Prove that $f$ is the identically zero function. Let $f : [0,1] \times [0,1] \longrightarrow [0,\infty)$ be a continuous function. Suppose that $$\int_{0}^{1} \left ( \int_{0}^{1} f(x,y)\ dy \right ) dx = 0.$$ Prove that $f$ is the identically zero function.
My attempt $:$ $f$ is a non-negative measurable function on $[0,1] \times [0,1].$ So by Tonelli's theorem we have $$\iint_{[0,1] \times [0,1]} f(x,y)\ dx\ dy = \int_{0}^{1} \left ( \int_{0}^{1} f(x,y)\ dy \right ) dx = 0.$$ So if there exists $\textbf {x}_0= (x_0,y_0) \in [0,1] \times [0,1]$ such that $f(\textbf {x}_0) > 0,$ by continuity of $f$ at $\textbf {x}_0$ there exists an open ball $B( \textbf {x}_0, \delta)$ of some radius $\delta > 0$ surrounding $\textbf {x}_0$ with $B(\textbf {x}_0, \delta) \subseteq [0,1] \times [0,1]$ such that for any $\textbf {x} = (x,y) \in B(\textbf {x}_0, \delta)$ we have $$f(\textbf {x}) \gt \frac {f(\textbf{x}_0)} {2 \pi {\delta}^2}.$$ Since $f$ is a non-negative function so we must have $$\frac {f(\textbf {x}_0 )} {2} \lt \iint_{B(\textbf {x}_0, \delta)} f(x,y)\ dx\ dy \leq \iint_{[0,1] \times [0,1]} f(x,y)\ dx\ dy = 0$$ a contradiction. This shows that $f \equiv 0$ on $[0,1] \times [0,1].$
Can anybody plaese check my proof to ascertain whether it holds good or not? Thanks in advance.
 A: Two problems with your proof :

*

*If $\frac{f(x_0)}{2} < f(x)$ on $B(x_0,\delta)$, then this does not imply that $\iint_{B(x_0,\delta)} f(x,y)dxdy > \frac{f(x_0)}{2}$ because the measure of $B(x_0,\delta)$ could be much smaller than $1$. For example : the function $f(x) =2$ is bigger than $1$, but $\int_0^{0.01} f(x)dx = 0.02 < 1$.

The true fact is that $\iint_{B(x_0,\delta)} f(x,y)dxdy > \color{blue}{m(B(x_0,\delta))}\frac{f(x_0)}{2}$ where $m(\cdot)$ denotes the Lebesgue measure of the given set. Since this set definitely has Lebesgue measure non-zero, we know that the above statement is still one of significance : when this is continued with the statements you present, things look good :
$$
0 < \frac{f(x_0)}{2} < \iint_{B(x_0,\delta)} f(x,y)dxdy < \iint_{[0,1]^2} f(x,y)dxdy  = 0
$$
Contradiction.


*

*There is a second problem : $[0,1]^2$ is not open, so if the above $x_0$ is on the boundary of $[0,1]^2$ (say $x_0 = (0,0)$ for example) then we have the problem that $B(x_0,\delta)$ lies outside $[0,1]^2$ for any $\delta$, and the integral over such a ball is undefined because $f$ is undefined there.

To work around this is simple : simply take $S=B(x_0,\delta) \cap [0,1]^2$. All you need to do is replace the ball by $S$. But you need to make sure $S$ has non-zero measure : I am sure you can do this, take it as an exercise (hint : try to find a ball contained in $S$). Of course, then we just have :
$$
0 < \frac{f(x_0)}{2} < \iint_{B(x_0,\delta) \cap [0,1]^2} f(x,y)dxdy < \iint_{[0,1]^2} f(x,y)dxdy  = 0
$$

Following OP edits, the first problem was solved by taking $f(x_0)\over 2 \pi \delta^2$ instead of $\frac{f(x_0)}{2}$, which then multiplies with $\pi \delta^2 = m(B(x_0,\delta))$ to make OP's first issue null.
