Lecture 2. Measure and Integration There are several ways of presenting the de nition of integration with respect to a measure. We will follow, more or less, the approach of Rudin’s ‘Real and Complex Analysis’ - this is probably the fastest route. Conventions concerning ±∞. In measure theory it is often essential to use ∞ and −∞. We shall view these as two objects and de ne the following relationship with real numbers (the set of all real numbers will be denoted by R) :
−∞ < c < ∞ for every c ∈ R
c + ∞ = ∞ if −∞ < c ≤ ∞, and c + (−∞) = −∞ if −∞ ≤ c < ∞ 0 · ∞ = 0 ±∞, if 0 < c ≤ ∞; c(±∞) = ∓∞, if −∞ ≤ c < 0; Note that ∞−∞ is not de ned! The convention 0·∞ = 0 might seem like an arbitrary choice, but it is the really the right one : it will be useful in integration and there it will appear as the natural choice. These operations have the usual properties on [0, ∞] : commutative, associative and distributive. Throughout our discussion (X, B, µ) will be a measure space. This means that X is a set, B is a collection of subsets of X forming a σ−algebra, and µ is a measure. Recall that to say that B is a σ−algebra means that : (S1) ∅ ∈ B; (S2) if A ∈ B then the complement Ac ∈ B; ∞ (S3) if A1, A2, A3, ... is a countable collection of sets with each Ai ∈ B then ∪i=1Ai ∈ B. Sets in B will be called B−measurable sets, or simply measurable sets.
Observations. (i) The intersection of a countable collection A1, A2, ... of measurable ∞ c c sets is also measurable (because ∩i=1Ai = (∪iAi ) is measurable by (S2) and (S3).) m (ii) If A1, ..., Am ∈ B then ∪i=1Ai ∈ B; this follows from (S1) and (S3) by taking An = ∅ m for all n bigger. Using the complements we obtain also that ∩i=1Ai ∈ B. (iii) Notice also that if A, B ∈ B then A \ B, being equal to A ∩ Bc, is also in B. To say that µ is a measure on (X, B) means that µ is a mapping B → [0, ∞] such that : (M1) µ(∅) = 0 (M2) if A1, A2, ... is a countable collection of mutually disjoint sets in B (i.e. each Ai ∈ B and Ai ∩ Aj = ∅ for every i 6= j) then:
∞ ∞ µ(∪i=1Ai) = µ(Ai). Xi=1
Taking Ai = ∅ for i bigger than some m, we see (from (M1) and (M2)) that µ is finitely-additive :
1 m m µ(∪i=1Ai) = µ(Ai) Xi=1
provided, of course, that the Ai are disjoint measurable sets. (Comment : It might seem that nite additivity implies the condition (M1): we have µ(∅) = µ(∅ ∪ ∅) = µ(∅) + µ(∅), \and so" µ(∅) = 0. What is wrong with this argument? Under what additional condition on µ would nite additivity imply (M1) ?) Exercise. If A and B are measurable sets with A ⊂ B, show that µ(A) ≤ µ(B). (Hint : write B as the union of A and B \ A; check that B \ A (which is B ∩ Ac) is measurable and use additivity.)
The simplest meaningful example of a measure is counting measure:
µ(A) = number of elements in A (taken to be ∞ if A is in nite)
2.1. De nition. A function f : X → [−∞, ∞] is measurable with respect to the σ−algebra B if the set f −1[a, b] (i.e. the set {x ∈ X : f(x) ∈ [a, b]} ) is in B, for every a, b ∈ [−∞, ∞]. Note that, in particular, if f is measurable then (taking b = a) every set of the type f −1(a) is measurable (a ∈ [−∞, ∞]). 2.2. Terminology. (i) If A ⊂ X, then the indicator function of A is the function 1A : X → R, de ned by : 1, if x ∈ A; 1A(x) = 0, otherwise. (ii) A simple function is a function s : X → R whose range consists of a nite set of points (i.e. the image s(X) is a nite subset of R). If a1, ..., an are all the distinct values of the simple function s then (verify that)
n
s = ai1Ai Xi=1 n where Ai = {x : s(x) = ai}. Notice (verify) that the Ai are disjoint sets and ∪i=1Ai = X. Notice also that a simple function can be written in di erent ways in the form
c cj1Cj . For instance, 1X is also equal to 1A + 1A , for any A ⊂ X. The represen- tationX described above in terms of the distinct values of s is, of course, unique; which is why we use it in the de nition (2.4) of s dµ below. Z 2.3. Exercise. Verify that s is measurable if and only if each Ai is. 2.4. De nition. (Integrating simple functions) If s : X → [0, ∞) is a measurable simple function of the form
n
s = ai1Ai Xi=1
2 −1 where a1, ..., an are all the distinct values of s (and Ai = s (ai)), then
n
s dµ = aiµ(Ai). Z Xi=1 Note that a term of the form 0 · ∞ might occur in the sum on the right side of the above equation; recall, for this purpose, that 0 · ∞ = 0. 2.5. Question. What is the relevance of Problem 2.3 to De nition 2.4 ? m
2.6. Note. If s = cj1Cj is a non-negative measurable simple function then a fact Xj=1 m that is not immediately obvious from De nition 2.4 is that s dµ = cjµ(Cj) (provided Z Xj=1 this sum makes sense). De nition 2.4 says that this is true only when the ci are all the distinct values of s. 2.7. Proposition. If s and t are nonnegative measurable simple functions then so is s + t and (s + t) dµ = s dµ + t dµ. Z Z Z If c ∈ [0, ∞) is a constant then :
cs dµ = c s dµ Z Z
n m
Proof. Write s = ai1Ai and t = bj1Bj , where a1, ..., an are all the distinct Xi=1 Xj=1 values of s and b1, ..., bm are all the distinct values of t. The function s + t takes the value ai + bj on the set Ai ∩ Bj (because on this set s takes the value ai and t takes the value bj). Moreover, (verify that) the sets Ai ∩ Bj are disjoint and cover all of X (i.e. 0 0 (Ai ∩Bj)∩(Ai0 ∩Bj0 ) = ∅ unless i = i and j = j ; and ∪i,j(Ai ∩Bj) = X.) This is because the Ai’s are disjoint and cover X and the Bj’s are disjoint and cover X. Therefore,
s + t = (ai + bj)1Ai∩Bj . 1≤i≤n,X1≤j≤m (To see this, consider any point x ∈ X. It must belong to exactly one of the sets Ai ∩ Bj; say x ∈ A1 ∩ B2. Then the function de ned by the sum on the right hand side of the above equation takes the value a1 + b2 at x. But this also the value of s(x) + t(x).) By Problem 2.3 we see that s + t is measurable. It is also clear that s + t is simple and non-negative. We would like to conclude, by De nition 2.4, that it follows that :
(s + t) dµ = (ai + bj)µ(Ai ∩ Bj) (2.1) Z 1≤i≤n,X1≤j≤m
3 However, there is no guarantee that the values ai + bj are distinct and therefore De nition 2.4 cannot be applied directly. Fortunately, equation (2.1) is still true because the sets Ai ∩ Bj are disjoint (this is the content of Lemma 2.8 below). So we accept equation (2.1). The right side of equation (2.1) can be split up into the sum of two sums :
n m m n (ai + bj)µ(Ai ∩ Bj) = ai µ(Ai ∩ Bj) + bj µ(Ai ∩ Bj) . 1≤i≤n,X1≤j≤m Xi=1 Xj=1 Xj=1 Xi=1
Now the sets Ai ∩ Bj are disjoint and, for xed i and variable j, they cover Ai ( for
∪j (Ai ∩ Bj) = Ai ∩ ∪j Bj = Ai ∩ X = Ai.) Therefore : m
µ(Ai ∩ Bj) = µ(Ai). Xj=1 Similarly, n
µ(Ai ∩ Bj) = µ(Bj). Xi=1 Putting all this together we obtain :
n m
(s + t) dµ = aiµ(Ai) + bjµ(Bj) Z Xi=1 Xj=1
and we recognize the the two sums on the right hand side to be s dµ and t dµ. Z Z Thus :
(s + t) dµ = s dµ + t dµ. Z Z Z
That cs dµ = c s dµ, for c any constant in [0, ∞), is veri ed easily by writing out Z Z s as a linear combination of indicator functions and using De nition 2.4.
2.8. Lemma. If C1, ..., Cn are disjoint measurable sets and c1, ..., cn non-negative real numbers then n n
( ci1C ) dµ = ciµ(Ci). Z i Xi=1 Xi=1 n Proof. We shall assume that ∪i=1Ci = X; for if this is not so then we can always n throw in another set Cn+1 = X \ Ci, and use cn+1 = 0, and this would not alter the i[=1
4 value of either side of the above equation (note : here we need to use 0 · ∞ = 0 in case n
µ(Cn+1) = ∞.) Let s = ci1Ci , and let a1, ...am be all the distinct values of s. Note that Xi=1 s is a non-negative measurable simple function. Moreover, (verify that) c1, ..., cn constitute all the values of s (i.e. {c1, ..., cn} = {a1, ..., am}). The trouble is that the ci may not all be distinct. For each i ∈ {1, ..., m}, set
Ji = {all j for which cj = ai}
−1 Then ∪j∈Ji Cj = Ai, where Ai = s (ai) is the set of all points where s takes the value ai. Then (verify that) the Ji are disjoint and their union is {1, ..., n}. Therefore,
n m
ciµ(Ci) = cjµ(Cj)
Xi=1 Xi=1 jX∈Ji m
= ai µ(Cj) Xi=1 jX∈Ji m
= aiµ(∪j∈Ji Cj) Xi=1 m
= aiµ(Ai) Xi=1 = s dµ. Z This completes the proof.
2.9. Comment. Suppose C1, ..., Cm are measurable sets and c1, ..., cm non-negative real numbers. Then Proposition 2.7 implies that
m m m cj 1C dµ = cj1C dµ = cjµ(Cj). Z j Z j Xj=1 Xj=1 Xj=1
(Check that De nition 2.4 does imply what we have used in the last equality above :
cj1C dµ = cjµ(Cj).) Compare this with Note 2.6. Z j We are now familiar with integrals s dµ where s is simple. The rst order of business Z now is to extend this to de ne the integral f dµ for a general measurable function f. First Z
5 a note of caution. We can’t expect f dµ to be meaningful for all measurable functions Z f; for if f is a simple function of the form 1A − 1B where both A and B are sets of in nite measure then f dµ has the nonsense value ∞ − ∞. In general we should expect that Z + − + f dµ should be de ned as f dµ − f dµ (here f = f1f≥0 is the function f where Z Z Z − f is non-negative and zero elsewhere, while f = −f1f≤0) and the integral f dµ being Z meaningful only when we don’t have an ∞ − ∞ situation. Thus if we can de ne f + dµ Z and f − dµ we should be done, i.e we should focus on de ning g dµ for non-negative Z Z measurable functions g. Since we know how to integrate simple functions, the rst idea is to pick a sequence of simple functions sn converging to g and then de ning g dµ to be Z
the limit of sn dµ. Of course there are a lot of things to check. As it turns out, it is a Z faster route if we de ne g dµ to be the supremum of s dµ for all non-negative simple Z Z s ≤ g, and this is what we do: 2.10. De nition. We know what s dµ is when s is a simple function. For a more Z general function f, it is reasonable to gure that f dµ can be obtained by taking the Z supremum of all s dµ for simple functions s ≤ f. Polishing this up, we de ne for any Z measurable function f : X → [0, ∞],
f dµ = sup s dµ Z Z where the supremum is over all measurable simple functions s satisfying 0 ≤ s ≤ f. (Such simple functions exist; s = 0 is one such function. Moreover, if f is itself a simple function then taking s = f we see that f dµ equals s dµ - in the sense of De nition 2.4 - and Z Z so there is no con ict between this new de nition for f dµ and the one in De nition 2.4 Z in the case f is simple.) 2.11. Basic observations about integration. Some simple facts that follow immediately from De nition 2.10 are : (a) if 0 ≤ f ≤ g then f dµ ≤ g dµ; Z Z (b) if f ≥ 0 and c is a constant with 0 ≤ c < ∞, then
(cf) dµ = c f dµ; Z Z
6 (c) if f = 0 except on a set of measure 0 then f dµ = 0 (verify this as follows: let E be Z the set of all points of X where f 6= 0; then check that any s with 0 ≤ s ≤ f satis es
s = s1E - write out s as a linear combination of indicator functions and show s dµ Z is 0 by using the fact that µ(E) is 0.)
Lecture 3.
As usual (X, B, µ) is a measure space.
3.1. Notation. If E ⊂ X, we write f dµ for f1E dµ. In particular, f dµ = ZE Z ZX f dµ. Z
3.2. Exercise. (a) If (1) A and B are measurable sets, (2) A ⊂ B, and if (3) µ(A) < ∞, then µ(B \ A) = µ(B) − µ(A). (b)Show that if E is a measurable subset of X, and f and g are measurable functions with 0 ≤ f ≤ g on E, then f dµ ≤ g dµ. ZE ZE Now assume (1) and (2) but instead of (3) assume that µ(A) = ∞. Show that then µ(B) = ∞. (Consult the Problem preceding De nition 2.1.) (Hint : Write B as the union of the disjoint measurable sets A and B \ A; then µ(B) = µ(A) + µ(B \ A). Proceed from here. Do you see why the hypothesis that A has nite measure is important (for the rst part above)?)
3.3. Proposition. If A1, A2, ... is a countable sequence of measurable subsets of X such that A1 ⊂ A2 ⊂ A3 ⊂ · · ·, then
∞ µ(∪i=1Ai) = lim µ(An) n→∞
If B1, B2, ... is a countable sequence of measurable subsets of X such that B1 ⊃ B2 ⊃ B3 ⊃ ..., and if, moreover, some µ(Bj) < ∞, then
∞ µ(∩n=1Bn) = lim µ(Bn) n→∞
Proof. Since we know how to calculate the measure of the (countable) union of disjoint sets we manufacture a sequence of disjoint sets C1, C2, ..., whose union is still ∞ ∪i=1Ai, according to the following prescription : set Cj = Aj \ Aj−1 for j > 1, and set C1 = A1. Then C1, C2, ... are measurable and mutually disjoint (verify). Note that ∞ ∞ µ(Ci) = µ(Ai)−µ(Ai−1), if Ai−1 has nite measure. Moreover, (verify) ∪j=1Cj = ∪j=1Aj. Therefore,
7 ∞ ∞ µ(∪i=1Ai) = µ(∪i=1Ci) ∞ = µ(Ci) Xi=1
= lim µ(C1) + µ(C2) + · · · + µ(Cn) n→∞ = lim µ(A1) + {µ(A2) − µ(A1)} + · · · + {µ(An) − µ(An−1)} n→∞ = lim µ(An). n→∞
This would conclude the proof except for the fact the fourth ‘=’ above is valid under the additional assumption that each µ(Ai) is nite. Thus the above chain of equalities prove the result if we also assume that each Ai has nite measure. If this assumption is not valid then some Ai has in nite measure. In that case (by Exercise 3.2 above) all the Aj, with j bigger than i, also have in nite measure. Therefore lim µ(An) would equal ∞. On the n→∞ ∞ other hand, the set ∪n=1An must also have in nite measure since it contains the set Ai as a ∞ subset. Thus, even if some Ai has in nite measure, the equality µ(∪n=1An) = lim µ(An) n→∞ still holds. The second part is left as an exercise.
Do Problem 2 in Problem Set 1 before attempting (3) of the following exercise. 3.4. Exercise (Some useful facts concerning inverse images) (1) Let f : X → Y be a mapping. Show that f −1 preserves all set-theoretic operations. That is, (a) f −1(∅) = ∅; −1 −1 (b) if A is any set of subsets of Y then f (∪E∈AE) = ∪E∈Af (E); (c) for any subsets E, and H of Y , f −1(H \E) = f −1(H)\f −1(E); (d) prove the analog of (b) with ∪ replaced by ∩. (2) Verify that every interval [a, b] ⊂ [−∞, ∞] can be obtained by using countable unions, intersections, and complements of sets of the form (x, ∞] ⊂ [−∞, ∞]. (Hint : if 1 −∞ < a < ∞ then [a, ∞] = ∩∞ (a − , ∞]; on the other hand [−∞, b] = (b, ∞]c. Note n=1 n that [a, b] = [a, ∞] ∩ [−∞, b]. What about the case when a = ∞ or a = −∞?) (3) Show that the σ−algebra of subsets of R generated by the intervals of the type (a, b) coincide with the one generated by those of the type [a, b] (and also with the σ−algebra generated by intervals of the type (a, ∞)). This σ−algebra is called the Borel σ−algebra of R. (Hint : Show that every interval of each type can be obtained by countable unions, complements, and countable intersections of intervals of any other type.) (4) Show that the σ−algebra of subsets of R generated by the intervals coincides with the σ−algebra generated by the open sets. (Hint : You will need the important fact that every open subset of R is the union of a countable collection of open intervals.) (5) If f : X → Y and g : Y → Z show that (g ◦ f)−1 = f −1 ◦ g−1 in the sense that for any subset E ⊂ Z, (g ◦ f)−1(Z) = f −1 g−1(E) . 3.5. Proposition (Manufacturing measurable functions I).
8 (1) A function f : X → [−∞, ∞] is measurable if and only if f −1(a, ∞] is a measurable set for every a ∈ [−∞, ∞] (one may replace (a, ∞] with any of the sets [a, ∞], [−∞, a), [−∞, a]). In case f : X → R, it is measurable if and only if f −1(a, b) is measurable for every a, b ∈ R. (2) A function f : X → R is measurable if and only if f −1(U) is measurable for every open set U ⊂ R. (3) If f : X → R is measurable and g : R → R is any continuous function then g ◦ f : X → R is also measurable. (4) If f, g : X → R are measurable functions and φ : R2 → R is continuous, then the function φ(f, g) : X → R : x 7→ φ f(x), g(x) is measurable. In particular, both f + g and fg are measurable (by taking φ to be (p,q) 7→ p+q and (p, q) 7→ pq, respectively.) Proof. (1), (2) : Use Exercise 3.4. (3) Recall that g is continuous if and only if g−1(U) is open for every open set U ⊂ R, and use part (2) above in combination with Ex 3.4(5). (4) Let h : X → R2 : x → f(x), g(x) . Then φ ◦ (f, g) = φ ◦ h. By part (3) it will su ce to show that (φ◦h)−1(U) is measurable for every open set U ⊂ R. Let U be an open subset of R. Since φ is continuous, φ−1(U) is an open subset of R2. Then, since every open subset of R2 is a countable union of open boxes of the form (a, b) × (a0, b0), it follows that φ−1(U) is a countable union of such boxes. Therefore, by Ex 3.4(1,5), (φ ◦ h)−1(U) is a countable union of sets of the form h−1 (a, b) × (a0, b0) . Thus it will su ce to show that this set h−1 (a, b) × (a0, b0) is measurable. Examination of the de nition of h shows that this set is equal to f −1(a, b) ∩ g−1(a0, b0). Since f and g are measurable these inverse images under f and g, and hence also their intersection, are measurable. This completes the proof. 3.6. Proposition (Manufacturing measurable functions II). (1) Constant functions are measurable. The sum and product of two measurable functions are measurable functions (assuming that the sum/product is de ned; i.e. we don’t have sums like ∞ + (−∞) or similar unde ned products). (2) If f1, f2, ... is a sequence of measurable functions X → [−∞, ∞] then the functions x 7→ sup fn(x) (and hence also the function x 7→ inf fn(x)) and x 7→ lim sup fn(x) (and n≥1 n≥1 n≥1 hence also x 7→ lim inf fn(x)) are measurable. In particular, if lim fn(x) exists for every n≥1 n→∞ x ∈ X, then the function x 7→ lim fn(x) is measurable. n (3) If f and g are measurable functions X → [−∞, ∞] then so are the functions max{f, g} and min{f, g}. In particular, the functions
f + = max{f, 0} and f − = − min{f, 0} are measurable. As a consequence (of this and (1) ) the absolute value |f|, being the sum f + + f −, is measurable. Proof. (1) It is easy to see (verify) that a constant function is measurable. Now suppose f and g are measurable functions and that f(x) + g(x) is de ned for every x ∈ X.
9 It will su ce (because of Ex 3.5(1) ) to show that that for every a < ∞, the set E(a) = {x ∈ X : f(x) + g(x) > a} is measurable. Now this set is the union of the measurable sets g−1{∞}, f −1{∞}, and the set F (a) = E(a) ∩ f −1(R) ∩ g−1(R). (The reason why we bring in F (a) is that we want to avoid dealing with points where f and g have in nite values.) Thus it will su ce to show that the set F (a) is measurable. However, for F (a) we are essentially dealing with nite-valued (i.e. real-valued) functions f and g and therefore the argument in Proposition 3.5 (4) works. The measurability of fg follows in exactly the same way. −1 ∞ −1 (2) If f = sup fn, then it may be veri ed that f (a, ∞] = ∪n=1fn (a, ∞] (verify n this). This, being a countable union of measurable sets, is measurable. Similarly (or by using negatives), inf fn is also measurable. To deal with lim sup fn note rst that n≥1 n lim sup fn is equal to inf {sup fk}. Therefore, by combining our preceding two statements, n n≥1 k≥n we see that lim sup fn is also measurable. If lim fn exists then it is equal to lim sup fn and n n n is therefore measurable. (3) This is immediate from (2) by taking f1 = f and f2 = f3 = ... = 0.
Problem Set 1. 1. If (a, b) and (c, d) are two points in R2, say that they are equivalent if both c − a and d − b are rational numbers. Verify that this is indeed an equivalence relation. Verify also that every point in R2 is equivalent to at least one point (in fact in nitely many points) in the unit square [0, 1] × [0, 1]. Form a set E ⊂ R2 by picking one point in this unit square from each equivalence class. (The formal existence of such a set is guaranteed by the Axiom of Choice.) The set of all points of [0, 1] × [0, 1] with rational coordinates is a countable set {r1, r2, ...}. Show that the sets E + rn are mutually disjoint and that
∞ ∪n=1 E + rn ⊂ [0, 2] × [0, 2]. Use this to show that there is no measure µ de ned on the set of all subsets of R2 which has the following properties : (1) the measure of the translate of a set is equal to the measure of the set (i.e. µ(E + r) = µ(E)) , and (2) the measure of the square [0, 2] × [0, 2] is a nite positive (> 0) number. 2. Let E be a collection of subsets of X, and let denote the family of all σ−algebras (of subsets of X) which contain the collection E. First show that is not empty by verifying that the set P(X) of all subsets of X is a σ−algebra. Then show that ∩ (i.e. the collection of all sets which belong to each of the σ−algebras in the family ) is a σ−algebra and that it is the smallest σ−algebra containing E (i.e. if B is a σ−algebra containg all the sets of E then B ⊃ ∩ . The σ−algebra ∩ is called the σ−algebra generated by E and is often denoted by σ(E).)
3. Let X be a set, and P(X) the σ−algebra of all subsets of X. Show that every function f : X → [−∞, ∞] is measurable with respect to P(X). Conversely, suppose B
10 is a σ−algebra of subsets of X such that every function f : X → [−∞, ∞] is measurable. Show that B = P(X). 4. Let (X, B) be a measurable space and b a xed point in X. De ne µ : B → [0, ∞] by :
1, if b ∈ A; µ(A) = 0, otherwise. Show that µ is a measure. This measure is called the ‘delta function’ at b. 5. (Hard) Let B be a σ−algebra of subsets of {1, ..., n}. Show that B has 2k elements, for some integer k ∈ {1, ..., n}. 6. Show that every continuous function R → R is measurable with respect to the Borel σ−algebra of R. Give some examples of discontinuous Borel measurable functions. Can you prove that there exist functions R → R which are not Borel measurable ? 7. If s and t are non-negative simple functions and t ≤ s then show that t dµ ≤ Z s dµ (with the integrals as de ned in De nition 2.4) ? (Hint : show that t − s is a Z non-negative simple function and use Proposition 2.7 on s = t + (s − t).) What is the relevance of this to the comments in De nition 2.10 ? 8. The set of all rotations in R3 is denoted SO(3); it is a group under composition of 1 rotations. Let a denote the rotation around the z−axis by the angle cos−1( ), and b the 3 rotation by the same angle around the x−axis. It can be shown by a clever computational argument that a and b are ‘independent’ rotations in the following sense : no product of r1 s1 rn sn the form a b ...a b with the ri and si being non-zero integers, or similar products (of non-zero powers of a and b, successively) starting/ending with non-zero powers of a or b, can be equal to the identity rotation. Rotations satisfying this independence property were constructed by Hausdor in 1914. Let G denote the set of all rotations which can expressed as ( nite) products of powers (positive and negative) of a and b. Check that G + r1 s1 is a subgroup of SO(3). Let A denote the set of all nite products a b ... with r1 > 0 (i.e the product starts with a positive power of a), let A− be the set of all nite products r1 s1 + s1 r2 a b ... with r1 < 0; and let B be the set of all nite products b a ... with s1 > 0, and − s1 r2 ± ± B the set of all nite products b a ... with s1 < 0. Thus the sets A and B form a partition of G \ {I} into four disjoint subsets. 2 3 2 2 2 3 Let S = {(x1, x2, x3) ∈ R : x1 + x2 + x3 = 1}, the unit sphere in R . (i) Show that there is a countable set D ⊂ S2 such that G acts on S2 \ D without xed points; i.e. if T ∈ G, and T 6= I, and x ∈ S2 \ D then T x ∈ S2 \ D and T x 6= x. (Hint : how many points on S2 does a rotation keep xed ?) We shall denote S2 \ D by X. The orbit of a point x ∈ X under G is the set {T x : T ∈ G}. Distinct orbits are disjoint sets. Picking one element from each such orbit (this is legal under the Axiom of Choice) we form a set M ⊂ X such that for every x ∈ X there is a unique element m ∈ M for which T x = m for some T ∈ G. + Let XA+ = {T m : m ∈ M, T ∈ A }, and de ne sets XA− , XB+ , XB− analogously. (ii) Show that the four sets XA± and XB± are disjoint and their union is X.
11 −1 −1 (iii) Show that X = a XA+ ∪ XA− , and X = b XB+ ∪ XB− . (iv) Suppose µ is a measure on the σ−algebra of all subsets of S2 which has the properties that : (1) it is invariant under rotations (i.e. µ T (E) = µ(E) for every T ∈ SO(3) and every E ⊂ S2); and (2) 0 < µ(S2) < ∞. Use (ii) and (iii) above to show that such a measure µ cannot exist ! Is it necessary to use the countable additivity of µ ? (v) Show that there is no measure ν de ned on the σ−algebra of all subsets of R3 which satis es the following conditions : (1) ν is invariant under rotations; (2) 0 < ν(Br) < ∞ for some ball Br, centered at the origin and of radius r(< ∞). (Hint : Suppose such a measure ν exists, and assume, for convenience and without loss of generality, that the ν−measure of the closed unit ball is nite and positive. De ne a measure µ on the set of all subsets of S2 as follows. Consider any Y ⊂ S2, and form the ‘cone’ Y 0 = {ry : y ∈ Y, r ∈ [0, 1]} ⊂ R3. De ne µ(Y ) to be ν(Y 0). Now use part (iv) above.) (vi) Let D be a countable subset of S2. Choose a line through the origin which does not pass through any point in D (why does such a line exist ?). Verify that the set of all rotations around the chosen line which carry some point of D into some point of D is countable (Hint : how many rotations around the line take a given point to another given point ?); conclude that there is a rotation T such that T (D) ∩ D = ∅. Let µ be a finitely additive (but not necessarily countably additive) measure on the set of all subset of S2 which is invariant under rotations and for which 0 < µ(S2) < ∞. Show that 1 µ(D) ≤ µ(S2) (Hint : consider D ∩ T (D).) Apply this to D ∪ T (D) in place of D and 2 1 thus show that µ(D) ≤ µ(S2). What can you say about µ(D) ? Can you now prove a 4 stronger form of the results in (iv) and (v) above ? Definition. A set F ⊂ R3 is ‘equidecomposable’ with a set H ⊂ R3 if F can be partitioned into sets F1, ..., Fm (for some integer m ≥ 1) and H can be partitioned into sets H1, ..., Hm such that each Fi can be rotated to yield Hi; in other words, F can be broken down and rearranged (by using rotations) and reassembled to form H. Verify that if E is equidecomposable with F , and F is equidecomposable with H, then E is equidecomposable with H. (vii) Let D be a countable subset of S2, and let T be the rotation described in (vi). Let D0 = D ∪ T (D) ∪ T 2(D) ∪ · · ·. Check that T j(D) ∩ T i(D) = ∅, unless j = i. Consider the partition of S2 into S2 \ D0 and D0. Consider, on the other hand, the partition of S2 \ D into the sets S2 \ D0 and T (D) ∪ T 2(D) ∪ · · ·. Using these partitions, show that S2 \ D is equidecomposable with S2. (viii) Recall the sets XA± and XB± . Let XA = XA+ ∪ XA− , and XB = XB+ ∪ XB− . 2 Combine some of the preceding results to show that XA is equidecomposable with S , and 2 that XB is also equidecomposable with S . (ix) Let B denote the unit ball, centered at the origin, minus the origin itself; i.e. 3 2 2 2 B = {(x1, x2, x3) ∈ R : 0 < x1 + x2 + x3 ≤ 1}. Use (viii) and the idea used in (v), to show that there are two disjoint subsets of B each of which can be broken down and rearranged to form the whole of B. In particular, by doing this we can create two copies of B by taking two (disjoint) pieces from B and rearranging (by cutting, rotating, and reassembling) each piece ! (It is possible to include the origin as well but we won’t pursue
12 this any further). This is one (somewhat weak) version of the Banach-Tarski Paradox. For more on this, see the book with this title by Stan Wagon, Cambridge University Press (1985), based on which the above exercises have been composed.
Lecture 4. Integration and the Convergence Theorems.
The main results here are the Monotone Convergence Theorem (4.6), Fatou’s Lemma (4.10), and the Dominated Convergence Theorem (4.15). As usual, (X, B, µ) is a measure space. We have already de ned f dµ for non-negative measurable functions f. In the Z following we extend to more general f : 4.1. De nition. (Integration) Let f : X → [−∞, ∞] be a measurable function. As we have seen before, f = f + − f − and both f + and f − are non-negative, measurable functions, and so f + dµ and f − dµ make sense. If at least one of these integrals is Z Z nite then we de ne :
f dµ = f + dµ − f − dµ Z Z Z
(Note that we have excluded the troublesome case ∞−∞ by not de ning f dµ for those Z f for which both f + dµ and f − dµ are ∞.) Z Z Note . Let f : X → [−∞, ∞] be measurable. Then f dµ is finite if and only if Z |f| dµ < ∞. (Recall that |f| is measurable and that |f| = f + + f −. The reason why Z this fact is useful is that |f| is non-negative and so |f| dµ always makes sense - in some Z situations it is possible to show that this is nite without having to deal with f + and f −.)
Notation. f dµ means f1E dµ. ZE Z Check that in case f is non-negative measurable then f dµ as de ned by De nition Z 2.10 equals f dµ as de ned in De nition 4.1 above. (Hint : what is f + and what is f − Z for non-negative f ?) Exercise. Here is something we forgot to verify earlier. In case f is a non-negative simple function we had two de nitions of f dµ: one according to De nition 2.4 (integral Z of non-negative measurable simple functions) and one according to De nition 2.10 (for non-negative measurable functions). Show that there is no con ict between these two
13 de nitions. (Hint : if f is non-negative measurable simple then in De nition 2.10 take s = f; you need also the fact that if s and t are non-negative simple functions and t ≤ s then t dµ ≤ s dµ - which follows by writing s = t + (s − t), checking that t − s is a Z Z non-negative measurable simple function, and using Proposition 2.7).
4.2. Exercise. (a) Let f = 1A − 1B, where A and B are disjoint measurable subsets of X. Under what conditions on µ(A) and µ(B) is f dµ well-de ned ? What is its value Z in this case ? (b) Check that if f is any non-negative measurable function and E is any measurable set
then f dµ ≤ f dµ. (Is f1E ≤ f?) ZE Z 4.3. Exercise. (Sets of measure zero aren’t important) Suppose f and g are measurable functions X → [−∞, ∞]. (a) Show that the set N = {x ∈ X : f(x) 6= g(x)} is measurable. (Hint : N is the union of Ns = {x ∈ X : f(x) < g(x)} and Nb = {x ∈ X : f(x) > g(x)}. A point x lies in Ns if and only if there is a rational number r such that f(x) < r and r < g(x). Thus Ns is the union, as r runs over the countable set of all rational numbers, of the sets {x ∈ X : f(x) < r} ∩ {x ∈ X : r < g(x)}. Verify that each of the two sets involved this intersection here is measurable. Proceed. Note that this argument shows that the sets {f > g} (which is short-hand for {x ∈ X : f(x) > g(x)}) and {f < g} is measurable.) Then show that the sets {f ≥ g} and {f ≤ g} are also measurable. (b) Suppose µ(N) = 0. That is, f and g are equal everywhere except on a set N of measure zero : we say then that f and g are equal almost everywhere (of course, with respect to the measure µ under consideration) - in short, f = g a.e. Show that then f + equals g+ almost everywhere, and f − = g− a.e. (Hint : if f(x) = g(x) then both f +(x) = g+(x) and f −(x) = g−(x); thus if at x, either f + doesn’t equal f −, or g+ doesn’t equal g− then f(x) 6= g(x) - i.e. {x : f +(x) 6= g+(x)} and {x : f −(x) 6= g−(x)} are subsets of N. Why are these sets measurable ?) (c) Show that if s is a non-negative simple function which is equal to 0 almost everywhere
then s dµ = 0. (Hint : write s = aj1Aj and verify that whenever aj 6= 0 the set Z X Aj has measure zero; then look at the formula de ning s dµ.) Z (d) If s is a non-negative simple function and N is a (measurable) set of measure zero,
show that s dµ = s1N c dµ. (Hint : you can do this directly from the De nition Z Z
of s dµ or by writing s = s1N c + s1N and using (c) on s1N dµ.) Z Z (e) Let h be a non-negative measurable function, and N a set of measure 0. Show that
h dµ = h1N c dµ. (Hint : Consult De nition 2.10, replace each s in that de nition Z Z by s1N c and use part (d). Check : (i) if 0 ≤ s ≤ h then 0 ≤ s1N c ≤ h1N c , and (ii) if 0 ≤ t ≤ h1N c then 0 ≤ t ≤ h.) (f) Let f, g : X → [−∞, ∞] be measurable, and suppose f = g a.e.; i.e. that the set N =
14 {f 6= g} has measure zero. Show by combining (b) and (e) that f + dµ = g+ dµ Z Z and that f − dµ = g− dµ. From this show that f dµ is de ned if and only if Z Z Z g dµ is de ned, and this case the two integrals are equal. Thus, if f and g are equal Z almost everywhere then f dµ = g dµ whenever either side is de ned. medskip Z Z 4.4. Excercise. (to be used ) Let s be a non-negative measurable simple function on X. De ne, for every measurable set E,
ν(E) = s dµ ZE
Show that ν is a measure in the following way. First check quickly that ν(E) makes sense and is non-negative for every measurable set E. Check that ν(∅) = 0 (hint : What is s1∅?) The non-trivial part is checking that ν is countably additive. For this write out s in the n
form ai1Ai , where a1, ..., an are all the distinct values of s. Consider now a countable Xi=1 collection of disjoint measurable sets E1, E2, .... Then examine carefully each step of the following chain of equalities and justify any step that is not obvious :
ν(∪j Ej) = s1∪ E dµ Z j j n
= ai1A ∩∪ E dµ (Hint : 1A1B = 1A∩B) Z i j j Xi=1 n = aiµ(Ai ∩ ∪jEj) Xi=1 n ∞ = ai µ(Ai ∩ Ej) Xi=1 nXj=1 o ∞ n = aiµ(Ai ∩ Ej) Xj=1 Xi=1 ∞ = ν(Ej). Xj=1
The following result as well as the construction given in its proof is very useful. 4.5. Proposition. Let f : X → [0, ∞] be measurable. Then there exist measurable simple functions sn on X such that : (1) 0 ≤ s1(x) ≤ s2(x) ≤ · · · ≤ f(x) for every x ∈ X; (2) sn(x) → f(x) as n → ∞, for every x ∈ X.
15 Proof. The idea is to ‘discretize’ the values of f and in this way approximate f by simple functions. For example, suppose we want a simple function sn which doesn’t di er 1 1 from f by more than 2n . For this we cut up [0, ∞) into pieces of length 2n ; the k-th piece k − 1 k k − 1 k would be [ , ). Let En denote the set of all point x ∈ X at which f(x) ∈ [ , ]. 2n 2n k 2n 2n k − 1 We set s (x) = , for x ∈ En. Thus for x ∈ En, both s (x) and f(x) belong to an n 2n k k n 1 1 interval of length 2n , and therefore f(x) and sn(x) di er by at most 2n . De ning sn in n this way for each Ek , we obtain the function sn approximating f pretty closely, and sn should converge to f everywhere. Notice also that the way we have de ned sn insures that sn(x) ≤ f(x) for every x. While this is the key strategy there are a number of small 1 aws. First, you can’t cut up [0, ∞] into bits of length 2n , for in this way you can never capture ∞. Secondly, the way we de ned sn above, the set of values of sn looks more like a countable set than a nite set. These problems can be xed in one stroke. De ne sn(x) as above for those points x where f(x) < n; while for points x where f(x) ≥ n, set sn(x) = n. By doing this we give up our initial goal of approximating f closely everywhere (we give up in the region where f is bigger than n - but as n → ∞ this works out all right). To be precise, de ne : n n 2 k − 1 − − sn = 1{ k 1 ≤f< k } + n1f 1[n,∞]. 2n 2n 2n kX=1 k − 1 k The sets f −1[ , ) and f −1[n, ∞] are measurable and mutually disjoint (for xed n 2n 2n and variable k). Thus sn is a measurable simple function, and is clearly non-negative. If x is a point where f(x) < ∞, then for large enough integer n we will have f(x) < n and for 1 all such integers n the values f(x) and sn(x) will di er by at most 2n . Thus sn(x) → f(x), as n → ∞, for such x. If, instead, f(x) = ∞, then x belongs to f −1[n, ∞] for every integer n, and so sn(x) = n for every n; as a result, sn(x) → ∞ as n → ∞. Thus in either case sn(x) → f(x) as n → ∞. Finally, it is easy to check that we still have sn(x) ≤ f(x) for every x ∈ X. One of the questions one meets over and over again in analysis is when, given a
sequence of functions fn converging to a function f, fn dµ converges to f dµ. There Z Z three important and versatile tools to handle such situations : (a) the Montone Convergence theorem (b) the Lebesgue Dominated Convergence theorem (c) Fatou’s Lemma. Of these, Fatou’s lemma is used less frequently but is nevertheless just the right tool to use in certain situations. We will use all three results over and over again. We turn rst to the monotone convergence theorem. 4.6. Monotone Convergence Theorem. Let (fn) be a sequence of functions on X such that : (1) each fn is measurable and non-negative; (2) 0 ≤ f1(x) ≤ f2(x) ≤ · · · ≤ ∞ for every x ∈ X;
16 (3) fn(x) → f(x) as n → ∞, for every x ∈ X. Then f is measurable and
fn dµ → f dµ as n → ∞. Z Z Proof. We shall go through a very descriptive account : after all the ideas have been digested, it is possible to condense them into a fairly short e cient proof. Note rst that
(1) guarantees that fn dµ exists for each n ≥ 1. Moreover, f being the limit of a sequence Z of non-negative measurable functions is itself a non-negative measurable function. Thus f dµ is also de ned. Z
Condition (2) implies that fn dµ ≤ fn+1 dµ, and so the sequence of numbers Z Z
fn dµ → α as n → ∞, for some α ∈ [0, ∞]. Clearly, fn(x) ≤ f(x), for every x ∈ X, and Z therefore fn dµ ≤ f dµ. Since this holds for every n, we can take n → ∞ to see that Z Z α ≤ f dµ. Z Now we need to prove the opposite inequality, that α is greater or equal to f dµ. Z This is the hard part. Recall that f dµ is the supremum of s dµ as s runs over measurable simple Z Z functions with 0 ≤ s ≤ f. So we want to show that α is greater or equal to this supremum. That is, we want to show that α is greater or equal to s dµ for every s of the type Z mentioned above. Since α is the limit of the sequence of fn dµ we would be done if some Z fn is greater or equal to s. But, although fn → f, and s ≤ f, we can’t conclude that some fn is greater or equal to s. All we can say is that if s(x) < f(x) (at some point x) then for n large enough fn(x) > s(x). So let us see if we can get anywhere with the set :
En = {x : fn(x) ≥ s(x)}.
This set is measurable (see Excercise 4.3 (a) ). Moreover, E1 ⊂ E2 ⊂ E3 ⊂ ... (because f1 ≤ f2 ≤ f3 ≤ · · ·.) The comments above show that {x : s(x) < f(x)} ⊂ ∪n≥1En. Also :
fn dµ ≥ s dµ ZEn ZEn
because - and check this - fn1En ≥ s1En . Now fn dµ is less or equal to the whole ZEn
integral fn dµ, since fn is non-negative (Excercise 4.2(b)). Therefore, Z
fn dµ ≥ s dµ. Z ZEn
17 Now we take n → ∞ to obtain :
α ≥ lim s dµ (4.1) n→∞ ZEn
Now since (by Excercise 4.4) E 7→ s dµ is a measure, and since E1 ⊂ E2 ⊂ · · ·, we ZE have s dµ → s dµ, as n → ∞. ZEn Z∪nEn Thus inequality (4.1) reads :
α ≥ s dµ (4.2) Z∪nEn
If ∪nEn were equal to X then we would be done. But in general, ∪nEn is not the whole space X. Let us see how big it is. As we have already seen it contains the set {x : s(x) < f(x)}. Now this set clearly excludes the points where f(x) = 0 (because s ≥ 0). But we can see that the set {f = 0} is contained in each En (because of the requirements 0 ≤ fn ≤ f, and 0 ≤ s ≤ f). Thus A = {f = 0} ∪ {s < f} ⊂ ∪nEn. So inequality (4.2) implies : α ≥ s dµ (4.3) ZA But still A might not equal all of X. Recalling that 0 ≤ s ≤ f, we see upon examining A that what we are missing is the set on which s = f 6= 0. The following trick xes the problem. Pick a fraction c close to 1 (any c ∈ (0, 1) will do) and replace s by cs. Now cs is still a measurable simple function satisfying 0 ≤ cs ≤ f. The nice thing about cs is that cs(x) < f(x) unless f(x) = 0. Thus if we rede ne A by replacing s in the de nition of A by cs we actually obtain all of X; i.e. {f = 0} ∪ {cs < f} = X. Therefore, inequality (4.3) reads (with s replaced by cs) :
α ≥ cs dµ ZX which is the same as : α ≥ c s dµ (4.4) Z Now we want to get rid of the c. This is easy : inequality (4.4) is valid for every c ∈ (0, 1) and so we can take c → 1, to obtain :
α ≥ s dµ (4.5) Z Recall that here s is any measurable simple function with 0 ≤ s ≤ f. Taking the supremum over all such s gives the desired inequality :
α ≥ f dµ Z
18 and this completes the proof.
Remarks : The hypothesis that each fn is non-negative is important. For consider an example where µ(X) = ∞ (for example, µ is counting measure on the set of all subsets of 1 an in nite set X), and set f = − . Then all the hypotheses, except for non-negativity, n n are satis ed but each fn dµ = −∞ while ( lim fn) dµ = 0 dµ = 0. Z Z n→∞ Z 4.7. Fact. Let f be a function X → [−∞, ∞] which can be written in the form g − h, where h and g are functions X → [0, ∞] (i.e. g, h ≥ 0). Then f + ≤ g and f − ≤ h. (Thus, the representation f = f + − f − is ‘minimal’ in a sense.) Proof . Consider any x ∈ X. If f(x) ≥ 0, then f +(x) = f(x) and so :
f +(x) = f(x) = g(x) − h(x) ≤ g(x).
If f(x) < 0, then f +(x) = 0 and so certainly f +(x) ≤ g(x). Thus f + ≤ g. Similarly, f − ≤ h. 4.8. Proposition. Let f and g be measurable functions X → [−∞, ∞]. Suppose that : (1) f dµ and g dµ are de ned, Z Z (2) the sum f dµ + g dµ is de ned, and Z Z (3) f(x) + g(x) is de ned for every x ∈ X. Then (f + g) dµ is de ned and : Z
(f + g) dµ = f dµ + g dµ (4.6) Z Z Z
If c ∈ R is a constant then cf dµ is de ned if and only if f dµ is de ned, and in this Z Z case : cf dµ = c f dµ (4.7) Z Z Proof. First suppose that f and g are non-negative measurable functions. Then by Proposition 4.5, there exist sequences of measurable simple functions (sn) and (tn) such that 0 ≤ sn ≤ f, 0 ≤ tn ≤ g, and sn(x) → f(x), and tn(x) → g(x) for every x ∈ X. Then sn + tn is a sequence of measurable simple functions satisfying 0 ≤ sn + tn ≤ f + g and sn(x) + tn(x) → f(x) + g(x), as n → ∞, for every x ∈ X. Then the monotone convergence theorem gives the following limits (as n → ∞):
(sn + tn) dµ → (f + g) dµ Z Z
sn dµ → f dµ Z Z
19 tn dµ → g dµ. Z Z
Now sn dµ + tn dµ = (sn + tn) dµ. So letting n → ∞ we obtain : Z Z Z
(f + g) dµ = f dµ + g dµ. Z Z Z
A similar argument (with sn and csn) proves that if f is non-negative measurable and c ∈ (0, ∞) then:
cf dµ = c f dµ. Z Z Now we deal with general measurable f, g : X → [−∞, ∞], for which the hypotheses (1),(2) and (3) hold. Let h = f + g. Now f = f + − f − and g = g+ − g−. So :
h+ − h− = h = f + g = f + − f − + g+ − g−.
(this is okay even if ±∞’s are involved.) We then have (the case when ±∞ are involved requires special checking):
h+ + f − + g− = f + + g+ + h− (4.8)
Since these are all non-negative measurable functions we can conclude that :
h+ dµ + f − dµ + g− dµ = f + dµ + g+ dµ + h− dµ (4.9) Z Z Z Z Z Z
From which we would like to conclude that :
h+dµ − h− dµ = f + dµ − f − dµ + g+ dµ − g− dµ Z Z Z Z Z Z and hence that
h dµ = h+dµ − h− dµ Z Z Z = f + dµ − f − dµ + g+ dµ − g− dµ Z Z Z Z = f dµ + g dµ Z Z which is what we wanted. The only problem with this argument is that the second line which is obtained by a rearrangement of equation (4.9) is a valid conclusion from (4.9) only if h+ dµ and h− dµ are not both ∞. Now, by Fact 4.7, h+ ≤ f + + g+, and Z Z
20 h− ≤ f − + g−. So if both h+ dµ and h− dµ were ∞ then both f + dµ + g+ dµ Z Z Z Z and f − dµ + g− dµ would be ∞ (here we have again used the fact, proved above, Z Z that the integral of the sum of two non-negative measurable functions is the sum of the integrals). But, as may be veri ed by checking each of the four possible cases here, this would contradict the hypotheses that f dµ, g dµ and their sum are de ned. Z Z This takes care of all the details in proving that (f + g) dµ = f dµ + g dµ. Z Z Z The proof of (cf) dµ = c f dµ for general f (i.e. not necessarily non-negative) is Z Z done similarly. n n
Exercise. Show that cj1C dµ = cj µ(Cj) if c1, ..., cn are real numbers and Z j Xj=1 Xj=1 n
C1, ..., Cn are measurable sets of nite measure. Is it possible for cj1C dµ to exist Z j Xj=1 n
even if cjµ(Cj) is unde ned ? Xj=1
4.9. Proposition. If fn : X → [0, ∞] (note fn ≥ 0) is measurable, for n = 1, 2, ..., and
∞ f(x) = fn(x) for every x ∈ X nX=1 then f is measurable and ∞ f dµ = fn dµ Z Z nX=1 (i.e. the sum and integral can be interchanged). Proof. Let gn = f1 + · · · + fn. Then each gn is measurable, 0 ≤ g1 ≤ g2 ≤ · · ·, and gn(x) → f(x) as n → ∞, for every x ∈ X. Therefore by the monotone convergence theorem, f is measurable and
gn dµ → f dµ as n → ∞. Z Z Now by Proposition 4.8 :
n gn dµ = fi dµ. Z Z Xi=1
Letting n → ∞ in this formula and comparing with the earlier stated limit of gn dµ we Z obtain the desired result.
21 4.10. Fatou’s Lemma. If fn : X → [0, ∞] (note : fn ≥ 0) is measurable, for n = 1, 2, ..., then
(lim inf fn) dµ ≤ lim inf fn dµ. Z n→∞ n→∞ Z Proof. This is obtained easily from the monotone convergence theorem by realizing that lim inf fn(x) equals lim gn(x), where gn(x) = inf fk(x), and that then g1, g2, ... are n→∞ n→∞ k≥n measurable functions with 0 ≤ g1 ≤ g2 ≤ · · ·. For, with this notation, we have rst (since, by de nition of gn, we have gn ≤ fn) :
gn dµ ≤ fn dµ (4.10) Z Z
By the monotone convergence theorem (and since lim inf fn = lim gn): n→∞ n→∞
gn dµ → lim inf fn dµ (4.11) Z Z n→∞ We might be tempted to let n → ∞ in equation (4.10) to conclude (using (4.11)) that
lim inf fn dµ is less or equal to lim fn dµ. However, there is no guarantee that Z n→∞ n→∞ Z
lim fn dµ exists. So the best we can do is to take lim inf in equation (4.10) and this n→∞ Z n→∞ gives the desired inequality. 4.11. Standard Convention. Consider a function f : E → [−∞, ∞] where E ⊂ X is measurable and µ(Ec) = 0. Suppose, moreover, that f is measurable in the sense that f −1[a, b] is a measurable subset of X (of course, it is in fact a subset of E) for every a, b ∈ [−∞, ∞]. Choose any function f 0 : X → [−∞, ∞] which is measurable and agrees with f on the set E. (For example, the function h de ned by h(x) = f(x) for x ∈ E and h(x) = 0 for x ∈ Ec is one such function - verify that h is measurable.) Then we de ne f dµ to be f 0 dµ. This value is independent of the choice of f 0, for if we use a di erent Z Z choice say f 00 then f 0 and f 00 are equal (on the set E) almost everywhere and hence have the same integral. 4.12. Chebyshev’s Inequality. Let f : X → [0, ∞] be a measurable function and let c ∈ (0, ∞) a positive number. Then : 1 µ({x ∈ X : f(x) ≥ c}) ≤ f dµ. c Z Proof. We have :
f dµ ≥ f dµ ≥ c dµ = cµ({f ≥ c}) Z Z{f≥c} Z{f≥c} where the second inequality follows from the simple observation that on the set {f ≥ c} the function f is ≥ c ! The desired result now follows by dividing by c.
22 The following observation will be used constantly. 4.13. Exercise. If f : X → [−∞, ∞] is measurable and |f| dµ < ∞ then f is Z finite-valued almost everywhere. 4.14. A Useful Basic Fact. If f : X → [−∞, ∞] is measurable and f dµ is defined Z then : | f dµ| ≤ |f| dµ. Z Z
Proof. Since f ≤ |f|, we have f dµ ≤ |f| dµ. Replacing f by −f this reads : Z Z − f dµ ≤ |f| dµ. These two inequalities are summarized in one in : Z Z
| f dµ| ≤ |f| dµ. Z Z
We come now to one of the most frequently used convergence theorem for integrals : 4.15. Lebesgue’s Dominated Convergence Theorem. Suppose (fn) is a sequence of measurable functions X → [−∞, ∞] (note that this time we don’t require that fn be non-negative) such that f(x) = lim fn(x) n→∞ exists for every x ∈ X. Suppose, moreover, that there is a measurable function g : X → [0, ∞] (the ‘dominating function) such that
|fn(x)| ≤ g(x) for every x ∈ X
and that g dµ < ∞ (4.12) Z Then |f| dµ < ∞, Z
lim |fn − f| dµ = 0 (4.13) n→∞ Z and
lim fn dµ = f dµ (4.14) n→∞ Z Z
(Note : There might be points where fn −f is of the form ∞−∞ and hence unde ned.
The set of such points has measure zero because f dµ (as will be shown) and fn dµ Z Z are nite and we can use the result 4.13 above. Therefore Convention 4.11 is applicable.)
23 Proof. Each |fn| dµ and |f| dµ (note that |f| ≤ g and g dµ < ∞) are nite. Z Z Z Therefore we may and will assume that each fn and f are nite valued; in particular, fn − f will then make sense everywhere. The following shows that equation (4.13) implies equation (4.14) :
| fn dµ − f dµ| = | (fn − f) dµ| ≤ |fn − f| dµ. Z Z Z Z Thus it will su ce to prove equation (4.13). Since there is no monotone sequence in sight we try to see if Fatou’s Lemma does anything for us here. The rst try would be with lim inf |fn − f| but as you can readily this doesn’t give anything useful. The key idea n→∞ is is to notice that equation (4.13) can be read as saying that lim sup |fn − f| dµ ≤ 0 n→∞ Z (for this would be mean that zero is the lim sup of the sequence of non-negative numbers
|fn − f| dµ - and this is the same as saying that the limit of this sequence is zero). To Z get lim inf the obvious thing to do is to switch signs (keeping in mind that lim inf an = n − lim sup(−an)). But then we would get into trouble with the requirement in Fatou’s n Lemma that the functions involved be non-negative. We can get around this problem by working with h−|fn −f|, for some suitable function h which is greater or equal to |fn −f|, and hopefully we can throw it out at the end of the argument. A choice for h is obtained easily by noting that : |fn − f| ≤ |fn| + |f| ≤ 2g. Thus our candidate for h can be 2g. So we apply Fatou’s Lemma to the sequence of functions 2g − |fn − f|. The lim inf of this sequence, as n → ∞, is the same as its limit - lim (2g − |fn − f|) = 2g − 0 = 2g. n→∞ Thus Fatou’s Lemma gives :
(2g) dµ ≤ lim inf (2g − |fn − f|) dµ Z n→∞ Z
= 2g dµ + lim inf(− |fn − f| dµ) Z m→∞ Z
= 2g dµ − lim sup |fn − f| dµ. Z n→∞ Z
(Notice that we have used the simple fact that lim inf(c + an) = c + lim inf an valid for n→∞ n→∞ any sequence of real numbers an, and any xed real number c). Now, because (2g) dµ is Z finite, we can subtract it o to obtain :
0 ≤ − lim sup |fn − f| dµ n→∞ Z which is the same as :
24 lim sup |fn − f| dµ ≤ 0 n→∞ Z and this, as noted earlier, is equivalent to
lim |fn − f| dµ = 0. n→∞ Z
The following result is of some interest but we probably won’t need it. Egorov’s Theorem. If (fn) is a sequence of measurable functions which converges to a measurable function f on a set E of finite measure, then for any > 0 there is a measurable set F ⊂ E with µ(E − F ) < (i.e. F can be as close to E in size as we want), such that (fn) converges uniformly to f on F . Proof. Try to prove it yourself. Or look up Folland’s ‘Real Analysis’ (or Halmos’ ‘Measure Theory’).
Problem Set 2. 1. (This requires knowledge of algebra) If A, B ⊂ X, then the symmetric difference A B is the set of all points of X which belong to either A or B but not to both; i.e A B = (A ∪ B) − (A ∩ B) = (A − B) ∪ (B − A). Let P(X) denote the set of all subsets of X. For this exercise we denote the elements of P(X) by lower case letters a, b, c, ... Also, for this exercise only, we will write a+b for a b, and a·b for a∩b. Note that ∅, X ∈ P(X). (1) Check that under + and ·, P(X) is a commutative ring with X as identity element under multiplication (‘·’). (That is : (i) a+∅ = a, (ii) a+b = b+a, (iii) a+(b+c) = (a+b)+c, (iv) a + a = ∅ - so each element is its own ‘negative’ -(v) a · X = a, (vi)a(bc) = (ab)c, (vii) ab = ba, (viii) a(b + c) = ab + ac.) (2) Verify that {∅, X} forms a subring of P(X) which is isomorphic to the eld Z2 = {0, 1} (∅ 7→ 0, X 7→ 1). Recall that Z2 is the set of integers modulo 2. (3) Verify that P(X) is a vector space over the eld Z2 under the following natural de nition of ‘multiplication by scalars ’ : (i) 0.a = 0 and 1.a = a for every a ∈ P(X). (You will need to use the fact - item (iv) in part (1) - that a + a = ∅ for every a ∈ P(X).) (Because of (1),(2),(3) one says that P(X), or any subset of P(X) which contains ∅ and X and is closed under ‘+’ and ‘·’, is an algebra over Z2.) (4) Suppose B ⊂ P(X) contains ∅ and is closed under ∪ (i.e. A ∪ B ∈ B whenever A, B ∈ B) and complements (i.e. Ac ∈ B for every A ∈ B). Show that then B is closed under ‘+’ and ‘·’ (i.e. a + b ∈ B and ab ∈ B whenever a, b ∈ B). Check that B is a vector subspace of the vector space P(X). (5) Let B be as described in (4). From the theory of vector spaces we know that every vector space has a basis (this uses the Axiom of Choice in case B is in nite). Let A ⊂ B be a basis of B. Thus every element x ∈ B can be written in a unique way as a linear combination (with coe cients 0 and 1) of elements of A. Show that if A is nite and contains n elements then B contains 2n elements.
25 (6) Use the ideas in (5) to show that no σ−algebra can have a countably in nite number of elements. (Hint : Use the fact that if A is an in nite set then the set of all mappings A → {0, 1} is uncountable.) (7) Now let µ be a measure on a σ−algebra B of subsets of X. Let N denote the collection of all sets N ∈ B for which µ(N) = 0. Show that N is an ideal in B : i.e. (i) ∅ ∈ N , (ii) if a, b ∈ N then a + b ∈ N , amd (iii) if a ∈ N and c ∈ B then ac ∈ N .
2. Let (X, B, µ) be a measure space. Let A1, A2, ... ∈ B. Show that
∞ ∞ µ(∪i=1Ai) ≤ µ(Ai). Xi=1
∞ (Hint: Let A = ∪i=1Ai. Verify that 1A(x) ≤ 1Ai (x) for every x ∈ X. Then apply Xi≥1 Proposition 4.9 (which relies on the monotone convergence theorem). This Problem can also be done directly without use of the monotone convergence theorem.) Show that if each Ai has measure zero then so does their union. 3. Let (X, B, µ) be a measure space. Intuitively, a subset of a set of measure zero should also have measure zero. However, a subset of a set of measure zero might not be measurable; i.e. might not be in B. This suggests that we remedy the situation by enlarging B to include all subsets of sets in B which have measure zero. But then we will also have to include all sets formed by the union of these newly measurable sets with the old ones along with complements also. With this in mind, we let N 0 denote the set of all N ⊂ X for which there exists M ∈ B such that N ⊂ M and µ(M) = 0, and we de ne B0 to be the collection of all sets of the form E N where E ∈ B and N ∈ N 0. Verify that : (i) B0 ⊃ B; (ii) F ∈ B0 if and only if there exist A, B ∈ B with A ⊂ F ⊂ B and µ(B − A) = 0. (Hint : If F does lie between two such sets A and B, then check that E = A (F − A) and that F − A ⊂ B − A. Next suppose F = E N where E ∈ B and N ∈ N 0. Then there is a set M ∈ B with N ⊂ M and µ(M) = 0. Put A = E − M and B = E ∪ M, and verify that these two sets satisfy all the necessary requirements.) (iii) If E ∈ B0 then Ec ∈ B0. (Hint : by (ii), there are A, B ∈ B with A ⊂ E ⊂ B with µ(B − A) = 0. Try Ac and Bc for the set Ec. You will nd it useful to verify that Ac − Bc = B − A - think of what it means for a point x to belong to Ac − Bc.) (iv) Show that if N1, N2, ... ∈ B is a countable collection of sets of measure zero then their union ∪iNi also has measure 0. 0 0 (v) Show that B is closed under countable unions. (Hint : Suppose E1, E2, ... ∈ B . Then there are sets A1, B1, A2, B2, ... ∈ B such that Ai ⊂ Ei ⊂ Bi and µ(Bi − Ai) = 0 for each i. Set E = ∪iEi, A = ∪iAi, and B = ∪iBi. To work with B − A check rst that B − A = ∪i(Bi − A) and next that Bi − A ⊂ Bi − Ai; thus (B − A) ⊂ ∪i(Bi − Ai) - now use (iv) to show that B − A has measure zero.) Summary : Parts (i) -(v) show that B0 is a σ−algebra with B ⊂ B0. (vi) Suppose that E ∈ B0 and A, A0, B, B0 ∈ B are such that A ⊂ E ⊂ B and A0 ⊂ E ⊂ B0 with µ(B − A) = 0 and µ(B0 − A0) = 0. Show that µ(A) = µ(B) = µ(A0) = µ(B0).
26 (Hint : That µ(A) = µ(B) and µ(A0) = µ(B0) is easy to see and is not the substantial part of this problem. To show that µ(A) = µ(A0), verify rst that A − (A ∩ A0) ⊂ B0 − A0 and next, as a consequence, that µ(A) = µ(A ∩ A0); switching the roles of A and A0, obtain µ(A0) = µ(A0 ∩ A). Proceed.) (vii) De ne µ0 : B0 → [0, ∞] by µ0(E) = µ(A) where A, B ∈ B, A ⊂ E ⊂ B and µ(B − A) = 0. By (vi) the value µ0(E) does not depend on the speci c choice of A and B (any other choice A0, B0 would lead to the same value for µ0(E).) Check rst that µ0(E) = µ(E) in case E ∈ B. (What would be your choice for A and B be in this case?) 0 0 Next show that µ is a measure. (Hint : Let E1, E2, ... be disjoint sets in B , and write E = ∪iEi. The goal is to show that µ(E) equals µ(Ei). Choose Ai,Bi,as before, for Xi≥1 each Ei. By (v), the sets A = ∪iAi and B = ∪iBi work for E. Now µ(Ei) = µ(Ai) by de nition. What is µ(E)? Are the sets Ai disjoint ? Proceed.) (viii) Show that if E ∈ B0, µ0(E) = 0, and F ⊂ E then F ∈ B0. Terminology : The measure µ0 is called the completion of µ. 4. Let X = {a, b}, and let µ be counting measure on the σ−algebra P(X) of all subsets of X. De ne a sequence of P(X)−measurable functions fn (recall that all functions on X are P(X)−measurable) as follows : for n odd, set fn(a) = 1 and fn(b) = 0; while for n even, set fn(a) = 0 and fn(b) = 1. Apply Fatou’s Lemma to this sequence and check that strict inequality ‘<’ holds. 5. Let X be the set of positive integers {1, 2, 3, ...} with counting measure µ on the σ−algebra of all subsets of X. Let fn = 1{n}; i.e. fn(x) is equal to 1 at x = n and is zero at all other values of x. Note that each fn is measurable (with respect to P(X)) and 0 ≤ fn ≤ 1; i.e. they are uniformly bounded. What is lim fn(x) at any x ∈ X? n→∞
Calculate lim fn dµ and lim fn dµ. Why is the dominated convergence theorem n→∞ Z Z n→∞ not applicable here? 6. Let µ be counting measure on the σ−algebra P(X) of all subsets of the set X = {1, 2, 3, ...}. Recall that this means that if A ⊂ X, then µ(A) is the number of elements in A, taken to be ∞ if A is in nite. Show that for any function f : X → [0, ∞] : j 7→ fj, ∞ the integral f dµ equals the sum fj. (Hint : rst verify this in the case fj is zero Z Xj=1 for all j bigger than some n; next, for arbitrary f, consider the sequence of functions (n) f = f1 ,...,n and use monotone convergence to deal with f dµ, and also the de nition {1 } Z ∞ n of the in nite sum, i.e. that fj = lim fj.) n→∞ Xj=1 Xj=1 7. If aij ≥ 0, for every i, j ∈ {1, 2, 3, ...}, show that :
aij = aij. Xi≥1 Xj≥1 Xj≥1 Xi≥1
(Hint : Use Proposition 4.9 and Problem 5.)
27 8. Let fn be a measurable function X → R for each n = 1, 2, 3, .... Suppose that there
is a number c ∈ R such that |fn| dµ ≤ c for every n. Suppose also that fn(x) → f(x), Z as n → ∞, for every x ∈ X. Show that |f| dµ ≤ c. Z 9. Let (X, B, µ) be a measure space. Verify that if f, g, f 0, g0 are measurable functions on X such that f = f 0 a.e and g = g0 a.e. then : f + g = f 0 + g0 a.e, fg = f 0g0 a.e. and f f 0 = a.e. (provided, of course, that these ratios are de ned almost everywhere). If (f ) g g0 n 0 0 and (fn) are sequences of measurable functions with fn = fn a.e. for every n = 1, 2, ..., then 0 0 0 lim inf fn = lim inf fn, lim sup fn = lim sup fn, and lim fn = lim fn, all a.e. (provided n→∞ n→∞ n→∞ n→∞ n→∞ n→∞ these two limts exist almost everywhere). (Hint : it is convenient to prove the last of these equalities rst and use these in proving the rst two.) 10. Suppose f1, f2, ... is a sequence of real-valued measurable functions and f : X → R is measurable. Suppose lim |fn − f| dµ = 0. Show that, for every > 0, lim µ{|fn − n→∞ Z n→∞ f| > } = 0 (this condition is stated as ‘fn converges to f in measure’.) (i) Consider a sequence of sets A1, A2, .... ⊂ X, and write Em = ∪n≥mAn. Verify that E1 ⊃ E2 ⊃ · · ·. De ne lim sup An = ∩nEn; i.e. lim sup An = ∩n ∪m≥n Am. Verify that n n x ∈ lim sup An if and only if x belongs to An for in nitely many values of n. (Hint : for n each m, there is an integer n ≥ m such that x ∈ An.) (For part (ii) below you will need the simple observation that if a1, a2, ... is a sequence ∞ of real numbers for which ai < ∞, then ai → 0 as n → ∞. Verify this (let Xi=1 Xi≥n n ∞ sn = ai, and s = ai; then, by de nition, sn → s, as n → ∞; what does this say Xi=1 Xi=1 about s − sn -assuming s < ∞ - as n → ∞?) (ii) Let A1, A2, ... be measurable subsets of X, where (X, B, µ) is a measure space. Prove the ( rst) Borel-Cantelli Lemma : If µ(An) < ∞ then µ(lim sup An) = 0. n Xn (Hint : lim sup An ⊂ ∪n≥mAn; so µ(lim sup An) ≤ µ(∪n≥mAn). Recall that µ(∪iBi) ≤ n n µ(Bi) and use the fact mentioned above about convergent sums.) Thus if the sum of Xi the measures of the sets An is nite, then almost every point belongs to at most nitely many An.
28