A crash course in Lebesgue measure theory, Math 317, Intro to Analysis II These lecture notes are inspired by the third edition of Royden s Real analysis. The Jordan content is an attempt to extend the obvious sense of the length of an interval to a way of measuring the size of any subset of R (or R n ). It has the upshot that it adds under disjoint union, it scales correctly when you scale up a set by a constant factor, it remains the same when you shift a set. There were faults, however. The Jordan content does not add under countable unions, and there are sets which it cannot measure. The Lebesgue measure manages to fix the first problem, and we will discover that there is no way to fix the second. The following definition should seem familiar to the definition of the Jordan content. Basically the only difference is that instead of using finite covers we use countable covers. recall that for an interval I = (a, b), l(i) = b a is the length of I. Definition 1. The Lebesgue outer measure of a subset X R is given by { } µ (X) := inf l(i k ) where X I k, I k an open interval Notice that is a map from subsets of R to non-negative number together with infinity µ : {subsets of R n } [0, ) { } The idea is as follows, for any set you can get an overestimate on what the measure of the set should be by covering it with countably many rectangles. by taking the infemum of the overestimates you should have a reasonable idea of how big the set is. Here is a good test of the reasonable-ness of this definition. Let s see if it actually does a good job on intervals. Proposition 2. If I = [a, b] is an interval, then µ (I) = l(i) Proof. There is a cover of I by countably many intervals, in fact we only need one interval. Consider any ɛ > 0. Let I 1 = (a ɛ/2, b + ɛ/2) I 2 = I 3 = =. Then l(i k) = l(i 1 ) = (b a) + ɛ. Since µ (I) is the infemum over all countable covers, µ (I) l(i) + ɛ. Since this is true for every ɛ > 0, µ (I) l(i). Now consider any countable cover of [a, b], J := {I 1, I 2,... }. By the Heine-Borel theorem any countable cover of a closed interval has a finite sub cover. Thus, there is a finite collection of intervals J = {I n1, I n2,..., I nk } such that I k I n i. If there is some j, k with I nj I ni then by removing I nj from J we can continue to have a finite cover. Thus we will assume that there are no j, k with I nj I ni. Similarly we can assume that there is no I nj with I I nj =. Since removing some terms from a sum of non-negative sequence can only make it smaller we see that k l(i n ) l(i ni ) n=1 Let I n = (a n, b n ). Since I = [a, b] k I n i a nj < a < b nj for some j. Lets reorder {I nj } so that a n1 < a < b n1. Notice, either a < b < b n1, I I 1 and we can take J = {I n1 } or b n1 I so that there must be a choice for n 2 with a n2 < b n1 < b n2. Continuing in this way, stopping once we have exhausted all of [a, b] we assume that: a n1 < a < b n1, a nj+1 < b nj < b nj+1 for 1 j < k, and a nk < b < b nk 1
2 so that there are positive numbers e 1,... e k such that so a n1 = a e 1 and a nj+1 = b nj e j for 1 j n k l(i ni ) = (b 1 a 1 ) + (b 2 a 2 ) +... (b k a k ) = (b 1 (a e 1 )) + (b 2 (b 1 e 2 )) +... (b k (b k 1 e k )) = a + b k + e 1 + e 2 + + e k > b a = l(i) Thus, for every countable cover {I 1,..., I n,... } of I = [a, b], k l(i n ) l(i nk ) > l(i) n=1 j=1 Thus, µ (I) l(i) and we conclude that µ (I) = l(i) Proposition 3. If A B then µ (A) mu (B) Proof. If µ (B) = then we are done, as µ (A) is true independent of µ (A). If µ (B) < then by the nature of the infemum, for any ɛ > 0 there is a collection J = {J 1, J 2,... } of open intervals such that B n=1 J and µ (B) n=1 l(j n) < µ (B) + ɛ. Since A B, we see that J is also a cover of A, so that µ (A) n=1 l(j n). Putting these inequalities together, µ (A) < µ (B) + ɛ. Since this is true for every ɛ > 0, it follows that µ (A) µ (B). You should use the previous two propositions to prove the following corollaries. Corollary 4. for a < b, µ ((a, b)) = µ ([a, b)) = µ ((a, b]) = µ ([a, b]) = b a. Corollary 5. If p R then µ ({p}) = 0. Proposition 6 (countable subadditivity). If {A 1, A 2,... } is a countable collection of subsets of R then µ ( A i ) µ (A i ) Proof. Let A = A i. If µ (A i ) = then the result is true independent of µ (A). Thus we assume that µ (A i ) < is a convergent series. Consider any ɛ > 0, there is a countable collection of intervals J i = {J1 i, J 2 i,... } such that A i n=1 J n i and µ (A i ) l(jn) i µ (A i ) + ɛ 2 i. Notice that J 1 J 2... is a countable cover of A. Thus, µ (A) l(jn) i (µ (A i ) + ɛ ) 2 i = ɛ + (µ (A i )). n=1 n=1 Since this is true for all ɛ > 0 we conclude that µ (A) (µ (A i )). The following proposition should be familiar from our work with Jordan content. For a set X R and c R (X + c) = {x + c : x C}. The proof is left to you.
3 Proposition 7. For any c R, and X R it follows that µ (X) = µ (X + c) Measurable sets. The outer measure is powerful and defined for every subset of R. It has a shortcoming, though. We want a theory which behaves well under countable disjoint unions: For any collection of disjoint sets, A 1, A 2,..., we want that µ ( A i ) = µ(a i ) Notice that I am leaving off the star. By using the axiom of choice it is possible to construct counterexamples. We will make the following definition: Definition 8 (The Carathéodory criterion for measurability). A set E R n is called measurable if for all X R µ (X) = µ (X E) + µ (X E) If X is measurable then we say that µ(x) = µ (X). Use Proposition 6 to prove the following Proposition 9. For every choice of E and X µ (X) µ (X E) + µ (X E). Thus, a set is measurable if and only if µ (X) µ (X E) + µ (X E) Proposition 10. If µ (E) = 0 then E is measurable. Proposition 11. If E is measurable, then so is is complement, R E. Proposition 12. If A and B are measurable then A B is measurable. Proof. Let X R be any set. We must show that µ (X) µ (X (A B))+µ (X (A B)). Since A is measurable, µ (X) = µ (X A) + µ (X A) On the other hand since B is measurable µ (X A) = µ (X B A) + µ (X A B) = µ (X B A) + µ (X (A B)) So that µ (X) = µ (X A) + µ (X B A) + µ (X (A B)) It remains to simplify the sum of the first two terms. Using the facts that A = (A B) A and that B A = (A B) A we see that µ (X) = µ (X (A B) A) + µ (X (A B) A) + µ (X (A B)) We use the measurability of A to combine the first two terms: as we claimed. µ (X) = µ (X (A B)) + µ (X (A B)) Definition 13. An algebra of sets A is a collection of subsets of R such that If A A then R A A. If A n A for n = 1, 2,..., n, then A 1 A 2 A 3 A n A.
4 Thus, we know at this point that measurable sets form an algebra. We want something better, though. Definition 14. A σ-algebra A is a collection of subsets of R such that If A A then R A A. If A n A for n = 1, 2,..., then the countable union A 1 A 2 A 3 A. Exercise 1. Use DeMorgan s law to prove that if A is a σ-algebra and A n A for n = 1, 2,..., then A 1 A 2 A 3 A Theorem 15. The collection of measurable sets is a σ algebra. Proof. We have already proven closure under complement and under finite unions (to get finite unions use induction and proposition 12.) All that remains is that the union of countable many measurable sets is measurable. Lemma 16. Let X be any set and {E 1,... E n } be a finite collection of disjoint measurable sets. Then n µ (X ( n E k)) = µ (X E k ) Proof of lemma 16. The lemma is proved by inducting on n. When k = 1 the two sides of the equality are identical. We now inductively assume that µ ( X ( n 1 n 1 E )) k = µ (X E k ) Since E n is measurable, and the E k are all disjoint, µ ((X ( n E k)) = µ ((X ( n E k) E n ) + µ ((X ( n E k) E n ) = µ ((X E n ) + µ ( ( X ( n 1 E )) k Now we use the inductive assumption to conclude that as we claimed. µ ((X ( n E k)) = µ ((X E n ) + n 1 µ (X E k ) Let {E 1,... E n,... } be a countable collection of measurable sets and E = E k. We need to show that for all X R, µ (X) µ (X E) + µ (X E) Let F 0 := and F k := E 1 E k for k 1. Let E k = F k F k 1. Notice that E = E k F n = n E k For k j E k and E j are disjoint and measurable. Now we can prove the theorem. Since F n is measurable, µ (X) = µ (X F n ) + µ (X F n ) We can ample Lemma 16 to replace the first term with a sum. n µ (X) = µ (X E k ) + µ (X F n )
Now, F n E so that X F n X E. Monotonicity implies that n µ (X) µ (X E k ) + µ (X E) Passing to the limit we see that µ (X) µ (X E k ) + µ (X E) By countable subadditivity, since E = E = E k, we see that We conclude that E is measurable. µ (X) µ (X E) + µ (X E). We wanted µ to be countable additive, let s prove it. Theorem 17 (countable additivity of Lebesgue measure). If {E k } is a sequence of disjoint measurable sets, then µ( E k) = µ(e k ) Notice the lack of stars. Everything in sight is measurable. Proof. Since we already have subadditivity, we need only show that µ( E k) µ(e k ) Well, for all n, µ( E k) µ( n E k) by monotonicity = n µ(e k) by Lemma 16 By passing to the limit we see that µ( E k) µ(e k ) But are there any non-measure-zero measurable sets? Lemma 18. For any a R, (a, ) is measurable. Corollary 19. All intervals are measurable, Any set that can be made by starting with intervals, taking complements and countable unions is measurable. Definition 20. The sets in the corollary are called Borel sets. Proof of Lemma 18. Let X be any set. Let X = X (a, ) and X = X (a, ) = X [a, ). We must show that µ (X) µ (X ) + µ (X ). If µ (X) = then we re done. So we can assume that µ (X) <. Fix ɛ > 0. Let {I k } be a cover of X by open intervals with the property that µ (X) l(i k ) µ (X) + ɛ. 5
6 Let I k = I k (a, ) and I k = I k [a, ). Then l(i k ) + l(i k ) = l(i k). All of these sets are intervals so that µ (I k ) = l(i k ) and µ (I k ) = l(i k ) Since X I k and X I k subadditivity impiles that µ (X ) µ (I k ) = l(i k ) and µ (X ) l(i k ) Adding these, µ (X ) + µ (X ) l(i k ) + l(i k ) = l(i k ) So that by the choice of {I k }, µ (X ) + µ (X ) µ (X) + ɛ. Since this is true for all ɛ > 0, µ (X ) + µ (X ) µ (X). As we needed. But do we need this restriction? Maybe the outer measure is already countably additive? This section justifies the creation of the condition of measurability. Proposition 21. There is no map M : {subsets of R} [0, ) which satisfies the conditions that For all X R and c R M(X) = M(X + c), For all sets X and Y, if X Y then M(X) M(Y ), For all intervals, I, M(I) = l(i), If X 1, X 2,... are disjoint then M(X 1 X 2... ) = M(X 1 ) + M(X 2 ) +.... Proof. This will mark the only use of the axiom of choice in this class: Axiom 1 (The axiom of choice). If X is any set and P(X) { } = {nonempty subsets of X} is the power set of X then there is a map C : P(X) { } X with the property that for all U, C(U) U. The map C is called a choice function for X. Think about the quotient of [0, 1] by the equivalence relation a b if a b Q. Each equivalence class U [0, 1]/Q is a subset of [0, 1]. Let C : P([0, 1]) { } [0, 1] be a choice function. Let X 0 = C[U]. For every q Q [ 1, 1] let X q = X 0 + q. Claim 1: For q r Q X q X r =. Suppose that y X q X r Then y q and y r are each in X 0 so that y q = C(u 1 ) and y r = C(u 2 ) for equivalence classes u 1 and u 2. By the definition of the choice function y q u 1 and y r u 2. But (y q) (y r) = r q Q. The equivalence classes u 1 and u 2 must the the same! But then q = r. This proves the claim. Claim 2: [0, 1] X q [ 1, 2]. Let x [0, 1]. Let u be the equivalence class containing x. Then C(u) X 0 and x are in the same equivalence class, so that q = x C(u) Q, since x and C(u) are each in [0, 1], 1 x C(u) 1. But then x = c(u) + q X q. This proves that [0, 1] X q. To see the remaining inclusion, notice that X 0 [0, 1] by the nature of the choice function. So that for all p [ 1, 1] X p [p, p + 1] [ 1, 2]. This proves the claim. By assumption 2 and claim 2, By assumption 3 and claim 1, M([0, 1]) M( X q ) M([ 1, 2]). 1 M(X 0 + q) 3.
But by the first assumption M(X 0 + q) = M(X 0 ). 1 M(X 0 ) 3. We have an infinite sum of positive terms which are either always zero, or do not converge to zero. M(X 0 ) = 0 or M(X 0 ) = Neither of these option are between 1 and 3. This is a contradiction. The set X 0 must not be measurable. Homework: Which I will not collect. Talk to me if you would like to about these problems. (1) Complete the proofs which I left out (2) Remember the exam in which I gave a definition of measure zero? Prove that such a set has zero Lebesgue measure. (3) What is the Lebesgue measure of the set of irrationals? (4) Suppose that X 1 X 2 X 3... is a sequence of measurable sets. Prove that µ(x 1 X 2... ) = lim µ(x k ). (Hint: Think about Y k = X k X k 1 ) 7