Measure and Integration: Lecture 11 Notes

MEASURE AND INTEGRATION: LECTURE 11 Principles of measure theory. (1) Every measurable set is nearly a Borel set: A = Fσ ∪N = Gδ ∪N (N is a null set: a set of measure zero). (2) Every measurable set is nearly an open set: λ(U ) < λ(E) + �. (3) Every measurable function is nearly continuous (Lusin’s theo­ rem). (4) Every convergent sequence of measurable functions is nearly uniformly convergent (Egoroff’s theorem). Lusin’s theorem. Theorem 0.1. Let f : X → R (or C) be a measurable function on a locally compact Hausdorff space X. Let A ⊂ X, µ(A) < ∞, and f (x) = 0 if x �∈ A. Given � > 0, there exists g ∈ Cc (X) such that � g(x)}) < �. µ({x | f (x) = Proof. Assume that 0 ≤ f ≤ 1 and A compact. Recall that if f : X → [0, ∞] is measurable, then there exist simple measurable functions si such that (a) 0 ≤ s1 ≤ · · · ≤ f and (b) si → f as i → ∞. (Proof: Let δ= 2−n and for t ≥ 0 define kn (t) such that kδn ≤ t < (k + 1)δn , and define � kn (t)δn 0 ≤ t < n; ϕn (t) = n n ≤ t ≤ ∞. Then ϕn (t) ≤ t and ϕn (t) → t as n → ∞. The function ϕ ◦ f is simple and ϕn ◦ f → f as n → ∞.) Next, let t1 = s1 , . . . , tn = sn − sn−1 . Claim: sn − sn−1 takes only values 0 and 2−n . Let Tn ⊂ A where Tn = {x | tn = 2−n }. Then ∞ � f (x) = s1 + (s2 − s1 ) + · · · = tn . n=1 Since X is locally compact, we may choose A ⊂ V , V open, and V compact. There exists Kn ⊂ Tn ⊂ Vn ⊂ V , Kn compact, Vn open, such there exist function that µ(Vn \ Kn ) < 2−n �. By Urysohn’s lemma, �∞ hn such that Kn � hn � Vn . Define g(x) = n=1 2−n hn (x). Since Date: October 9, 2003. 1 2 MEASURE AND INTEGRATION: LECTURE 11 this series converges uniformly on X, g is continuous and supp f ⊂ V . f (x) except on But 2−n hn (X) = tn except on Vn \ Kn . Thus, g(x) = � ∞ n ∪∞ V \ K , and µ of this set is less than or equal to n n=1 n n=1 �/2 = �. Thus, we have proved the case where 0 ≤ f ≤ 1 and A is compact. Thus, it is true when f is a bounded measurable function and A is compact. Now look at f f= . (sup f ) + 1 If A is not compact and µ(A) < ∞, then there exists K ⊂ A such that µ(A \ K) < � for any �. Let g = 0 on A \ K. For f not bounded, let Bn = {x | |f (x)| > n}. Then ∩n Bn = ∅, so µ(Bn ) → 0. Then f agrees with (1 − χBn )f except on Bn , and we can let g = 0 on Bn . � Corollary 0.2. Let f : X → R, A ⊂ X, µ(A) < ∞, f (x) = 0 if x �∈ A, and |f (x)| < M for some M < ∞. Then there exists a sequence gn ∈ Cc (X) such that |gn (x)| < M and f (x) = limn→∞ gn (x) almost everywhere. Proof. By the theorem, for n > 0, there exists gn ∈ Cc (X) such that −n . Let En = {x | f �= gn }. Then sup |gn | ≤ sup |f | < M�and µ(En ) < 2� ∞ ∞ En is measurable and n=1 µ(En ) = n=1 2−n < ∞. Claim: � almost all x ∈ X lie in at most finitely many Ek . Proof: Let g(x) = ∞ k=1 χEk (x). Then x is in infinitely many Ek ⇐⇒ g(x) = ∞. We have � ∞ � � g(x) dµ = χEk (x) dµ < ∞ X k=1 X by monotone convergence. Thus, g(x) < ∞ almost everywhere, which � implies that lim gn = f a.e. Vitali­Caratheodory theorem. Theorem 0.3. Let f : X → R and f ∈ L1 (µ). Given � > 0, there exists functions u, v : X → R such that�u ≤ f ≤ v, u is upper semicontinuous, v is lower semicontinuous, and (v − u)dµ < �. Proof. Assume f ≥ 0. Choose 0 ≤ s1 ≤ · · · ≤ f simple and measurable sn − sn−1 . Then tn has only finitely such that lim sn = f . �Let tn = � ∞ many values and f = ∞ t = n k=1 k=1 ck χEk . We have � ∞ � ci µ(Ei ) < ∞. f dµ = X i=1 MEASURE AND INTEGRATION: LECTURE 11 3 Choose Ki ⊂ Ei ⊂ Vi , Ki compact, Vi open, such that ci µ(Vi \ Ki ) < 2−i−1 �. Let ∞ N � � v= ci χVi , u = ci χKi , i=1 i=1 � where N is chosen so that ∞ c µ(E ) i < �/2. Then v is lower semi­ N +1 i continuous, u is upper semicontinuous, and u ≤ f ≤ v. Also, N � v−u= ≤ ≤ i=1 ∞ � ci (χVi − χKi ) + ci (χVi − χKi ) + i=1 ∞ � ∞ � ci χVi N +1 ∞ � ci (χVi − χVi + χKi ) i=N +1 ci χKi , i=N +1 and so � (v − u)dµ ≤ X ∞ � ci µ(Vi \ Ki ) + i=1 ∞ � ci χEi N +1 < �/2 + �/2 = �. �