Question

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

Answer 1

Is \(D \subset \mathbb{R}^2\) where the measure space is \((\mathbb{R}^2, \mathcal{L}, \mu_{\text{Lebesgue}})\)?

Answer 2

D is any bounded set in a plane, so yep maybe D⊂R2. we still haven't studied Lebesgue integrals though.

Answer 3

\[D \subset \mathbb{R} ^{2}\]

Answer 4

If \(f(p) \neq 0\) for some \(p \in \mathbb{R}^2\) then by continuity, we can find some nbhd \(U\) of \(p\) on which \(f(U) > 0\). From there it is not difficult to show that this means that the integral will be nonzero (since the nbhd will contain some rectangle and the function is non-zero and positive on that rectangle).

Answer 5

I am currently stuck at that. Can I say that
since the nbhd will contain some rectangle, the area will never be zero
to show a contradiction?
I'm sorry. I suck in proving some math problems.

Answer 6

It is true that every open set will contain a rectangle in \(\mathbb{R}^2\). It is even easier than that since you can explicitly find an open ball on which \(f > 0\) and certainly there exists a rectangle of non-zero area in a ball.

Answer 7

Anyway, since there is a rectangle on which \(f > 0\), using the definition of Riemann integration you can show that the integral must be non-zero

Answer 8

thanks :)))