Fubini for continuous functions over intervals

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1 Fuini for ontinuous funtions over intervls We first prove the following theorem for ontinuous funtions. Theorem. Let f(x) e ontinuous on ompt intervl =[, [,. Then [, [, [ [ f(x, y)(x, y) = f(x, y)y x = f(x, y) x y. () Proof. As f(x, y) is ontinuous, for every fixe x n fixe y, f(x, y) n f(x, y ) re ontinuous. Furthermore, f(x, y)y n re lso ontinuous. Thus ll the ove integrls re well-efine. f(x, y)x (2) We prove [, [, [ f(x, y)(x, y)= f(x, y)y x (3) n the other equlity is similr. Wlog ssume = =, = =. Fix ny ε >. Sine f is ontinuous on [, [, it is uniformly ontinuous n there is δ > suh tht for ny x y <δ, f(x) f(y) < ε. (4) Now tke n N suh tht /n < δ/ 2 n ivie [, [, into squres of the form ij 8 [i h, (i+)h) [j h, (j + ) h) for i, j Z. Then we hve i, j, sup ij f inf ij f < ε. (5) Now efine g(x, y) 8 sup ij f, h(x, y) 8 inf ij f, (x, y) ij. (6) We hve g(x, y) f(x, y) h(x, y), g(x, y) h(x, y) <ε. (7) Now it n e heke through iret lultion tht g(x, y) = [ g(x, y) y x, h(x, y)= [ h(x, y)y x. (8)

2 Fix x=x. We hve g(x, y) f(x, y) h(x, y) (9) therefore g(x, y)y f(x, y)y h(x, y)y () for every x [,. Consequently This gives [ g(x, y)y x [ f(x, y)y x [ h(x, y)y x. () f(x, y) [ f(x, y)y x <ε (2) n the onlusion follows from the ritrriness of ε. Theorem 2. Let R N, J R M e ompt intervls n let f(x, y) e ontinuous on J. Then J [ [ f(x, y) (x, y)= f(x, y) y x= f(x, y) x y. (3) J J Proof. The proof is similr n is left s exerise. Corollry 3. Let f(x) e ontinuous on 8 [, [ N, N. Then we hve [ [ ( 2 ) N f(x)x = f(x,, x N )x N x 2 x (4) 2 N n the orer of the integrtion n e ritrrily hnge. Proof. Exerise. Exmple 4. Let A 8 [, [,. Clulte Solution. We write A A x e xy x y = xe xy xy. (5) = [ x [ x e xy y e z z x x 2

3 = (e x ) x = e 2. (6) Exerise. Let f C 2. Let = [, [,. Clulte 2 f xy. (7) x y Exerise 2. Clulte the followng. e x+y x y, = [, 2 ; (8) x 2 + y 2 x y, = [,2 ; (9) x sin(x y)xy, = [, π/2 [,. (2) sin(x + y) x y, = [, π/2 2 ; (2) y x 2 xy, = [, [,2; (22) Exerise 3. Let = [, 2. Clulte f(x, y)xy for the following f(x, y): y x 2 y > x 2 ; (23) x y x + y x + y > ; (24) x + y x 2 y 2 x 2. (25) elsewhere. Prolem. (USTC2) Construt B R 2 suh tht the following re stisfie.. for every R, B x = } n B y = } oth onsist of t most one single point. 2. B = R 2. Now efine Prove tht (x, y) B elsewhere. (26) ) Both exist n equl ; [ f(x, y)y x n [ f(x, y)x y (27) ) f is not integrle on [, 2. 3

4 Fuini: The generl se (Optionl) Two-vrile Fuini We still strt from the two-vrile se. Theorem 5. Let f(x, y): R 2 R e integrle on 8 [, [,. Further ssume tht for every x [,, f(x, y) s funtion of y is Riemnn integrle on [,. Then the funtion F(x) 8 is Riemnn integrle on [, n furthermore f(x, y) y (28) [ f(x, y)(x, y)= F(x) x= f(x, y)y x. (29) f furthermore for every y [,, f(x, y) s funtion of x is Riemnn integrle on [,, then we n swith the orer of integrtion: [ f(x, y)y [ x = f(x, y) x y = f(x, y) (x, y). (3) This theorem follows immeitely from the following result whih revels wht is relly going on here. Theorem 6. Let f(x, y): R 2 R n 8 [, [,. Define two funtions Φ(x) n φ(x) s follows: Φ(x) 8 U(f(x, ), [, ), φ(x) 8 L(f(x, ), [, ). (3) Here U(f(x,,[,)) n L(f(x, ),[,) enote the upper n lower integrls for the funtion f(x, y) trete s funtion of y lone (with x fixe). Then U(Φ(x), [, ) U(f(x, y), ); (32) L(φ(x), [, ) L(f(x, y), ). (33) Exerise 4. Prove Theorem 5 using Theorem 6. Remrk 7. From this theorem we see tht two imensionl Riemnn integrility puts strong restrition on the ehvior of the funtion long every slie. Exerise 5. Let f(x, y): R 2 R e integrle on 8 [, [,. For ny ε >, Let S ε 8 x [,P f(x, y) s funtion of y is not Riemnn integrle on [, n U(f,[,) L(f,[,)>ε}. Then µ (S ε )= where µ is the one-imensionl Jorn mesure. n other wors, if f(x, y) is integrle on, then most of its slies re Riemnn integrle. Remrk 8. Note tht in the ove exerise we nnot reple S ε y S 8 x [, P f(x, y) s funtion of y is not Riemnn integrle on [, }. See the prolem elow. 4

5 Prolem 2. Let x Q, y Q p x= r, (p, r) o-prime; y Q p f(x, y) 8 q x Q, y = s. (34), (s, q) o-prime q x Q, y Q Prove tht f(x, y) is Riemnn integrle on [, [,. But for every x [, Q, f(x, y) is not Riemnn integrle on [,. Proof. (of Theorem 6) Rell our results regring uniform prtition. For ny n N, set h 8 n n h 2 8. Let n ij,h 8 [ +(i )h, + ih [ + (j )h 2, + jh 2. Then we know tht lim f ij h h 2 = lim F ij h h 2 = f(x, y)(x, y) (35) n n i,j= i,j= where f ij 8 Now for eh (i, j), we hve inf f(x, y), F ij 8 sup f(x, y). (36) (x,y) ij,h (x,y) ij,h (x, y) ij,h, f ij f(x, y) f ij h 2 L(f(x, y), [ +(j )h 2, + jh 2 ); (37) Sine this is true for ll x [ + (i )h, +ih, we hve f ij h h 2 L(L(f(x, y), [ + (j )h 2, + jh 2 ), [ + (i )h, + ih ). (38) Now summing over i, j we hve f ij h h 2 L(φ(x), [, ). (39) i,j= The other inequlity n e prove similrly. Exerise 6. Let f: R R, < <. Prove tht Generl ses The proof for the generl se is similr. L(f, [, )+L(f,[, ) = L(f, [, ). (4) Theorem 9. (Fuini) Let f(x, y) (x R M, y R N ) e integrle on 8 2 where R M, 2 R N. Assume tht for every x the funtion f(x, y) s funtion of y only is integrle on 2, then [ f(x, y)(x, y)= f(x, y)y x. (4) 2 f furthermore for every y 2 the funtion f(x, y) s funtion of x only is integrle on, then [ 2 f(x, y)y [ x = f(x, y)x y = f(x, y)(x, y). (42) 2 Exerise 7. Let A:=(x, y, z)p x 2 + y 2 + z 2 }. Prove tht [ µ(a) = z 2 z 2 ( z 2 y 2 z 2 y 2 x ) y z. (43) 5