Measure and Integration: Lecture 2 Notes

MEASURE AND INTEGRATION: LECTURE 2 Proposition 0.1. Let M be a σ­algebra on X, let Y be a topological space, and let f : X → Y . (a) Let Ω be a collection of sets E ⊂ Y such that f −1 (E) ∈ M. Then Ω is a σ­algebra on Y . (b) If f is measurable and E ⊂ Y is Borel, then f −1 (E) ∈ M. (c) If Y = [−∞, ∞] (with open sets along with [−∞, a) and (b, ∞] with a, b ∈ R) and f −1 ((α, ∞]) ∈ M for all α ∈ [−∞, ∞], then f is measurable. Proof. (a) Since f −1 (Y ) = X ∈ M, we have Y ∈ Ω. Also f −1 (E c ) = (f −1 (E))c ∈ M ⇒ E c ∈ M. Lastly, ∞ −1 f −1 (∪∞ (Ei ) ∈ M. i=1 Ei ) = ∪i=1 f (b) Because f is measurable, all open sets are in Ω. Since Ω is a σ­algebra, we have B ⊂ Ω. (c) Recall Ω = {E | f −1 (E) ∈ M}. Given α ∈ R, choose αn < α so that αn → α as n → ∞. By assumption (αn , ∞] ∈ Ω. Then ∞ c [−∞, α) = ∪∞ n=1 [−∞, αn ] = ∪n=1 (αn , ∞] . Thus [−∞, α) ∈ Ω. Then (α, β) = [−∞, β) ∩ (α, ∞] ∈ Ω. Hence, since Ω is a σ­algebra, Ω contains all open sets. It follows that f is measur­ able. � Remark. All of these are equivalent: f −1 ([−∞, α)) ∈ M ⇐⇒ f −1 ([−∞, α]) ∈ M ⇐⇒ f −1 ([α, ∞]) ∈ M ⇐⇒ f −1 ((−∞, α)) & f −1 ({−∞}) ∈ M. Limits. Let {an } be a sequence in R or [−∞, ∞]. Set bk = sup{ak , ak+1 , . . .}, k = 1, 2, . . . . Then inf bk = lim supn→∞ an . As k gets larger, the sup is being taken over a smaller set, so bk ≥ bk+1 ≥ . . .. Thus bk is a (weakly) decreasing sequence and so lim bk = inf bk exists. In other words, Date: September 9, 2003. 1 2 MEASURE AND INTEGRATION: LECTURE 2 lim sup is the largest limit point of the sequence (there exists a sub­ sequence which converges to lim sup). Similarly, we could instead set bk = inf{ak , ak+1 , . . .} and then sup bk = lim inf n→∞ an , that is, lim inf is the smallest limit point of the sequence. Note the relation lim inf an = − lim sup{−an }. Proposition 0.2. A sequence {an } converges if and only if lim inf an = lim sup an = lim an . Limits of functions. Let {fn } : X → R be a sequence of functions. (sup fn )(x) = sup{fn (x)} (lim sup fn )(x) = lim fn (x). n→∞ If, for each x ∈ X, the sequence {fn (x)} converges, then f (x) = lim fn (x) is the pointwise limit. This works for X = [−∞, ∞] (conver­ gence to ±∞ is obvious). Theorem 0.3. If, for each i = 1, 2, . . ., the function fi : X → [−∞, ∞] is measurable, then g = sup fi and h = lim sup fn n→∞ i>1 are both measurable. Proof. NTS g −1 ((α, ∞]) ∈ M for all α. We have g −1 ((α, ∞]) = ∞ fn−1 ((α, ∞]). n=1 If x is a member of the LHS, then g(x) > α. Thus, some fn > α from the definition of sup. If x is a member of the RHS, then fi (x) > α for some i, so g ≥ fi (x) > α. Thus, x ∈ g −1 ((α, ∞]). Since fn−1 ((α, ∞]) ∈ M, and since countable unions are in M, g ∈ M. But then lim sup is measurable as well since by definition � � lim sup fk = inf sup fk . j≥1 k≥j � Corollary 0.4. (a) Pointwise limits of measurable functions are measurable. (b) If f and g are measurable, then max{f, g} and min{f mg} are measurable. MEASURE AND INTEGRATION: LECTURE 2 3 Define “f plus” and “f minus” as follows: f + = max{f, 0}, f − = − min{f, 0}. Then f = f + − f − and |f | = f + + f − . Simple functions. A simple function is a function that takes only finitely many values in R and does not take values ±∞. Let α1 , . . . , αn be the values and Ai = {x ∈ X | s(x) = αi }. Then s= n � αi χAi . i=1 Note: χAi is measurable ⇐⇒ Ai ∈ M. Constant function αi is measurable ⇒ product αi χAi is measurable. Since sums of measurable functions are measurable, s is measurable ⇐⇒ all Ai are measurable. Theorem 0.5. Let f : X → [0, ∞] be measurable. Then there exists a sequence 0 ≤ s1 ≤ s2 ≤ . . . of measurable simple functions such that limn→∞ sn = f . Proof. Partition [0, n] into intervals of length 2−n . Define ϕn : [0, ∞] → [0, ∞) as follows. Let δn = 2−n . For each t, choose kn (t) such that kδn ≤ t < (k + 1)δn . Put � kn (t)δn 0 ≤ t < n; ϕn (t) = n n ≤ t ≤ ∞. Note that each ϕn is Borel measurable, ϕ1 ≤ ϕ2 ≤ · · · ≤ t and limn→∞ ϕn (t) = t. Let sn = ϕn ◦ f . Then for any open set U , −1 −1 ϕn (U ) = f −1 (Borel set) ∈ M. Thus sn is measurable, s−1 n (U ) = f increasing, and its limit is f . � A positive measure is a mapping µ : M → [0, ∞] which is countably additive: ∞ � ∞ µ (∪i=1 Ai ) = µ(Ai ) for disjoint Ai . i=1 Lebesgue Integral. Let (X, M, µ) be a measure space. Let s : X → � [0, ∞) be the simple function s = N i=1 αi χAi , where the Ai are disjoint. For each E ∈ M, define the integral � N � sdµ = αi µ(Ai ∩ E). E i=1 4 MEASURE AND INTEGRATION: LECTURE 2 For more general measurable functions f : X → [0, ∞], then �� � � f dµ = sup sdµ | s simple, 0 ≤ s ≤ f . E E If f : X → [−∞, ∞], then � � � + f dµ = f dµ − f − dµ, E E E provided both terms on the right­hand side are finite.