Measure and Integration: Lecture 18 Notes

MEASURE AND INTEGRATION: LECTURE 18 Fubini’s theorem Notation. Let � and m be positive integers, and n = � + m. Write Rn as the Cartesian product Rn = R� + Rm . We will write points in Rn as z ∈ Rn ; x ∈ R� ; y ∈ Rm ; z = (x, y). If f is a function on Rn and y ∈ Rm is fixed, then fy is the function on R� defined by fy (x) = f (x, y). The function fy is called the section of f determined by y. In particular, if A ⊂ Rn and f = χA , then � 1 if (x, y) ∈ A; fy (x) = 0 if (x, y) �∈ A. In this case, fy is the characteristic function of a subset of R� , and a point x ∈ R� is in this set if and only if (x, y) ∈ A. This set will be denoted by Ay = {x ∈ R� | (x, y) ∈ A}, and is called the section of A determined by y. Now let f be any function on Rn . For a fixed y ∈ Rm , it may be that the function fy on R� is integrable. In this case, let � F (y) = fy (x) dx. R� Of course, fy must be L­measurable, but there are two ways F (y) could exist: (1) fy ≥ 0, in which case 0 ≤ F (y) ≤ ∞, and (2) fy ∈ L1 (R� ), in which case −∞ < F (y) < ∞. We eventually want to prove the equation � � F (y) dy = f (z) dz. Rm Rn To show this, we assume f is L­measurable and integrable, and prove that F (y) exists for a.e. y ∈ Rm and that F is L­measurable and integrable on Rm . 1 2 MEASURE AND INTEGRATION: LECTURE 18 However, it cannot be expected that fy is an L­measurable function for all y ∈ Rm . Indeed, let E ⊂ R� be a non­measurable set, fix � y0 but Ay0 = E. y0 ∈ Rm , and let A = E × {y0 }. Then Ay = ∅ if y = The set A is measurable with λ(A) = 0. But Ay0 is not measurable. Fubini I: Non­negative functions. Theorem 0.1. Assume that f : Rn → [0, ∞] is L­measurable. Then for a.e. y ∈ Rm , the function fy : R� → [0, ∞] is L­measurable, and so � F (y) = fy (x) dx R� exists. Moreover, F is L­measurable on Rm , and � � F (y) dy = f (x) dz. Rm Rn The second equation will be abbreviated � � �� � f (x, y) dx dy = f (x, y) dx dy, Rm R� Rn and the LHS of this equation will often be written � � � � f (x, y) dx dy or dy f (x, y) dx. Rm R� Rm R� Proof. The proof is long, and is broken into 10 steps. (1) Let J be a special rectangle. Then J = J1 × J2 , with J1 and J2 special rectangles in R� and Rm . Then for any y ∈ Rm , � J1 if y ∈ J2 ; Jy = ∅ if y �∈ J2 . Thus, λ(Jy ) = λ(J1 )χJ2 (y), and so � λ(Jy ) dy = λ(J1 )λ(J2 ) Rm = λ(J ). (2) Let G ⊂ Rn be open, and write G = ∪∞ k=1 Jk , with each Jk a disjoint rectangle. Thus, Gy = ∞ k=1 Jk,y MEASURE AND INTEGRATION: LECTURE 18 3 � is a disjoint union, and so λ(Gy ) = λ(Jk,y ). Thus, � ∞ � � λ(Gy ) dy = λ(Jk,y ) dy Rm = k=1 ∞ � Rm λ(Jk ) k=1 = λ(G). (3) Let K ⊂ Rn be compact, and choose G ⊃ K open and bounded. Apply (2) to G \ K: � λ(Gy \ Ky ) dy = λ(G \ K); Rm � � λ(Gy ) dy − λ(Ky ) dy = λ(G) − λ(K). Rm Rm Thus, applying (2) to G gives � λ(Ky ) dy = λ(K). Rm (4) Let K1 ⊂ K2 ⊂ · · · be compact. Let B = ∪k Kj . Then for all y ∈ Rm , ∞ By = Kj,y . j=1 So By is measurable, λ(By ) = limj → ∞λ(Kj,y ) is increasing, so by monotone convergence, � � λ(By ) dy = lim λ(Kj,y ) dy j→∞ Rm Rm = lim λ(Kj ) j→∞ by (3) = λ(B). (5) Let G1 ⊃ G2 ⊃ · · · be open and bounded. Let C = ∩j Gj and let K ⊃ G1 . Applying (4), ∞ K \C = (K \ Gj ), j=1 and so � λ(Ky \ Cy ) dy = λ(K \ C). Rm Since � λ(Ky ) dy = λ(K), Rm 4 MEASURE AND INTEGRATION: LECTURE 18 the result follows for C. (6) This step is the most important. Let A be bounded and mea­ surable. By the approximation theorem, there exist compact sets Kj and bounded open sets Gj such that K1 ⊂ K 2 ⊂ · · · ⊂ A ⊂ · · · ⊂ G 2 ⊂ G 1 and lim λ(Kj ) = λ(A) = lim λ(Gj ). j→∞ j→∞ Let B= ∞ Kj and C = j=1 ∞ � . j=1 Then B ⊂ A ⊂ C and λ(B) = λ(A) = λ(C). Thus, by (4) and (5), � (λ(Cy ) − λ(By )) dy = 0, Rm and so λ(Cy ) − λ(By ) = 0 for a.e. y ∈ Rm . This means that Cy \ By has measure zero in R� for a.e. y ∈ Rm , and for these y, By ⊂ Ay ⊂ Cy ⇒ Ay = By ∪ N , where N is a null set. Hence, Ay is measurable for a.e. y, λ(Ay ) is a measurable function of y, and � � λ(Ay ) dy = λ(By ) dy Rm Rm = λ(B) = λ(A). (7) Observe that if the theorem is valid for each function 0 ≤ f1 ≤ f2 ≤ · · · , then it is valid for f = lim fj . This is due to monotone convergence, (2) and (4). Since fj,y is measurable for a.e. y, fy is L­measurable for a.e. y, and thus for a.e. y ∈ Rm , � F (y) = fy (x) dx R� � = lim fj,y (x) dx j→∞ R� = lim Fj (y). j→∞ MEASURE AND INTEGRATION: LECTURE 18 5 Since this is an increasing limit and Fj is measurable, so is F , and by monotone convergence, � � F (y) dy = lim Fj (y) dy j→∞ Rm Rm � = lim fj (z) dz j→∞ Rn � = f (z) dz. Rn (8) Let fj be the characteristic function of the bounded set A ∩ B(0, j). Then the theorem is valid for the characteristic func­ tion of any measurable set by (6) the observation in (7). (9) Since non­negative measurable simple functions are (finite) lin­ ear 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 non­negative measurable function. � Fubini II: Integrable functions. Theorem 0.2. Assume that f ∈ L1 (Rn ). Then for a.e. y ∈ Rm , the function fy ∈ L1 (R� ), and � F (y) = fy (x) dx R� exists. Moreover, F ∈ L1 (Rm ), and � � F (y) dy = Rm f (z) dz. Rn Proof. Write f = f + − f − and apply Fubini I. Define � � − fy dx, H(y) = fy+ dx, G(y) = R� R� so that � � G dy = Rm f Rn − � dz, � H dy = Rm f + dz. Rn Because the integrals are finite, G(y) < ∞ and H(y) < ∞ a.e. and thus fy ∈ L1 (R� ). Also, F (y) = H(y) − G(y) a.e., and so F ∈ L1 (Rm ) 6 MEASURE AND INTEGRATION: LECTURE 18 and � � � H dy − F dy = Rm �R m �R n f + dz − = = G dy �R m f − dz Rn f dz. Rn � 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 func­ tion, it is L­measurable. We have by integration by parts � ∞ F (y) = y sin xe−xy dx 0 = y2 y . +1 Thus, � 1 1 log 2. 2 0 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 0 0 � 1 = dy F (y) dy = 0 = 1. Doing integration with respect to y first yields � � � 1 sin x 1 − e−x −x −xy . −e y sin xe dy = x x 0 Thus, Fubini’s theorem shows that � � � ∞ 1 sin x 1 − e−x −x dx = log 2. −e x x 2 0