MEASURE AND INTEGRATION: LECTURE 18

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1 MEASURE AND INTEGRATION: LECTURE 18 Fubini s theorem Notation. Let l and m be positive integers, and n = l + m. Write as the Cartesian product = +. We will write points in as z ; x ; y ; z = (x, y). If f is a function on and y is fixed, then f y is the function on defined by f y (x) = f (x, y). The function f y is called the section of f determined by y. In particular, if A and f = χ A, then { 1 if (x, y) A; f y (x) = 0 if (x, y) A. In this case, f y is the characteristic function of a subset of, and a point x is in this set if and only if (x, y) A. This set will be denoted by A y = {x (x, y) A}, and is called the section of A determined by y. Now let f be any function on. For a fixed y, it may be that the function f y on is integrable. In this case, let F (y) = f y (x) dx. Of course, f y must be L measurable, but there are two ways F (y) could exist: (1) f y 0, in which case 0 F (y), and (2) f y L 1 ( ), in which case < F (y) <. We eventually want to prove the equation F (y) dy = f (z) dz. To show this, we assume f is L measurable and integrable, and prove that F (y) exists for a.e. y and that F is L measurable and integrable on. Date: November 4,

2 2 MEASURE AND INTEGRATION: LECTURE 18 However, it cannot be expected that f y is an L measurable function for all y. Indeed, let E be a non measurable set, fix y 0, and let A = E {y 0 }. Then A y = if y = y 0 but A y0 = E. The set A is measurable with λ(a) = 0. But A y0 is not measurable. Fubini I: Non negative functions. Theorem 0.1. Assume that f : [0, ] is L measurable. Then for a.e. y, the function f y : [0, ] is L measurable, and so F (y) = f y (x) dx exists. Moreover, F is L measurable on, and F (y) dy = f (x) dz. The second equation will be abbreviated ( ) f (x, y) dx dy = f (x, y) dx dy, and the LHS of this equation will often be written f (x, y) dx dy or dy f (x, y) dx. Proof. The proof is long, and is broken into 10 steps. (1) Let J be a special rectangle. Then J = J 1 J 2, with J 1 and J 2 special rectangles in and. Then for any y, { J 1 if y J 2 ; J y = if y J 2. Thus, λ(j y ) = λ(j 1 )χ J2 (y), and so λ(j y ) dy = λ(j 1 )λ(j 2 ) = λ(j ). (2) Let G be open, and write G = k=1 J k, with each J k a disjoint rectangle. Thus, G y = k=1 J k,y

3 MEASURE AND INTEGRATION: LECTURE 18 is a disjoint union, and so λ(g y ) = λ(j k,y ). Thus, λ(g y ) dy = λ(j k,y ) dy k=1 = λ(j k ) k=1 = λ(g). (3) Let K be compact, and choose G K open and bounded. Apply (2) to G \ K: λ(g y \ K y ) dy = λ(g \ K); R m λ(g y ) dy λ(k y ) dy = λ(g) λ(k). Thus, applying (2) to G gives λ(k y ) dy = λ(k). (4) Let K 1 K 2 be compact. Let B = k K j. Then for all y, B y = K j,y. j=1 So B y is measurable, λ(b y ) = lim j λ(k j,y ) is increasing, so by monotone convergence, λ(b y ) dy = lim λ(k j,y ) dy = lim λ(k j ) by (3) = λ(b). (5) Let G 1 G 2 be open and bounded. Let C = j G j and let K G 1. Applying (4), K \ C = (K \ G j ), and so Since j=1 λ(k y \ C y ) dy = λ(k \ C). λ(k y ) dy = λ(k), 3

4 4 MEASURE AND INTEGRATION: LECTURE 18 the result follows for C. (6) This step is the most important. Let A be bounded and measurable. By the approximation theorem, there exist compact sets K j and bounded open sets G j such that and K 1 K 2 A G 2 G 1 lim λ(k j ) = λ(a) = lim λ(g j ). Let B = K j and C =. j=1 j=1 Then B A C and λ(b) = λ(a) = λ(c). Thus, by (4) and (5), (λ(c y ) λ(b y )) dy = 0, and so λ(c y ) λ(b y ) = 0 for a.e. y. This means that C y \ B y has measure zero in for a.e. y, and for these y, B y A y C y A y = B y N, where N is a null set. Hence, A y is measurable for a.e. y, λ(a y ) is a measurable function of y, and λ(a y ) dy = λ(b y ) dy = λ(b) = λ(a). (7) Observe that if the theorem is valid for each function 0 f 1 f 2, then it is valid for f = lim f j. This is due to monotone convergence, (2) and (4). Since f j,y is measurable for a.e. y, f y is L measurable for a.e. y, and thus for a.e. y, F (y) = f y (x) dx = lim f j,y (x) dx = lim F j (y).

5 MEASURE AND INTEGRATION: LECTURE 18 5 Since this is an increasing limit and F j is measurable, so is F, and by monotone convergence, F (y) dy = lim F j (y) dy Theorem 0.2. Assume that f L 1 ( ). Then for a.e. y, the function f y L 1 ( ), and F (y) = f y (x) dx exists. Moreover, F L 1 ( ), and F (y) dy = = lim f j (z) dz R n = f (z) dz. (8) Let f j be the characteristic function of the bounded set A B(0, j). Then the theorem is valid for the characteristic function of any measurable set by (6) the observation in (7). (9) Since non negative measurable simple functions are (finite) linear combinations of functions in (8), the theorem follows for them. (10) The theorem follows from the theorem that states that there exists a sequence of simple measurable functions converging to any non negative measurable function. Fubini II: Integrable functions. f (z) dz. Proof. Write f = f + f and apply Fubini I. Define G(y) = f y dx, H(y) = f y + dx, so that G dy = f dz, H dy = f + dz. Because the integrals are finite, G(y) < and H(y) < a.e. and thus f y L 1 ( ). Also, F (y) = H(y) G(y) a.e., and so F L 1 ( )

6 6 MEASURE AND INTEGRATION: LECTURE 18 and F dy = H dy G dy Rm Rm = f + dz f dz Rn = f dz. Example of Fubini s theorem. Let us calculate the integral y sin xe xy dx dy, E where E = (0, ) (0, 1). Since the integrand is a a continuous function, it is L measurable. We have by integration by parts F (y) = y sin xe xy dx 0 y =. y Thus, 1 1 F (y) dy = log Now, since f (x, y) ye xy, we may apply Fubini I to see that f (x, y) dx dy ye xy dx dy E E 1 = dy ye xy dx = dy 0 = 1. Doing integration with respect to y first yields 1 ( ) sin x y sin xe xy 1 e x dy = e x. 0 x x Thus, Fubini s theorem shows that ( ) sin x 1 e x 1 e x dx = log 2. 0 x x 2

Annalee Gomm Math 714: Assignment #2

Annalee Gomm Math 714: Assignment #2 3.32. Verify that if A M, λ(a = 0, and B A, then B M and λ(b = 0. Suppose that A M with λ(a = 0, and let B be any subset of A. By the nonnegativity and monotonicity