How to prove: double integral equals zero

Question

Let $f$ be continuous, $f\ge0$, on the rectangle $R$.

If $$\iint_R f(x,y) dA=0,$$ prove that $f=0$ on $R$.

Answer 1

Assume $|f(x_{0},y_{0})|>0$, say, $f(x_{0},y_{0})>0$, then there is a box $D\subseteq R$ such that, $f(x,y)>f(x_{0},y_{0})/2$ for $(x,y)\in D$. Then $0=\displaystyle\int_{R}f(x,y)dA\geq\int_{D}f(x,y)dA\geq\int_{D}(f(x_{0},y_{0})/2)dA=|D|f(x_{0},y_{0})/2>0$, a contradiction.

Answer 2

Nerdy topologist's overkill answer. This is really the same answer, but accents some features of the question which might let you apply it to other cases.

Since $f$ is continuous, the set $A_{\epsilon}=f^{-1}\left(\left(\epsilon,+\infty\right)\right)$ is open for any $\epsilon>0$ Non-empty open sets have non-zero measure, so $$\epsilon\mu\left(A_{\epsilon}\right)=\iint_{A_{\epsilon}} f(x,y)\,dA\leq \iint_{R}f(x,y)\,dA.$$

So if, for some $\epsilon>0,$ you have $A_{\epsilon}\neq\emptyset,$ then $\iint_R f(x,y)\,dA>0.$

On the other hand, if every $A_{\epsilon}$ is empty, then $A_{0}=\bigcup_{\epsilon>0} A_{\epsilon}=\emptyset,$ so $f(x,y)=0$ for all $x,y\in R.$

---

More generally, then, for any measure $\mu$ on a topology $X$, and any non-negative function $f:X\to\mathbb R$, such that $\int_X f(x)\,d\mu=0$ we have that $\{x\mid f(x)>0\}$ is an open subset of $X$ of measure $0.$