$\iint_D f = 0$ implies $f(p)=0$ for all $p$.

Question

If $D$ is open, and if $f$ is continuous, bounded, and obeys $f(p) \ge 0$ for all $p \in D$, then $\iint_D f = 0$ implies $f(p)=0$ for all $p$.

The hint in the back of my book says that

There's a neighborhood where $f(p) \ge \delta$.

From the definition in my book, it states that

The double integral $\iint_R f$ exists and has value $v$ if and only if for any $\epsilon > 0$ there is a $\delta > 0$ such that

$|S(N,f,\{p_{ij}\})-v| < \delta$.

There's another theorem in the book that states that

If $f(p) \ge 0$ for all $p\in D$, $\iint_D f \ge 0$.

So far, from the information I've been given, that would mean that I could use the contrapositive to prove that if $f(p) \ne 0$ for all $p$, then $\iint_D f \ne 0$.

Since $f(p) \ne 0$ I could use cases.

Case 1: If $f(p) > 0$, then I could use the aforementioned theorem to show that $\iint_D f > 0$ which is not equal to $0$.

Case 2: If $f(p) < 0$, then $-f(p)=g(p)>0$, so that $\iint_D g(p) > 0$. Then $\iint_D g(p) > 0$ is equivalent to -$\iint_D f(p) > 0$ which is not equal to $0$.

I would like to know if I'm approaching this proof the correct way? Also, would it be difficult to instead try a direct proof using the hint and the definition that I stated above?

Answer 1

Extended hint.

Assume $\exists p\in D$ such that $f(p)>0$. To prove $\iint_D f>0$ follow the steps:

1. Prove that there exists a neighborhood $N$ where $f(x)\ge\delta=\frac{f(p)}{2}$. One can even choose $N$ as a disc around $p$.
2. Estimate $\iint_N f\ge\ ?$.
3. Compare $\iint_D f=\iint_N f+\iint_{D\setminus N}f$ and $\iint_N f$.

Answer 2

Using the hint from your book, assume, toward a contradiction, that $f(p)=r >0$ for some $p\in D$. Then continuity of $f$ at $p$ implies that there is a square $[a,b]\times [c,d]$ containing $p$ such that $f(x)>r/2$ whenever $x\in [a,b]\times [c,d].$ (Drawing a picture and picking a suitable $\epsilon$ for the standard $\epsilon-\delta$ argument will help here).

But then, using an elementary property of the Riemann integral, we can write

$\int_D f\ge \int_{[a,b]\times [c,d]}f>\frac{r}{2}(b-a)(d-c)>0$ and we have our contradiction.