Defining the Integral

Transcription

1 Defining the Integral In these notes we provide a careful definition of the Lebesgue integral and we prove each of the three main convergence theorems. For the duration of these notes, let (, M, µ) be a measure space. Simple Functions Definition: Simple Function A simple function s: R is any function of the form s = c 1 χ 1 + c 2 χ c n χ n where c 1,..., c n R are distinct and { 1,..., n } is a partition of into nonempty measurable sets. quivalently, a simple function is any measurable function s: R whose range is a finite set. In particular, if s = c 1 χ 1 + c 2 χ c n χ n, then the range of s is {c 1,..., c n }, and i = s 1 (c i ) for each i. Definition: Integral of a Simple Function The Lebesgue integral of a non-negative simple function s = n i=1 c i χ( i ) is defined as follows: n s dµ = c i µ( i ). i=1

2 Defining the Integral 2 Here and in the future, we adopt the convention that 0 = 0 = 0. In particular, if c i = 0 and µ( i ) = for some i, then the corresponding term c i µ( i ) in the definition of the integral should be interpreted as 0. Note that we have required the coefficients c 1,..., c n in an expansion s = c 1 χ 1 + c 2 χ c n χ n of a simple function to all be distinct. The purpose is to avoid any ambiguity: if we require c 1,..., c n to be distinct, then any simple function has only one such expansion, so the integral is well-defined. However, it is often helpful to express a simple function as a sum s = c 1 χ 1 + c 2 χ c n χ n where the coefficients c 1,..., c n are not necessarily distinct. The following lemma lets us evaluate the integral in this case. Lemma 1 Integrating Simple Functions Let s = c 1 χ c n χ n be a non-negative simple function, where c 1,..., c n [0, ) are not necessarily distinct, and { 1,..., n } is a partition of into nonempty measurable sets. Then s dµ = c 1 µ( 1 ) + + c n µ( n ). PROOF We proceed by induction on n. The base case is that n = range(s), in which case c 1,..., c n are all distinct, and the lemma holds by definition. Now suppose that n > range(s). Then there must exist two coefficients that are the same, say c n 1 and c n. Then s = c 1 χ c n 2 χ n 2 + c n 1 χ n 1 n so by our induction hypothesis s dµ = c 1 µ( 1 ) + + c n 2 µ( n 2 ) + c n 1 µ( n 1 n ). But µ( n 1 n ) = µ( n 1 )+µ( n ), so c n 1 µ( n 1 n ) = c n 1 µ( n 1 )+c n µ( n ), and hence s dµ = c 1 µ( 1 ) + + c n 2 µ( n 2 ) + c n 1 µ( n 1 ) + c n µ( n ).

3 Defining the Integral 3 The next lemma is extremely helpful for proving things about simple functions. Lemma 2 Common Partitions Let s, t: R be simple functions. There there exists a partition { 1,..., n } of into nonempty measurable sets such that s = c 1 χ c n χ n and t = d 1 χ d n χ n for some constants c 1,..., c n, d 1,..., d n R. PROOF Suppose that s = c 1 χ c m χ m and t = d 1 χ F1 + + d n χ Fn for some constants c 1,..., c m, d 1,..., d n R and some partitions { 1,..., m } and {F 1,..., F n } of into nonempty measurable sets. Let P be the set of all pairs (i, j) for which i F j. Then { i F j } (i,j) P is a partition of into nonempty measurable sets which satisfies the given criterion. In particular, s = c i χ i F j and t = d j χ i F j. (i,j) P (i,j) P As a first application of these two lemmas, we give a quick proof of the following proposition. Proposition 3 Monotone Property for Simple Functions If s, t: [0, ) are simple functions and s t, then s dµ t dµ. PROOF By the common partitions lemma, there exists a partition { 1,..., n } of into nonempty measurable sets so that s = c 1 χ c n χ n and t = d 1 χ d n χ n for some constants c 1,..., c n, d 1,..., d n [0, ). Then c i d i for each i, so n n s dµ = c i µ( i ) d i µ( i ) = t dµ. i=1 i=1

4 Defining the Integral 4 The Lebesgue Integral We are now ready to define the Lebesgue integral for arbitrary measurable functions. Definition: The Lebesgue Integral (Non-Negative Case) If f : [0, ] is a non-negative measurable function, the Lebesgue integral of f is defined by { } f dµ = sup s dµ s: R is a simple function and 0 s f. Note that if f is itself a simple function, then the definition above agrees with our earlier definition for the integral of f. Specifically, since f is simple and 0 f f, we know that f dµ f dµ. (new) But by Proposition 3, we know that s dµ (old) f dµ (old) for any simple function s: R such that 0 s f, and hence f dµ f dµ (new) (old) which proves that the two integrals are equal. Of course, the definition above only works for non-negative measurable functions, but we can extend to measurable functions f : [, ] by the formula f dµ = f + dµ f dµ, where f + and f are the positive and negative parts of f. We can also integrate on any measurable set by the formula f dµ = f χ dµ.

5 Defining the Integral 5 lementary Properties We now turn to proving some of the elementary properties of the Lebesgue integral. As a general rule, there are three steps to proving anything about the Lebesgue integral: 1. Prove the proposition for non-negative simple functions. 2. Use the definition of the Lebesgue integral to extend to non-negative measurable functions. 3. Use positive an negative parts to extend to all measurable functions. The following proposition illustrates this technique. Proposition 4 Monotone Property If f, g : [, ] are Lebesgue integrable functions and f g, then f dµ g dµ. PROOF We already proved this for non-negative simple functions is Proposition 3. In the case where f and g are non-negative measurable functions, observe that any simple function s satisfying 0 s f also satisfies 0 s g, and hence f dµ = sup 0 s f s dµ sup 0 s g s dµ = g dµ. Finally, suppose that f, g : [, ] are Lebesgue integrable functions and f g. Then f + g + and g f, so f dµ = f + dµ f dµ g + dµ g dµ = g dµ. x Proposition 5 qual Almost verywhere Let f, g : [, ] be measurable functions, where f is Lebesgue integrable, and suppose that f = g almost everywhere. Then g is Lebesgue integrable, and f dµ = g dµ.

6 Defining the Integral 6 PROOF Suppose first that f and g are non-negative simple functions. By the common partitions lemma, there exists a partition { 1,..., n } of into nonempty measurable sets so that f = c 1 χ c n χ n and g = d 1 χ d n χ n for some constants c 1,..., c n, d 1,..., d n [0, ). Then for each i either c i = d i or µ( i ) = 0, and in either case c i µ( i ) = d i µ( i ), so f dµ = n c i µ( i ) = i=1 n d i µ( i ) = i=1 g dµ. Next, suppose that f and g are non-negative measurable functions, and let be a measurable set such that f = g and µ( c ) = 0. If s: R is any simple function for which 0 s f, then sχ is a simple function, 0 sχ g, and sχ = s almost everywhere, so s dµ = sχ dµ g dµ. Since s was arbitrary, it follows that f dµ and a similar argument shows the opposite inequality. Finally, in the general case observe that f + = g + almost everywhere and f = g almost everywhere, so f dµ = f + dµ f dµ = g + dµ g dµ = g dµ. g dµ x Proposition 6 Scalar Multiplication If f : [, ] is a Lebesgue integrable function and k R, then kf is Lebesgue integrable, and kf dµ = k f dµ. PROOF This proposition is trivial in the case where k = 0. For k > 0, it clearly holds for simple functions, so suppose that f is a non-negative measurable function.

7 Defining the Integral 7 If s: R is a simple function and 0 s f, then ks is simple and 0 ks kf, so kf dµ ks dµ = k s dµ. Since s was arbitrary, it follows that kf dµ k f dµ, and a similar argument proves the opposite inequality. If k > 0 and f : [, ], then (kf) + = kf + and (kf) = kf. If f is Lebesgue integrable, it follows that kf is Lebesgue integrable, with kf dµ = kf + dµ kf dµ = k f + dµ k f dµ = k f dµ. Finally, for k < 0 observe that (kf) + = k f and (kf) = k f +, so kf dµ = k f dµ k f + dµ = k f dµ. x The Monotone Convergence Theorem We are now ready to prove our first major convergence theorem. We begin with a special case involving simple functions. Lemma 7 xpanding the Domain Let s: [0, ) be a non-negative simple function, let be a sequence of measurable subsets of, and let = n N n. Then s dµ = lim s dµ. n PROOF Suppose that s = m i=1 c i χ Fi, where c 1,..., c m [0, ) and {F 1,..., F m } is a partition of into nonempty measurable sets. Then for any measurable set G, the product sχ G is a simple function, with sχ G dµ = c 1 µ(f 1 G) + + c m µ(f m G).

8 Defining the Integral 8 In particular s dµ = sχ n dµ = c 1 µ(f 1 n ) + + c m µ(f m n ). n for each n and s dµ = sχ dµ = c 1 µ(f 1 ) + + c m µ(f m ). But µ(f i n ) µ(f i ) for each i as n, so s dµ n s dµ. Theorem 8 Lebesgue s Monotone Convergence Theorem Let 0 f 1 f 2 f 3 be a sequence of measurable functions on. Then f n dµ = lim lim f n dµ. PROOF Let f be the limit of {f n }. Since f n f for each n, we know that f n dµ f dµ for each n, and hence f n dµ f dµ. lim For the opposite inequality, let s be a non-negative simple function for which s f. It suffices to prove that lim f n dµ s dµ. ( ) Let ɛ > 0, and for each n let n = {x f n (x) > (1 ɛ)s}. Note that n N n =, and that each n is measurable, being the preimage of (0, ] under the measurable function f n (1 ɛ)s. But f n dµ f n dµ (1 ɛ) s dµ n n for each n. Taking the limit as n and applying the lemma, we conclude that f n dµ (1 ɛ) s dµ. lim Since ɛ was arbitrary, the desired inequality ( ) follows. x

9 Defining the Integral 9 Corollary 9 Infinite Sums Let {f n } be a sequence of non-negative measurable functions on. Then f n dµ. n N f n dµ = n N Corollary 10 Integral on an Ascending Union Let f : [0, ] be a non-negative measurable function, let be a sequence of measurable subsets of, and let = n N n. Then f dµ = lim f dµ. n Corollary 11 Integral on a Disjoint Union Let f : [0, ] be a non-negative measurable function, let { n } be a sequence of pairwise disjoint measurable subsets of, and let = n N n. Then f dµ = f dµ. n N n Approximation by Simple Functions Among other consequences, the monotone convergence theorem allows us to use sequences of simple functions to evaluate certain Lebesgue integrals. In particular, suppose that f : [0, ] is a measurable function, and suppose that 0 s 1 s 2 s 3 is a sequence of simple functions that converges pointwise to f. Then according to the monotone convergence theorem, f dµ = lim s n dµ.

10 Defining the Integral 10 As we will see, this technique can be used with any non-negative measurable function. Theorem 12 Uniform Approximation by Simple Functions Let f : [0, ) be a bounded function. Then f is measurable if and only if f is a uniform limit of simple functions. PROOF We have already proven that any pointwise limit of measurable functions is measurable, and hence any uniform limit of simple functions is measurable. For the converse, let {s n } be the sequence of functions That is, for each integer k, s n = nf n. s n (x) = k n if and only if k n f(x) < k + 1 n. Since f is bounded, each s n has finite range. Since f is measurable, each s 1 n (k/n) is a measurable set, and hence s n is a simple function. But clearly s n f < 1/n for each n, so s n f uniformly. Theorem 13 Pointwise Approximation by Simple Functions If f : [, ], then f is measurable if and only if f is a pointwise limit of simple functions. PROOF Again, we have already proven that any pointwise limit of meaurable functions is measurable, and hence any pointwise limit of simple functions is measurable. For the converse, let {f n } be the sequence of functions n if x < n, f n (x) = f(x) if f(x) n, n if f(x) > n, and note that each f n is bounded and measurable. By the previous theorem, there exists for each n a simple function s n so that s n f n < 1/n. Then it follows easily that s n f pointwise.

11 Defining the Integral 11 Corollary 14 Combining Measurable Functions If f 1, f 2 : R are measurable and G: R 2 R is continuous, then the function h: R defined by h(x) = G(f 1 (x), f 2 (x)) is measurable. In particular, the sum, difference, product, maximum, or minimum of any two measurable functions is measurable. PROOF Let {s n } be a sequence of simple functions that converges pointwise to f 1, and let {t n } be a sequence of simple functions that converge pointwise to f 2. Then by the common partitions lemma, for each n the function u n (x) = G(s n (x), t n (x)) is simple. Since G is continuous, {u n } converges pointwise to h, and hence h is measurable. We can improve on Theorem 13 in the case of non-negative functions. Theorem 15 Pointwise Approximation by Simple Functions Let f : [0, ] be a measurable function. Then there exists a sequence 0 s 1 s 2 s 3 of simple functions that converges pointwise to f. PROOF For any real number x, let x denote the greatest integer less than or equal to x, and define x n = 2 n 2 n x for each n. That is, x n is the number obtained by truncating the binary expansion of x after n digits. It is clear that x 1 x 2 x 3 and that x n x as n. Now, it is easy to check that f is measurable for any measurable function f, and hence f n is measurable for any measurable function f and any n N. Let

12 Defining the Integral 12 f : [0, ] be a measurable function, and let {s n } be the sequence s n = min ( f n, n ). Then each s n is measurable, with range contained in the finite set { } k 2 n 0 k n2, so each s n is simple. Then {s n } is the desired sequence of simple functions. We now have the tools to prove the following important property of Lebesgue integrals. Proposition 16 Integral of a Sum Let f, g : [0, ] be Lebesgue integrable functions. Then (f + g) dµ = f dµ + g dµ, assuming the sum on the right is defined. PROOF The proposition clearly holds for non-negative simple functions by the common partitions lemma. For non-negative measurable functions, let {s n } and {t n } be non-decreasing sequences of non-negative simple functions that converge pointwise to f and g, respectively. Then {s n + t n } is a non-decreasing sequence of non-negative simple functions that converges pointwise to f + g, so by the monotone convergence theorem (f + g) dµ = lim (s n + t n ) dµ = lim s n dµ + lim t n dµ = f dµ + g dµ. Finally, for the general case, let h = f + g and observe that h + h = f + f + g + g. Rearranging gives h + + f + g = f + + g + + h

13 Defining the Integral 13 and hence h + dµ + It follows that f dµ + h dµ = f dµ + g dµ = g dµ. f + dµ + g + dµ + h dµ. The Bounded Convergence Theorem As a prelude to the bounded convergence theorem, we prove the following theorem named for Russian mathematician Dmitri gorov. gorov published his proof in 1911, but in fact this theorem first appears in a 1910 publication by Italian mathematician Carlo Severini, though this was not noticed until some time later. Theorem 17 gorov s Theorem Suppose that µ() <, and let {f n } be a sequence of measurable functions converging pointwise to a measurable function f. Then for every ɛ > 0 there exists a measurable set with µ( c ) < ɛ such that f n f uniformly on. PROOF For each N and k, let N,k = { x f n (x) f(x) < 1/k for all n N }. Note that each N,k is measurable, and that 1,k 2,k. Since f n f pointwise, we know that N N N,k = for each k, so choose an N k N such that µ( Nk,k) µ() ɛ 2 k. Let = k N N k,k. Then c = k N c N k,k, so µ( c ) µ ( ) N c ɛ k,k 2 = ɛ. k k N k N But f n f < 1/k on for all n N k, and thus f n f uniformly on.

14 Defining the Integral 14 Theorem 18 Lebesgue s Bounded Convergence Theorem Suppose that µ() <, and let {f n } be a uniformly bounded, pointwise convergent sequence of measurable functions on. Then lim f n dµ = lim f n dµ. PROOF Let ɛ > 0, and let M > 0 so that f n M for all n. By gorov s theorem, there exists a measurable set so that µ( c ) < ɛ/(4m) and f n f uniformly on. Let N N so that f n f < ɛ/ ( 2µ() ) on for all n N. Then for all n N, we have f n dµ f dµ f n f dµ = f n f dµ + f n f dµ c ɛ 2µ() µ() + 2Mµ(c ) = ɛ. The Dominated Convergence Theorem To prove the dominated convergence theorem, we first introduce an important family of measures. Proposition 19 Weighted Measures Let (, M, µ) be a measure space, and let g : [0, ] be a measurable function. Then the function ν : M [0, ] defined by ν() = g dµ is a measure on. PROOF Clearly ν( ) = 0. For countable additivity, let { n } be a sequence of pairwise disjoint measurable sets, and let = n=1 n. Then ν() = g χ dµ = g χ n dµ = ν( n ). n N n N g χ n dµ = n N

15 Defining the Integral 15 The measure ν defined above is sometimes called the weighted measure obtained from µ with density function g. The relationship between ν, µ, and g is usually denoted dν = g dµ. The following proposition justifies this notation. Proposition 20 Integration Using Weighted Measures Let (, M, µ) be a measure space, let g : [0, ] be a measurable function, and let dν = g dµ. Then for any measurable function f : [, ], f dν = fg dµ. (That is, the integral on the left exists if and only if the integral on the right exists, in which case their values are the same.) PROOF Suppose first that s = n k=1 a k χ k is a non-negative simple function. Then n n n s dν = a k ν( k ) = a k g dµ = a k χ k g dµ = sg dµ. k k=1 k=1 Next, if f : [0, ] is a non-negative measurable function, let s 1 s 2 be a sequence of simple functions that converge to f pointwise. Then s 1 g s 2 g and s n g fg pointwise as n, so by the monotone convergence theorem f dν = lim s n dν = lim s n g dµ = fg dµ. Finally, if f : [, ] is any measurable function, then f + dν = f + g dµ = (fg) + dµ and similarly f dν = (fg) dµ, so the desired conclusion follows. k=1 Theorem 21 Lebesgue s Dominated Convergence Theorem Let {f n } be a pointwise convergent sequence of measurable functions on, and suppose that there exists a non-negative L 1 function g on such that f n g for all n. Then lim f n dµ = lim f n dµ.

16 Defining the Integral 16 PROOF Note first that g(x) < almost everywhere, so we may assume that g : [0, ). Let dν = g dµ, and note that ν() = g dµ <. Let f be the pointwise limit of the sequence {f n }, let = {x g(x) > 0}, and note that f n = 0 on c for each n, and f = 0 on c as well. Then f n lim f n dµ = lim f n dµ = lim g dν. But f n /g 1 for all n, so by the bounded convergence theorem we can switch the limit and the integral to get f n lim f n dµ = lim g dν = f g dν = f dµ = f dµ.