1,A2 = E2 \ E1,A3 = E3 \ E2,...,Ak = Ek \ Ek−1 for k ≥ 2. Then {Ak} is a disjoint family in F, and furthermore ∞ ∞ [ [ Ek = Ak. k=1 k=1 Hence ∞ ! ∞ ! ∞ n n [ [ X X X m Ek = m Ak = m(Ak) = lim m(Ak) = lim m(Ek \ Ek−1). n→∞ n→∞ k=1 k=1 k=1 k=1 k=1 We note that the last sum is telescoping. So, it follows that ∞ ! [ m Ek = lim m(En). n→∞ k=1

If there exists an n0 such that m(En0 ) = ∞, then ∞ ∞ ! [ [ En0 ⊂ En ⇒ m En ≥ m(En0 ) = ∞. n=1 n=1

When n ≥ n0, we have En0 ⊂ En, which implies that limn→∞ m(En) = ∞.

ii) Let Bk = E1 \Ek for k ≥ 1. Then, since {Ek} is decreasing, {Bk} is an increasing collection in F. Also ∞ ∞ ∞ [ [ \ Bk = (E1 \ Ek) = E1 \ Ek. k=1 k=1 k=1 Therefore, since m(E1) is ﬁnite, we now have ∞ ! ∞ ! ∞ ! [ \ \ m Bk = m E1 \ Ek = m(E1) − m Ek . k=1 k=1 k=1

By (i) and the fact that m(Ek) < ∞ for each k, we get ∞ ! [ m Bk = lim m(Bn) = lim [m(E1) − m(Ek)] = m(E1) − lim m(Ek). n→∞ n→∞ n→∞ k=1 T∞ Hence m( k=1 Ek) = limn→∞ m(En). 14

Remark. The condition m(E1) < ∞ in (ii) above cannot be removed. Exercise: Give an example that if this condition is removed than the assertion of (ii) is no longer valid.

At this point, we would like to define a set operation that will be studied below in a special ∞ setting. Given a collection of subsets {En}n of a set X, similarly to the way we defined the lim sup an and lim inf an of a sequence, we define ∞ ∞ lim sup An = ∩n=1 ∪k=n Ak, and n ∞ ∞ lim inf An = ∪n=1 ∩k=n Ak. n

Exercises. Let {An}n ⊂ F.

1. m(lim infn An) ≤ lim infn m(An). ∞ 2. If m(∪n=1An) < ∞, then m(lim supn m(An) ≤ m(lim supn An).

Deﬁnition. We say that a property P holds almost everywhere on a measurable set E if there exists E0 ⊂ E with m(E0) = 0 such that P holds for all x ∈ E \ E0.

∞ 0 Observation: Let {Ek}k=1 ⊂ F. A point x ∈ R belongs to inﬁnitely many Eks ⇔ for all S n ≥ 1 there exists k ≥ n such that x ∈ Ek ⇔ for all n ≥ 1, x ∈ k≥n Ek ⇔ ∞ ! \ [ x ∈ Ek (= lim sup En). n n=1 k≥n

∞ Theorem.4 (Borel-Cantelli Lemma) Let {Ek}k=1 ⊆ F be a countable collection such that ∞ X m(Ek) < ∞, k=1 0 then m(lim supn En) = 0; i.e., almost every x ∈ R belongs to at most ﬁnitely many Eks. Proof. By the above observation, we need to only show that the set of x ∈ R which belong to 0 inﬁnitely many Eks has measure 0. Equivalently, ∞ !! \ [ m Ek = 0. k=1 k≥n First, notice that ∞ ! ∞ [ X m Ek ≤ m(Ek) < ∞. k=1 k=1 Lets “decreasify” by letting ∞ ∞ ∞ [ [ [ A1 = Ek,A2 = Ek,...,An = Ek,... k=1 k=2 k=n

Then Ak ↓, and m(A1) < ∞. Therefore, by continuity of m, ∞ !! ∞ ! \ [ \ m Ek = m An = lim m(An) n→∞ n=1 k≥n n=1 ∞ ! " ∞ # [ X = lim m Ek ≤ lim m(Ek) = 0. n→∞ n→∞ k=n k=1 15

Exercises

1. Show that m∗(A) = m∗(α + A) for all A ⊂ R, α ∈ R. 2. Calculate m∗(A) if

(i) A is a singleton or a ﬁnite set, (ii) A = Q ∩ [0, 1]. 3. Show that

(i) For any a, b ∈ R, a < b, the sets (a, b), [a, b), (a, b], and [a, b] are measurable. (ii) Q and Qc are measurable, and calculate m(Qc ∩ [0, 1]).

4.Let E ⊂ R with m∗(E) > 0. Show that there exists a bounded subset of E that also has positive outer measure. 5. Show that if A, B ∈ F, then m(A ∪ B) + m(A ∩ B) = m(A) + m(B). 6. Prove (i) ⇔ (iii) ⇔ (v) in Theorem.2; that is, TFAE:

(i) E ∈ F (iii) For all ε > 0, there exists C closed such that C ⊂ E and m∗(E \ C) < ε. ∗ (v) There exists F ⊂ Fσ with F ⊂ E such that m (E \ F ) = 0. [Hint: Use the fact that (i) ⇔ (ii).] 7. Prove that A ∈ F if and only if for any > 0 there is a closed set C and an open set O with C ⊂ A ⊂ O such that m∗(O \ C) < .

8. Let E ⊂ R have ﬁnite outer measure. Prove that E ∈ F if and only if for each bounded interval (a, b), b − a = m∗((a, b) ∩ E) + m∗((a, b) \ E).

∞ 9. Given a collection of subsets {En}n of a set X, similarly to the way we deﬁned the lim sup an and lim inf an of a sequence, we deﬁne ∞ ∞ lim sup An = ∩n=1 ∪k=n Ak, and n ∞ ∞ lim inf An = ∪n=1 ∩k=n Ak. n

10. Show that for {An}n ⊂ F.

(i) m(lim infn An) ≤ lim infn m(An). ∞ (ii) If m(∪n=1An) < ∞, then m(lim supn m(An) ≤ m(lim supn An).

11. Let A ∈ F and r ∈ R. Show that the sets r + A and rA are also measurable and m(r + A) = m(A), and m(rA) = |r|m(A).

12. Show that if A, B ∈ F, then m(A ∪ B) + m(A ∩ B) = m(A) + m(B).

II.3. Non Measurable Sets As we promised in section I.1 above, we will now “construct ”a non-measurable set; hence, proving that not all subsets of R belong to F (equivalently, not every subset of R is constructible). 16

Lemma. Let E ∈ F be a bounded set with m(E) ≥ 0. If Λ is a countably inﬁnite bounded set of real numbers such that {λ + E}λ∈Λ is disjoint, then m(E) = 0. Proof. Exercise. For a set E ∈ F let a relation ∼ on E be deﬁned by

x ∼ y if and only if x − y ∈ Q. It is easy to show that ∼ is an equivalence relation on E (Exercise). Recall that an equivalence relation partitions a set into equivalence classes, the equivalence classes of E with respect to ∼ are either disjoint or the same. Namely, E = {[x]: [x] is the equivalence class of x with respect to ∼}; and, for all x, y ∈ E, either [x] ∩ [y] = ∅ or [x] = [y]. Now, let

CE = {set of single element representatives from each E/ ∼}. Then we observe that:

(i) For all x ∈ E, there exists c ∈ CE such that x = c + q for some q ∈ Q. c (ii) For all x, y ∈ CE, x 6= y, x − y ∈ Q .

Hence, for any pair of distinct rationals, say p, q, we have (p+CE)∩(q +CE) = ∅, for otherwise, it follows that p + x = q + y for some x, y ∈ CE, which would imply x − y = p − q ∈ Q, contradicting (ii) above.

Theorem.5 (Vitali) If E ⊂ R with m∗(E) > 0, then there exists A ⊂ E such that A 6∈ F. ∗ Proof. Since m (E) > 0, we can assume that E is bounded. So consider CE for this set E.

Claim. CE 6∈ F This claim proves the assertion by letting A = CE. Hence, all we need is to prove the Claim.

For, assume for a contradiction that CE ∈ F. Since E is bounded, E ⊂ [−m, m] for some m ∈ Z.

Pick Λ0 = [−2m, 2m] ∩ Q. Then Λ0 is countably inﬁnite, bounded, and dense in [−2m, 2m]. Then, for any x ∈ E, there exists c ∈ CE such that x = c + q for some q ∈ Q. Since x, c ∈ [−m, m], q ∈ [−2m, 2m], [ (λ + CE)

λ∈Λ0 is bounded. By previous Lemma, m(CE) = 0. Hence ! ∗ ∗ [ X ∗ X ∗ 0 < m (E) ≤ m (λ + CE) ≤ m (λ + CE) = m (CE) = 0.

λ∈Λ0 λ∈Λ0 λ∈Λ0 This is a contradiction! Hence the claim must hold.

Corollary. There exists A, B ⊂ R, A ∩ B = ∅, such that m∗(A ∪ B) < m∗(A) + m∗(B).

Proof. For otherwise Theorem.5 would be false.

Remark. It follows from the Corollary that F ( P(R). 17

∗ Exercise. Let E ⊂ R be a non-measurable set with m (E) < ∞. Show that there is a Gδ-set G with E ⊂ G such that m∗(E) = m∗(G) and m∗(G \ E) > 0.

II.4 The Miraculous Cantor Set First, let’s recall the construction of the Cantor set:

C0 = [0, 1] 1 2 C = 0, ∪ , 1 1 3 3 1 2 1 2 7 8 C = 0, ∪ , ∪ , ∪ , 1 2 9 9 3 3 9 9 . .

T∞ Let C = n=1 Cn. Then we have Cn ⊃ Cn+1, for all n ≥ 1. Thus by the Nested Set Theorem, C 6= ∅. The set C is called the Cantor Set. Fact.8 (i) C is closed. (ii) m(C) = 0. (iii) C is uncountable.

2 Proof. By construction C is closed. Next, C ∈ F (why?). Also, since m(C1) = 3 , m(C2) = 4 8 2 n 9 , m(C3) = 27 ,..., we have m(Cn) = ( 3 ) for n ≥ 1. Hence by the continuity of m, ∞ ! \ m(C) = m Cn = lim m(Cn) = 0, n→∞ n=1 proving (i). Now, observe that if x ∈ [0, 1] can be expressed in ternary form as ∞ X ak x = , where a ∈ {0, 2}, 3k k k=1 then x ∈ C. The converse is true as well. Recall that this (ternary) expansion is unique with the convention that, if ∞ X ak x = , where a = 1 and a = 0 for k ≥ m + 1, 3k m k k=1 then instead take its equivalent m−1 ∞ X ak X 2 x = + . 3k 3k k=1 k=m Therefore, it follows that C ∼ {0, 2}N; and consequently |C| = |{0, 2}N| = c.

Exercise. Prove that C ⊕ C = [0, 2] and C C = [−1, 1]. Now we will construct a function that will be instrumental in many examples/constructions; namely, the Cantor-Lebesgue Function. 18

Let ϕ1 ∈ (C[0, 1], || · ||∞) be deﬁned as   0 if x = 0  1 if x = 1 ϕ (x) = 1 1 if x ∈ 1 , 2  2 3 3  interpolate linearly on the rest.

So ϕ1 is continuous, non-decreasing, and ϕ1([0, 1]) = [0, 1]. Next deﬁne   0 if x = 0   1 if x = 1  1 if x ∈ 1 , 2 ϕ (x) = 2 3 3 2 1 if x ∈ 1 , 2  4 9 9  3 if x ∈ 7 , 8  4 9 9  interpolate linearly on the rest.

Again, ϕ2 is continuous, non-decreasing, and ϕ2([0, 1]) = [0, 1]. Continue constructing ϕn, n ≥ 3, in this manner. So {ϕn} ⊂ C[0, 1] for each n ≥ 1, deﬁned as   0 if x = 0  1 if x = 1 ϕ (x) = n 2k−1 if x ∈ In, 1 ≤ k ≤ 2n−1  2n k  interpolate linearly on the rest, n where {Ik }1≤k≤2n−1 , are the open intervals removed from the set Cn−1 to obtain Cn in the construction of Cantor set. Now, for m < n are integers, we have 1 |ϕ (x) − ϕ (x)| < n m 2m 1 for all x ∈ [0, 1] (notice that |ϕn+1 − ϕn| ≤ 2 |ϕn − ϕn−1| for n ≥ 2). Hence 1 ||ϕn − ϕm||∞ = max |ϕn(x) − ϕm(x)| < ; x∈[0,1] 2m and hence, {ϕm} is Cauchy in C[0, 1]. Since C[0, 1] is a complete metric space, {ϕn} converges uniformly to a function ϕ, that is continuous and non-decreasing with ϕ([0, 1]) = [0, 1]. Thus ϕ : [0, 1] → [0, 1] is an onto function such that

• ϕ is non-decreasing, k n−1 k k th • ϕ(x) = 2n , n ≥ 1. 1 ≤ k ≤ 2 when x ∈ In, where In’s are intervals removed at the n stage of the Cantor set construction.

Let O = [0, 1] \C, which is measurable. Then [0, 1] = O ∪ C and [0, 1] = ϕ([0, 1]) = ϕ(C) ∪ ϕ(O). Since ϕ(O) is a countable set, m∗(ϕ(O)) = 0. Hence, ϕ(O) ∈ F. Consequently, ϕ(C) is also measurable with measure m(ϕ(C)) = 1. Thus by Vitali’s Theorem, there exists P ⊂ ϕ(C) such that P 6∈ F. On the other hand ϕ−1(P ) ⊂ C. Since m(ϕ−1(P )) = 0, it follows that ϕ−1(P ) ∈ F. Thus, we proved that Fact.9 The function ϕ maps a measurable set (a subset of C) onto a non-measurable set.

Next let f : [0, 1] → [0, 2] by f(x) = x + ϕ(x). Then it follows from the properties of ϕ and the function x that

• f is continuous and onto [0, 2]. • f is strictly increasing. 19

Since f is strictly increasing, it maps Borel sets onto Borel sets. (Why?) Note that if A ∈ F such that ϕ(A) 6∈ F, then f(A) 6∈ F either. Then A 6∈ B holds, for otherwise if A ∈ B, so would f(A). This implies that f(A) ∈ F which is a contradiction. Hence: Corollary. There exists a nonempty set A ∈ F \ B.

Exercise. Show that there exists a continuous, strictly increasing function f : [0, 1] → R that maps a set of positive measure onto a set of measure zero.

II.5. General Measure Spaces (A Brief Survey)

In this section, following similar steps as in Section II.2, we will construct measures on sets not necessarily equal to R, generalizing the Lebesgue measure. Deﬁnition. A nonempty collection G of subsets of a set X is called an algebra if

i) X ∈ G, n n ii) if {Ai}i=1 ⊂ G, then ∪i=1Ai ∈ G, and iii) A ∈ G implies X \ A ∈ G.

n n Remark. It follows that ∅ ∈ G and that if {Ai}i=1 ⊂ G, then ∩i=1Ai ∈ G. Deﬁnition. A nonempty collection G of subsets of X is called a σ-algebra if it is an algebra 0 ∞ ∞ and (ii ) if {Ai}i=1 ⊂ G, then ∪i=1Ai ∈ G. Remark. The collection G = {∅,X} and P(X) are σ-algebras for any X, where {∅,X} is the smallest and P(X) is the largest σ-algebra.

It is easy to see that the intersection of a family {Aα} of σ−algebras of a set X is another σ−algebra (Exercise). As before, we actually have a more general statement: Fact.10 Given a collection E of subsets of X, then there exists a smallest σ−algebra A of subsets of X such that E ⊂ A. Proof. Exercise. [Hint: Mimic the proof of Fact.1 in Section II.1]

Notation: A := σ(E) is called as the σ−algebra generated by E.

Exercise. If G is a σ−algebra and E ∈ G, then the collection A = {A ∩ E : A ∈ G} is also a σ−algebra.

Deﬁnition. Let X be a (non-empty) set and A be a σ−algebra of subsets of X, then the pair (X, A) is called a measurable space. Examples. 1. Let X be any uncountable set and let c A = {A ⊂ X : |A| ≤ ℵ0 or |A | ≤ ℵ0}. Then A is a σ-algebra and (X, A) is a measurable space.

2. Let X = R and A = {ﬁnite, disjoint union of sets [a, b), where − ∞ ≤ a < ∞, −∞ < b ≤ ∞}. Then (X, A) is not a measurable space. Note that A is an algebra, but not a σ−algebra. On the other hand, σ(A) = B, which is a σ-algebra. 20

Deﬁnition. Let (X, A) be a measurable space. A set function µ : A → [0, ∞] is called a measure (on A) if:

i) µ(∅) = 0. S∞ P∞ ii) µ( k=1 Ek) = k=1 µ(Ek) for any disjoint collection {Ek} ⊂ A. The triple (X, A, µ) is called a measure space. If µ(X) < ∞, then (X, A, µ) is called a ﬁnite measure space; if, in particular, µ(X) = 1, then (X, A, µ) is called a probability space. If S∞ X = n=1 Xn such that µ(Xn) < ∞ for each n, then (X, A, µ) is called a σ−ﬁnite measure space.

Deﬁnition. A set function µ : A → [0, ∞] satisfying

i) µ(∅) = 0. Sn Pn ii) µ( i=1 Ei) = i=1 µ(Ei)

n for any disjoint collection {Ei}i=1 ⊂ A, is called a ﬁnitely additive measure. Examples. 1. (R, F, m) and (R, B, m) are both measure spaces.

2. For any a, b ∈ R, a < b, the triple ([a, b], F|[a,b], m), where F|[a,b] = {A ∩ [a, b]: A ∈ F}, is a finite measure space; in particular, ([0, 1], F|[0,1], m) is a probability space. 3. Let X be any infinite set and let A = P(X). Define µ : A → [0, ∞] by 0 if |A| < ∞ µ(A) = ∞ if A is infinite. Then µ is not a measure, but is a finitely additive measure.

4. Let X = Z, A = P(Z), and define c : A → [0, ∞] by c(A) = |A|. Then c is a measure on Z (called the counting measure). 5. Let X be any uncountable set and let A = {A ⊂ X : A or Ac is countable}. If a set function µ : A → [0, ∞] is defined by 0 if A is countable µ(A) = , 1 if Ac is countable then µ is a measure. Remark. Sets of measure zero and almost all properties are defined in any measure space exactly the same as in the case of the Lebesgue measure. Theorem.6 Let (X, A, µ) be a measure space. Then

i) For all A, B ∈ A such that A ⊂ B, µ(A) ≤ µ(B). ∞ ii) If {Ai}i=1 ⊂ A, then ∞ ! ∞ [ X µ Ai ≤ µ(Ai) (Subadditivity). i=1 i=1

iii) If {Ai} ⊂ A, and Ai ⊂ Ai+1 for all i ≥ 1, then ∞ ! [ µ Ai = lim µ(An) (Continuity from below). n→∞ i=1 21

iv) If {Ei} ⊂ A, and Ei ⊃ Ei+1, for all i ≥ 1 and µ(E1) < ∞, then ∞ ! \ µ Ei = lim µ(En) (Continuity from above). n→∞ i=1 v) Borel-Cantelli Lemma holds.

Proof. Similar to the proofs of the analogous properties in the case of m. Discussion: (R, B, m) vs. (R, F, m). It is known that C ∈ B, but C contains an E ⊂ C such that E is not a Borel set. Thus, in (R, B, m), C has a subset, say E, such that E 6∈ B. However, E ∈ F. Notice m(C) = 0, hence there exists a measure space with the property that not every subset of measure zero is measurable. Such measure spaces are called incomplete. Definition. A measure space (X, A, µ) is called complete if E ∈ A where µ(E) = 0 and F ⊂ E, then F ∈ A. Theorem.7 Let (X, A, µ) be a measure space and N = {N ∈ A : µ(N) = 0}. Let A = {E ∪ F : E ∈ A,F ⊂ N for some N ∈ N }, and define µ : A → [0, ∞] by µ(E ∪ F ) = µ(E). Then A is a σ−algebra and µ is a measure on A such that (X, A, µ) is a complete measure space. Proof. First we will show that A is a σ−algebra. Let E = A ∪ F ∈ A. Without loss of generality we can assume that A ∩ F = ∅ (otherwise replace F by F \ A and N by N \ A). Observe that A ∪ F = (A ∪ N) ∩ (N c ∪ F ). Then Ec = (A ∪ N)c ∪ (N c ∪ F )c = (A ∪ N)c ∪ (N \ F ). | {z } | {z } ∈A ∈N c Thus E ∈ A. If {Ai ∪ Fi} ⊂ A, then ∞ ∞ ∞ [ [ [ (Ai ∪ Fi) = ( Ai) ∪ ( Fi). i=1 i=1 i=1 | {z } | {z } ∈A ∈N Thus A is a σ−algebra. It is left as an exercise to show that µ satisfies all the properties of being a measure.

Now, all we need to show is that µ is well deﬁned. Let E ⊂ A, and A1 ∪ F1 = E = A2 ∪ F2. Then A1 ⊂ A2 ∪ F2 ⊂ (A2 ∪ N2). Hence

µ(A1) ≤ µ(A2 ∪ N2) = µ(A2).

Similarly we get µ(A1) ≥ µ(A2). Thus µ(A1) = µ(A2), so µ is well deﬁned.

Question: How can we construct measures? We will answer this question, as in the case of Lebesgue measure, via “outer measures.”

Deﬁnition. Let X 6= ∅ be a set. A set function µ∗ : P(X) → [0, ∞] is called an outer measure (on X) if :

i) µ∗(∅) = 0. ii) µ∗(A) ≤ µ∗(B) when A ⊂ B. ∗ S∞ P∞ ∗ iii) µ ( k=1 Ak) ≤ k=1 µ (Ak). 22

Exercise. If an outer measure is is ﬁnitely additive, then it is a measure.

Given an outer measure on X, then we deﬁne µ∗−measurable sets as: Deﬁnition. A set E ⊂ X is called a µ∗−measurable set if for all A ⊂ X, µ∗(A) = µ∗(A ∩ E) + µ∗(A ∩ Ec).

Examples. 1. Let X be an infinite set and define µ∗(E) = |E| if E is finite and µ∗(E) = ∞ if E is infinite. Then µ∗ is an outer measure (Show). Exercise: Determine the µ∗-measurable sets. 2. Let X be any set and define µ∗(∅) = 0 and µ∗(E) = 1 if E 6= ∅. Then µ∗ is an outer measure (Show). Exercise: Determine the µ∗-measurable sets. 3. 1. Let X be an uncountable set and define µ∗(E) = 0 if E is countable and µ∗(E) = 1 if E is uncountable. Then µ∗ is an outer measure (Show). Exercise: Determine the µ∗-measurable sets.

Theorem.8 (Carath`eodory) Given a set X and an outer measure µ∗ on X, the collection A := {E ⊂ X : E is µ∗ − measurable}

∗ is a σ−algebra of subsets of X, and µ |A = µ is a (complete) measure on X. Proof. Similar to the corresponding proofs for m (Theorem.1, Theorem.2 and its Corollary in Section II.2).

Question: How can we construct outer measures?

Deﬁnition. Let X 6= ∅ be a set. A collection κ of subset of X is called a (sequential) ∞ S∞ covering class for X if for all A ⊂ X, there exists {Ei}i=1 ⊂ κ such that A ⊂ i=1 Ei.

Examples.

1. In R, κ = {∅} ∪ {(n, n + 2)}n∈Z is a covering class. 2. In R, κ = {All intervals with rational endpoints} ∪ {∅} is a covering class (for m). .

Exercise. Find a suitable sequential covering class κ for X in each of the examples 1-3 above.

∗ Let λ : κ → [0, ∞] be a set function such that λ(∅) = 0. For any A ⊂ X, let µλ(A) be the set function (induced by λ) on P(X) deﬁned as

( ∞ ∞ ) ∗ X [ µλ(A) = inf λ(Ek): {Ek} ∈ κ, Ek ⊃ A . k=1 k=1

∗ Theorem.9 The set function µλ deﬁned above is an outer measure on X. ∗ ∗ ∗ ∗ Proof (sketch). µλ(∅) = 0, µλ(A) ≥ 0 for all A ∈ X, and A ⊂ B ⇒ µλ(A) ≤ µλ(B) is left as an exercise to the reader. We need to prove the countable subadditivity property. Let 23

{Ak} ⊂ P(X). Given > 0, ∞ ∞ [ X A ⊂ E1 such that {E1} ⊂ κ with λ(E1) < µ∗(A ) + 1 k k k 1 2 k=1 k=1 ∞ ∞ [ X A ⊂ E2 such that {E2} ⊂ κ with λ(E2) < µ∗(A ) + 2 k k k 2 22 k=1 k=1 . . . . ∗ S∞ P∞ ∗ Now show that µλ( m=1 Am) ≤ m=1 µλ(Am) + . Since > 0 is arbitrary, we get the desired result. ∗ ∗ Examples. 1. If X = R, κ = {all open intervals}, and λ = length, then we obtain µλ = m which gives the Lebesgue measure m.

2. (Lebesgue Stieltjes measure) Let X = R, κ = {(−∞, a]: a ∈ R} or κ = {(a, b]: a, b ∈ R#}. Let f : R → R be such that f is monotone increasing and right continuous (i.e. limx→a+ f(x) = ∗ f(a)). Deﬁne λ(a, b] = f(b) − f(a) with the convention that λ(∅) = 0. Let µf denote the set ∗ function “induced by λ” (via f). Then by Theorem.9, µf is the outer measure induced by f, ∗ and by Theorem.8, µf = µf |A=F = is the induced measure, called the Lebesgue-Stieltjes measure induced by f.

Exercises

1. If µ1, µ2, . . . , µn are measures on a measurable space (X, A) and α1, α2, . . . , αn ∈ [0, ∞), Pn then µ = i=1 αiµi is also a measure on (X, A).

2. If (X, A, µ) is a measure space and E ∈ A, then the set function µE : A → [0, ∞], deﬁned by µE(A) = µ(E ∩ A) for all A ∈ A, is also a measure on A.

3. Let µ be a ﬁnitely additive measure on a σ-algebra A of subsets of a set X. µ is a measure if and only if it is continuous from below.

4. Let µ be a ﬁnitely additive measure on a σ-algebra A of subsets of a set X with µ(X) < ∞. µ is a measure if and only if it is continuous from above.

5. Let (X, A, µ) be a measure space.

a) If A, B ∈ A and µ(A4B) = 0, then µ(A) = µ(B). b) Deﬁne: A ∼ B if and only if µ(A4B) = 0. Then “∼” is an equivalence relation on A. c) For A, B ∈ A, deﬁne ρ(A, B) = µ(A4B). Then ρ is a metric on A/ ∼. (Hence, (A/ ∼ , ρ) is a metric space).

6. If an outer measure µ∗ on a set X 6= ∅ is ﬁnitely additive, then it is a measure.

7. Let X 6= ∅ and κ = {∅,X, one-point sets}. Deﬁne λ(X) = ∞, λ(∅) = 0, and λ(E) = 1 if ∗ ∅= 6 E ( X. Describe the outer measure µλ.

8. Let X be an uncountable set and κ = {∅,X, one-point sets}. Deﬁne λ(X) = 1 and λ(E) = 0 ∗ if E ( X. Describe the outer measure µλ. 24

9. Let X 6= ∅, κ be a covering class for X and λ be a set function on P(X). If E ∈ κ, then ∗ µλ(E) ≤ λ(E). Give an example where strict inequality occurs. ∗ 10. Let X 6= ∅, κ be a covering class for X, λ be a set function on P(X) and µλ be the outer measure induced by κ and λ.

∗ a) If κ is a σ-algebra and λ is a measure, then µλ(E) = λ(E) for all E ∈ κ. ∗ b) If κ is a σ-algebra and λ is a measure, then every set in κ is µλ-measurable.

∗ [The measure µλ given by µλ is called an extension of λ.]

11. Let X 6= ∅, κ = {∅,X} and λ be a set function on P(X) given by λ(E) = 0 if E = ∅ ∗ ∗ and λ(E) = 0 if E = X. Determine the outer measure µλ and describe the σ-algebra of µλ- measurable sets. The same question when X = [0, 1].

12. Let X = R, κ = P(R) and deﬁne λ(E) = |{k ∈ Z : k ∈ E}|. Determine the outer measure ∗ ∗ µλ and describe the σ-algebra of µλ-measurable sets.

13. Let F be a real-valued, increasing and right continuous function on R. If µF is the associated Lebesgue-Stieltjes measure, then prove that, for any a, b ∈ R,

(i) µF ({a}) = F (a) − F (a−), (ii) µF ([a, b)) = F (b−) − F (a−), (iii) µF ([a, b]) = F (b) − F (a−), and (iv) µF ((a, b)) = F (b−) − F (a).