# Equality of functions

1. Apr 2, 2009

### jj1986

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 2, 2009

anyone?

3. Apr 3, 2009

### HallsofIvy

Staff Emeritus
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.