We have seen that Fubini's Theorem holds for sufficiently nice functions. The purpose of this exercise is to show that in some cases, Fubini's Theorem fails to be true. Consider the function f(x, y) defined on [0, 1] [0, 1] by:

f(x, y) = ? { y ?2 if 0

?x ?2 if 0

0 otherwise. }

(a) Compute ?? ????01??01?f(x,y)dydx?

(b) Compute ??01??01?f(x,y)dxdy?

(c) Show that f(x, y) is not continuous at the point (0, 0). Hence, this does not contradict Fubini's Theorem

1. We have seen that Fubini's Theorem holds for sufficiently nice functions. The purpose of this exercise is to show that in some cases, Fubini's Theorem fails to be true. Consider the function f(r, y) defined on [0, 1] x [0, 1] by: 2 if 0