Proving f(p) = 0 for all p when double integral over D of f is equal to 0

[cookiesyum]

Homework Statement

If D is open, and if f is continuous, bounded, and obeys f(p)>or=0 for all p in D, then the double integral over D of f is equal to 0 implies f(p) = 0 for all p.

Homework Equations

Hint: There is a neighborhood where f(p)>or=d.

The Attempt at a Solution

The integral is equal to the sum of f(p)*A(Rij) for some Rij partition of f(x). Since Rij > 0, for the whole thing to be 0, f(p) has to be 0.

Not really sure how to start my proof.

f is continuous and m<or= f(p) <or= M so mA(D) <or= double integral <or= MA(D)

since the double integral is = 0,

mA(D) <or= 0 <or= MA(D) which means m=M=0 and f(p) = 0 for all p.

Is this correct?

 

[clamtrox]

To my eye, an indirect proof looks tempting. D is open so it's measure is not zero. Since f is continuous, for it to be nonzero at a point p means also that there exists a neighbourhood where f is nonzero which is a subset of D (since D is open), and since the neighbourhoods are open sets, they have nonzero measure and therefore the integral can't be zero.