Proving Equality of Functions with Double Integrals

[jj1986]

Homework Statement

Let U \subset \Re^{2} be open, and f,g: U \rightarrow \Re are continuous, and \int^{b}_{a} ( \int^{d}_{c} f(x,y) dy ) dx = \int^{b}_{a} ( \int^{d}_{c} g(x,y) dy ) dx for every rectangle [a,b] x [c,d] in U. Show that f = g.

Homework Equations

The Attempt at a Solution

Can someone tell me if I'm approaching this problem correctly? I know both integrals exist because f,g are assumed to be continuous. If I apply the fundamental theorem of calculus twice to each integral I get F(b,d) - F(a,d) - F(b,c) + F(a,c) = G(b,d) - G(a,d) - G(b,c) + G(a,c) for every (b,d),(b,c),(a,d),(a,c) in U (so does this imply that F(x,y) = G(x,y)?) where d/dx(d/dy F(x,y)) = f(x,y) and d/dx(d/dy G(x,y) = g(x,y). Because F(x,y) = G(x,y) I can conlude that f(x,y) = g(x,y). Is this correct or am I way off?

 

[jj1986]

anyone?

 

[HallsofIvy]

Suppose there exist (x_0,y_0) at which f and g are not equal:f(x_0,y_0)\ne g(x_0,y_0) Take \epsilon= (1/2)|f(x_0,y_0)- g(x_0,y_0)| and show that there exist some neighborhood of (x_0,y_0) in which |f(x,y)- g(x,y)|> \epsilon. Integrate over a rectangle inside that neighborhood.