(U) =, if 0 U, 1 U, (U) = X, if 0 U, and 1 U. (U) = E, if 0 U, but 1 U. (U) = X \ E if 0 U, but 1 U. n=1 A n, then A M.

Transcription

1 1. Abstract Integration The main reference for this section is Rudin s Real and Complex Analysis. The purpose of developing an abstract theory of integration is to emphasize the difference between the Riemann and Lebesgue integrals and show that one can define an integral of a real valued function defined on very general sets σ-algebras and measurable sets. Definition 1.1. A collection M of subsets of a set is called a σ-algebra in if it has the following properties: 1) M. 2) if A M then \ A = A c M. 3) If A n M, n = 1, 2, 3... and A = A n, then A M. The pair (, M) is called a measurable space and the elements of M are called measurable sets. xamples: 1) The family of all subsets of is a σ-algebra. Proposition 1.1. If F is any collection of subsets of, then there exists the smallest σ-algebra M F in such that F M F. Proof. Let Ω be the family of all σ-algebras M in which contain F. The example above shows that this family is not empty. Let M F = M Ω M. Obviously, F M F. Now we just need to verify that M F is a σ-algebra. Since M for all M Ω, we deduce that M F. If A j M F, j = 1, 2,... it follows that If A j M, j = 1, 2,..., for all M Ω. Since each M is a σ-algebra, A n M, and thus If A j M F. Property 2 can be verified in the same way and we leave it as an exercise. Now consider the case where is a metric space (or more generally a topological space, if you know what that is). Let F consist of the family of all open subsets of. The family F is certainly not a σ-algebra, but the Proposition 1.1 states that there exists the smallest σ-algebra containing F. So we define Definition 1.2. Let be a metric (topological) space. The smallest σ-algebra that contains all open subsets of is called the Borel σ-algebra and will be denoted by B() measurable functions. Now we introduce the concept of a measurable function. Definition 1.3. Let (, M) be a measurable space and let Y be a metric (topological) space. A function f : Y is measurable if f 1 (U) is measurable for every open set U Y. xample 1.1. If is measurable, χ is measurable. Indeed, if U R is empty then one of the following possibilities hold. χ 1 (U) =, if 0 U, 1 U, χ 1 (U) =, if 0 U, and 1 U χ 1 (U) =, if 0 U, but 1 U χ 1 (U) = \ if 0 U, but 1 U. 1

2 2 In all cases χ 1 (U) is measurable. Notice that we did not really use that U is open. xample: If and Y are metric spaces and f : Y is constinuous, then f is Borel measurable. In fact, since for U open, f 1 (U) is open, it is Borel measurabe. Theorem 1.1. Let (, M) be a measure space, and let Y be a metric (topological)space. Let f : Y be a map. Then 1) If Ω = { Y : f 1 () M} then Ω is a σ-algebra in Y. 2) If f is measurable and Y is a Borel set, f 1 () M. 3) If f is measurable, Z is a metric space and g : Y Z a Borel mapping, and if h = g f, then h is measurable. 4) If Y = [, ] then f is measurable if and only if f 1 ((α, ]) M for all α R. Similarly, f is measurable if and only if f 1 ([, α)) M for all α R. Proof. To prove 1) first observe that f 1 (Y ) =, so Y Ω. Let A Ω and we want to show that Y \ A Ω. By definition of Ω, f 1 (A) M. Then f 1 (Y \ A) = f 1 (Y ) \ f 1 (A) = \ f 1 (A) M. Hence Y \ A Ω. Finally notice that f 1 (A 1 A 2 A 3...) = f 1 (A 1 ) f 1 (A 2 ) f 1 (A 3 )... So if A j Ω, j 1, 2,..., A 1 A 2... Ω. Item 2) is a consequence of item 1). To see that just observe that, since f is measurable, Ω contain all open subsets of Y. Since it is a σ-algebra, Ω must contain the Borel sets. Therefore if B(Y ) then Ω and therefore f 1 () M. To prove 3) notice that for any subset V Z, h 1 (V ) = f 1 (g 1 (V )). Let V be open. Since g is Borel measurable, then g 1 (V ) is a Borel set. By item 2) f 1 (g 1 (V )) M. This proves that h is measurable. Finally we prove 4). First notice that the open intervals of Y = [, ] are of the form [, α), (α, β), and (α, ]. So if f is measurable, then by definition f 1 ([, α) M for all α R. Since every open subset of Y is a countable union of disjoint intervals of these types, we just need to show that if f 1 ([, α) M for all α R, f 1 ((α, ]) and f 1 ((α, β)) M for all α, beta R. Notice that \ f 1 ([, α)) = f 1 ([α, ]) M. Since one can write f 1 ((α, ]) = f 1 ([α + 1/n, ]) M, andf 1 ((α, β)) = f 1 ((α, ]) f 1 ([, β)) M. it follows that f 1 ([, α) M, f 1 ((α, ]), and f 1 ((α, β)) M for all α, beta R. Therefore f is measurable. Item 4) will be very useful throughout this chapter lim sup and lim inf. Before proceeding, we would like to review two important concepts in the convergence of sequences.

3 3 Definition 1.4. Let {a n [, ], n N}. For each k N let b k = sup{a k, a k+1,...} c k = inf{a k, a k+1,...} Then b 1 b 2 b 3... c 1 c 2 c 3,,,, We define lim sup a n = inf{b 1, b 2,...} = lim b k k lim inf a n = sup{c 1, c 2,...} = lim c k. k Proposition 1.2. The lim sup a n and lim inf a n satisfy the following: lim sup a n = sup{β : there exists a subsequence a nj lim inf a n = inf{β : there exists a subsequence a nj of a n such that β = lim j a n j }. of a n such that β = lim j a n j }. Proof. First of all notice that a nj b nj and hence β = lim j a nj lim j b nj lim sup a n. Now we need to show that there exists a subsequence {a nj } of {a n } which converges to lim sup a n. Let A = {a n } be the closure of the set formed by the elements of the sequence {a n }. Then, by definition of closure b k A for k = 1, 2,... Hence, again by definition of closure, lim sup a n A. Therefore there exists a subsequence {a nj } of {a n } which converges to lim sup a n. The same proof works in the case of lim inf. Theorem 1.2. A sequence {a n } converges if and only if lim sup a n = lim inf a n. Proof. Suppose {a n } converges, and L = lim a n. Then the set {β : there exists a subsequence a nj of a n such that β = lim j a n j } = L. Hence lim sup a n = lim inf a n = L. Reciprocally, suppose that lim sup a n = lim inf a n = L. Then it is obvious that {a n } converges and its limit is equal to L. Definition 1.5. Let f n : [, ], n = 1, 2,... Then we define ( lim sup (sup f n )(x) = sup f n (x), n ) f n (x) = lim sup f n (x), and (inf f n)(x) = inf f n (x), n ( ) lim inf f n (x) = lim inf f n(x). The following is an important application of item 4) of Theorem 1.1. Theorem 1.3. If f n : [, ] is measurable for n = 1, 2,... then are measurable. g s = sup f n, n h s = lim sup f n g I = inf n f n, h I = lim inf f n

4 4 Proof. The main step in the proof is to observe that gs 1 ((α, ]) = g 1 I ([, α) = fn 1 ((α, ]), fn 1 ([, α)). The conclusion of the first part of the theorem follows from item 4) of Theorem 1.1. Indeed, if x gs 1 ((α, ]) then g s (x) > α. Therefore f n (x) > α for some n and hence, x f n 1 ((α, ]). If x f n 1 ((α, ]), then f n (x) > α for some n and hence, g s (x) > α. The proof of the measurability of g I is identical. To prove that h s and h I are measurable, one just has to use the definition of lim sup and lim inf and that g s and g I are measurable. Just recall that ( ) h s (x) = inf sup f i (x) k 1 i k and ( ) h I (x) = sup inf f i(x). k 1 i k We have just shown that g s,k = sup i k f i is measurable and that h s = inf k 1 g s,k is also measurable Simple functions. Definition 1.6. A function f : [0, ] is simple if its range consists of finitely many points {α 1,..., α N }. One can write f(x) = α j χ j, j = f 1 (α j ). Notice that = N. In view of xample 1.1, f is measurable if and only if j is measurable for j = 1, 2,..., N. Next we show that any function can be approximated by step functions. Theorem 1.4. Let f : [0, ] be measurable. There exist simple measurable functions s j : [0, ) such that 1) 0 s 1 s 2 s 3... f 2) lim s n(x) = f(x), x 3) If f is bounded then the convergence is uniform. Proof. For n = 1, 2,... and 1 i n2 n, let ([ )) i 1 n,i = f 1 2 n, i 2 n, F n = f 1 ([n, )) and s n = n2 n i 1 2 n χ n,i + nχ Fn. If f(x) <, then f(x) < n for some n and moreover f(x) [ i 1 2 n, i 2 n ), for some i with 1 i n2 n. In that case, (1.1) 0 f(x) s n (x) 1/2 n. If f is bounded, this shows that one can choose n such that (1.1) is satisfied for every x. If f(x) = then s n (x) = n and s n (x) f(x) as n.

5 measures. Definition 1.7. Let (, M) be a measurable space. A positive measure µ on M is a function µ : M [0, ] such that µ( ) = 0 and if A j M, j = 1, 2,..., are such that A j A k =, j k, then µ = µ(a j ), A j Theorem 1.5. Let (, M) be a measurable space and let µ be a positive measure on M. Then 1) If A j M, j = 1, 2,..., n are such that A j A k =, if j k, then µ(a 1 A 2... A n ) = µ(a 1 ) + µ(a 2 ) µ(a n ). 2) If A B then µ(a) µ(b). 3) If A j M, j = 1, 2,..., satisfies A 1 A 2 A 3... Let A = A j, then µ(a) = lim µ(a n ). 4) If A j M, j = 1, 2,... satisfy A 1 A 2 A 3... and µ(a 1 ) <. Let A = A j, then µ(a) = lim µ(a n ). Proof. Item 1) follows directly from the definition by just taking A n+j =, j = 1, 2,... To prove 2) just write B = A (B \ A). Then µ(b) = µ(a) + µ(b \ A). Since µ(b \ A) 0, the result follows. To prove 3) let B 1 = A 1 and B j = A j \ A j 1, j 2. Then B j M and B i B j = if i j. Now notice that A n = n B j and therefore A = B j. Hence µ(a n ) = n µ(b j ), and µ(a) = µ(b j ). Therefore µ(a) = lim µ(a n ). The proof of 4) is an application of 3). Let C j = A 1 \ A j. Then C 1 C 2 C 3... and A 1 \ A = C j. Since A j A 1, j = 2,..., µ(c j ) = µ(a 1 \ A j ) = µ(a 1 ) µ(a j ). Therefore from 3), µ(a 1 ) µ(a) = µ(a 1 \ A) = lim j µ(c j) = µ(a 1 ) lim j µ(a j). This proves 4) Integration. Throughout this section (, M) is a measurable space and µ is a positive measure on M. We consider the integration of positive functions, but keeping in mind that any function f can be written as f(x) = f + (s) f ( x), f + (x) = max{f(x), 0}, f (x) = min{f(x), 0}, where f ± are integrable, and in view of Theorem 1.3 they are also measurable. First we define the integral of a simple function.

6 6 Definition 1.8. Let s : [0, ] be a non-negative measurable simple function s(x) = α j χ j, α j 0. Let be a measurable set. Then s dµ def = α j µ( j ). The convention here is that if α j = 0 and µ( j ) =, α j µ( j ) = 0. The purpose of this convention is that the integral of the function zero, even if taken over a set of infinite measure, is equal to zero. We then define the integral of positive functions: Definition 1.9. Let f : [0, ] be a measurable non-negative function and let be a measurable set. Then f dµ def = sup s dµ, where the sup is taken over all measurable step functions which are less than or equal to f. Notice that if f is a simple function, the two definitions coincide. The following properties of the integral are immediate consequences of the definition: Proposition 1.3. The integral satisfies the following properties: 1) If f g, then f dµ g dµ, 2) If A B and f 0, then A f dµ B f dµ, 3) A cf dµ = c f dµ, for any constant c, A 4) If f)x) = 0 for all x, then f dµ = 0, even if µ() =, 5) If µ() = 0, then f dµ = 0 even if f(x) = 0 for all x, 6) If f 0, then f dµ = fχ dµ. The proof is left as an exercise. First we notice s f Theorem 1.6. Let f : [0, ] be measurable. If f dµ = 0 for some M, then µ{x : f(x) > 0} = 0. Theorem 1.7. Let f : [0, ] be measurable. If f dµ = 0, then µ{x : f(x) > 0} = 0. Proof. Notice that {x : f(x) > 0} = n, n = {x : f(x) > 1/n}, and that n µ( n) f dµ f dµ = 0. n Thus µ( n ) = 0. But µ({x : f(x) > 0}) = lim µ( n ) = 0. If this definition of integral is good, it must be linear. So we need to prove Theorem 1.8. Let f : [0, ], and g : [0, ] be measurable functions. Then (1.2) (f + g) dµ = f dµ + g dµ. This is not so easy to establish. First we prove it for simple functions:

7 7 Proposition 1.4. Let s and t be positive measurable simple functions on. Then ν() = s dµ is a measure (s + t) dµ = s dµ + t dµ. Proof. Let M s = α j χ Aj, t = β j χ Bj. Notice that it is implicit that = N A j and = M B j and that A i A j =, i j, B i B j =, i j. Let = m=1 m with m N =. Then A j = m A j, and µ( A j ) = µ( m A j ). Therefore ν() = m=1 α j µ(a j ) = α j m=1 k=1 µ(a j m ) = This proves the first claim. To prove the second claim, let Then ij = A i B j. m=1 m=1 ij (s + t) dµ = (α i + β j )µ( ij ). On the other hand it is easy to verify that s dµ + ij t dµ = (α i + β j )µ( ij ). ij α j µ(a j m ) = ν( m ). Since = i, ij and ij are pairwise disjoint sets, and s + t is a simple measurable function, it follows from the fisrt claim that (s + t) dµ = s dµ + t dµ. To use this result to prove Theorem 1.8 we need a convergence theorem Convergence Theorems. As in the previous section, (, M) is a measurable space and µ is a positive measure on M. We will prove three convergence theorems. The first and most important one is the monotone convergence theorem. The other two are applications of this theorem. Theorem 1.9. (The monotone convergence theorem) Let f n be a sequence of measurable functions on, and suppose that 1) 0 f 1 (x) f 2 (x) f 3 (x)... for every x 2) lim f n(x) = f(x), x. m=1

8 8 Then f is measurable and lim f n dµ = f dµ. Proof. Since f n dµ f n+1 dµ f dµ, the limit f n dµ = α lim f dµ. Let s be a measurable simple function 0 s f and let c (0, 1). For n N, let n = {x : f n (x) cs(x)} = (f n cs) 1 ([0, ]), n = 1, 2,... Then n is a msasurabel set. Since f n+1 f n, Finally we have = n. To verify this, let x. If f n (x) cs(x) < 0, then f(x) cs(x) 0, but this is absurd. f n (x) cs(x) 0, for some n and thus x n for some n. Therefore we have f n dµ f n dµ c s dµ, n = 1, 2,... n n Since by Proposition 1.4 ν() = is a measure on M, , and n =, it follows from item 3) of Theorem 1.5 that s dµ = lim s dµ. n Therefore we conclude that s dµ lim f n dµ = α c s dµ. for every simple measurable function s f and for every c (0, 1). Therefore α f dµ. This proves the theorem. Hence, To prove Theorem 1.8 we just need to recall Theorem 1.4. There exist sequences of simple functions s n and t n such that Then 0 s 1 s 2... f, lim n(x) = f(x), 0 t 1 t 2... f, lim n(x) = g(x). 0 s 1 + t 1 s 2 + t 2... f + g, lim s n(x) + t n (x) = f(x) + g(x).

9 9 By Proposition 1.4 (s n + t n ) dµ = s n dµ + t n dµ Then (1.2) follows from Theorem 1.9. The following are consequences of Theorem 1.8 and Theorem 1.9 is Corollary 1.1. Let f n : [0, ], n1, 2,... be measurable functions and let f(x) = f n (x), x. Then f dµ = f n dµ. Proof. Just apply Theorem 1.8 and Theorem 1.9 to the sequence F N (x) = N f n(x). Corollary 1.2. Let f : [0, ] be measurable, and let (1.3) ν() = f dµ, M. Then ν is a positive measure on M and g dν = gf dµ. Proof. Let = j, j k = if j k. One can write χ f = χ j f, where each term χ j f of the series is positive. Then by Corollary 1.1 ν() = f dµ = χ f dµ = χ j f dµ = ν( j ). This shows that ν is a measure. To prove the second part, notice that (1.2) holds for g = χ, for any M. Hence (1.2) holds for any measurable simple function. For an arbitrary measurable g, let s n, n N be a monotone sequence of simple functions that converge to g. Since f is positive, s n f is monotone and converges to fg. Hence the result follows from the monotone convergence theorem. Next we prove Fatou s lemma, which really is an application of the monotone convergence theorem Theorem (Fatou s lemma) Let f n : [0, ], n = 1, 2,... be measurable functions. Then (lim inf f n dµ lim inf f n dµ. Proof. Put g k (x) = inf{f k, f k+1,...}. Then g 1 g 2... and, by definition lim k g k (x) = lim inf f n (x). So Theorem 1.9 implies that lim g k dµ = lim g k dµ. k k

10 10 We do not know what lim k g k dµ is equal to. However, since g k f k, g k dµ f k dµ and we can estimate lim g k dµ lim inf g n dµ. k This proves the theorem The last of the three theorems is Theorem (The dominated convergence theorem) Let f n : [0, ], n = 1, 2,... be a sequence of measurable functions. Suppose that lim f n (x) = f(x) and that there exists a function g(x) such that f n (x) g(x), n = 1, 2,... and g dµ <. Then lim f n dµ = f dµ. Proof. Fatou s lemma applied to the sequence f n gives that lim inf f n dµ = f dµ lim inf f n dµ. On the other hand, Fatou s lemma applied to the sequence g n = g f n gives (1.4) lim inf (g f n) dµ lim inf (g f n ) dµ. But lim inf (g f n ) = g lim sup f n and lim inf (g f n) dµ = g dµ lim sup f n dµ. Therefore (1.5) lim sup f n dµ lim sup f n dµ = f dµ. Putting (1.4) and (1.5) together we deduce that This proves the theorem. f dµ lim inf f n dµ lim sup f n dµ 1.8. Integration of Complex Functions. In this section we study the integration of complex valued functions. There is nothing new here, but instead of leaving it all up to the students, I will say something about it. The first thing to notice is Proposition 1.5. A function f : C is measurable if and only if u = Rf : R and v = If : R are measurable. f dµ. Proof. Let R = (a, b) (c, d) be a rectangle with sides parallel to the axes. Then Let f 1 (R) = u 1 ((a, b)) v 1 ((c, d)). π R : C R π R (z) = Rz π I : C R π R (z) = Iz

11 11 Be the projections onto the real axis and the imaginary axis respectively. These projections are continuous functions. Notice that u = π R f, v = π I f Now let U R be an open subset. Then u 1 (U) = f 1 ( π 1 R (U)). Since π 1 R (U) is open, and f is measurable, u 1 (U) M. As mentioned before, one can write u = u + u, u + = max{u, 0}, and u = min{f, 0}. So if f is complex valued and measurable, we define ( ) ( ) f dµ = u + dµ u dµ + i v + dµ v dµ. Proposition 1.6. If f : C is measurable, then f = u + iv, then f = u 2 + v 2 is measurable. Proof. Let : C [0, ) z is the absolute value of z. This is a continuous function. So f 1 (U) = f ( 1 1 (U) ) is measurable, if f is measurable and U is open. It follows from Theorem 1.6 that if f, g : C are measurable and f g dµ = 0, M, then {x : f(x) g(x)} has measure zero. We say that this property holds almost everywhere. Let f : C be a measurable function. We define the relation f g if µ({x : f(x) g(x)}) = 0. It is very easy to see that is an equivalence relation and we denote the equivalence class of a function f by [f]. In view of Theorem 1.6 we define L 1 (, dµ) = {[f] : C : f dµ < } Notice that if f g then f g dµ = 0. We end this section with the following Theorem Let {f n } be a sequence of complex valued measurable functions defined almost everywhere on such that f n dµ <. Then the series f(x) = f n (x) converges almost everywhere on and f L 1 (, dµ), and f dµ = f n dµ.

12 12 Proof. Let S n be the domain of f n. By assumption µ( \ S n ) = 0. Let φ(x) = f n (x), x S = S n. Notice that \ S = ( \ S n). So µ( \ S) = 0. By Corollary 1.1 (1.6) φ dµ <. Let = {x : φ(x) < }. Then (1.6) implies that µ( \) = 0. Hence the series f n(x) converges absolutely on and f(x) φ(x) on. So f L 1 (, dµ). Now if g N = N f n, then g N φ. Since µ( \ ) = 0, the dominated convergence theorem implies that f dµ = f n dµ.