Measure and Integration: Lecture 13 Notes

MEASURE AND INTEGRATION: LECTURE 13 Egoroff ’s theorem (pointwise convergence is nearly uniform. Theorem 0.1. Suppose µ(X) < ∞. Let fn : X → C be a sequence of measurable functions such that fn → f a.e. For all � > 0, there exists a measurable subset E ⊂ X with µ(X \ E) < � and such that fn → f uniformly on E. Proof. Let S(n, k) = � {x | |fi (x) − fj (x)| < 1/k}. i,j>n Clearly, S(n, k) is measurable, since it is the countable intersection of measurable sets. Note that � � S(n, k) = {· · · } ∩ {· · · } i,j>n+1 i=n,j>n+1 = S(n + 1, k) ∩ {stuff}. Thus, S(n, k) ⊂ S(n + 1, k), that is, for each k, we have an ascending sequence of sets. Claim: for each k, X = ∪∞ n=1 S(n, k). Given k, and x ∈ X, we know fi (x) → f (x). Thus there exists N such that |fi (x) − fj (x)| < 1/k for all i, j > N since any convergent sequence is Cauchy. Thus, x ∈ S(N, k). Obviously ∞ X⊃ S(N, k) ⇒ X = n=1 ∞ S(n, k) n=1 for each k. So we have lim µ(S(n, k)) → µ(X) n→∞ for any k. For each k = 1, 2, . . ., choose nk so that |µ(S(nk , k) − µ(X)| < �/2k . Date: October 16, 2003. 1 2 MEASURE AND INTEGRATION: LECTURE 13 ∞ (Recall µ(X) < ∞.) Let E = ∩∞ k=1 S(nk , k). Then X \ E = ∪k=1 (X \ S(nk , k)). Thus, ∞ � µ(X \ E) ≤ µ(X \ S(nk , k)) = k=1 ∞ � µ(X) − µ(S(nk , k)) k=1 ≤ ∞ � |µ(X) − µ(S(nk , k))| k=1 < ∞ � �/2k = �. k=1 Claim: fn → f uniformly on E; that is, given any δ > 0, there exists N such that |fi (x) − fj (x)| < δ for all i, j > N and every x ∈ E. Choose k such that 1/k < δ. If x ∈ S(n, k), by defi­ nition |Fi (x) − fj (x)| < δ for all i, j > n. In particular, for x ∈ S(nk , k), |fi (x) − fj (x)| < δ for all i, j > nk . But S(nk , k) ⊃ E, so |fi (x) − fj (x)| < δ for all i, j > nk and all x ∈ E. � The theorem is not necessarily true if µ(X) = ∞. For example, if µ is Lebesgue measure on R and fn = χ[n,n+1] . Then fn → 0 pointwise, but for any n �= m, |fn (x) − fm (x)| = 1 on a set of measure 2. Convergence in measure. Here is an example. Let fn : X → R and � |fn | → 0. Then if � > 0, X � intX |fn | ≥ |fn | > �µ({x | fn (x) > �}). {x | fn >�} � So, given � > 0, choose N such that for all n > N , X |fn | < �2 . Then � ≥ µ({x | fn (x) > �} for all n > N . We say that fn → f in measure if given � > 0 there exists N such that, for all n ≥ N , µ({x | |f (x) − fn (x)| > �}) < �. Convergence almost everywhere implies convergence in mea­ sure. Theorem 0.2. If fn → f a.e. and µ(X) < ∞, then fn → f is measure. Proof. Let A = {x | fn (x) → f (x)}. Then µ(X \ A) = 0. Since µ(X) < ∞, µ(A) < ∞ and we may apply Egoroff’s theorem. Thus, there exists a set E such that µ(A \ E) < � and fn → f uniformly on E. Given � > 0, there exists N such that |f (x) − fn (x)| < � for all MEASURE AND INTEGRATION: LECTURE 13 3 n > N and all x ∈ E. So, for n > N , |f (x) − fn (x)| can be greater than � only on (A \ E) ∪ (X \ A). This means that µ({x | |fn (x) − f (x)| > �}) ≤ µ(A \ E) + µ(X \ A) <�+0=� for all n > N . � However, if fn → f in measure, then it is not true � 1 that fn → f a.e. For example, fn : [0, 1] → [0, 1] such that limn→∞ 0 fn (x) dx = 0 but fn (x) → 0 for no x. Convergence in measure implies some subsequence conver­ gence almost everywhere. Theorem 0.3. If fn → f in measure, then fn has a subsequence fnk such that limk→∞ fnk = f a.e. Proof. Let � = 2−k . Given k, there exists N (k) such� that for n ≥ N (k), � µ({x | |f (x) − fn (x)| > 2−k } < 2−k . Let Ek = {x | �fN (k) (x) − f (x)� > c ∞ ∞ c 2−k }. Then µ(Ek ) < 2−k . If x �∈ ∪∞ i=k Ei , then x ∈ (∪i=k Ei ) = ∩i=k Ei . Then � � �fN (i) (x) − f (x)� < 2−i for every i ≥ k ⇒ fN (i) (x) → f (x). Let ∞ ∞ � A= Ei . k=1 i=k So if x �∈ A, then fN (i) (x) → f (x). For any k, µ(A) ≤ µ (∪∞ i=k Ei ) ≤ ∞ � 2−i = 2−k+1 , i=k so µ(A) = 0. � Dominated convergence theorem holds for convergence in mea­ sure. We know that dominated convergence and monotone conver­ gence still hold if we replace convergence with convergence almost ev­ erywhere. Now we show that the theorems are valid if we replace convergence by convergence in measure. Theorem 0.4. Let fn : X → C be a sequence of measurable functions defined a.e. Suppose fn → f in measure and |fn | ≤ |g| a.e. with g ∈ L1 (µ). Then � f dµ = lim fn dµ. X n→∞ 4 MEASURE AND INTEGRATION: LECTURE 13 Note that the pointwise limit of fn may not exist. Proof. Take any subsequence fnk . Clearly, fnk → f in measure. There exists a subsequence f(nk )� such that f(nk )� → f pointwise a.e. Apply dominated convergence to this subsequence. Then � � f dµ = lim f(nk )� dµ. X �→∞ X � Lemma 0.5. Let an be a sequence. If every subsequence has a subse­ quence which converges to α, then limn→∞ an = α.