Is $f(x,y)=0~ \mu$-a.e.?

Question

Let $f:\mathbb{R_{2}^{+}}\to \mathbb{R}$ be a measurable function such that \begin{equation} \int\limits_{0}^{x}\int\limits_{0}^{y}f(u,v)\,\mu(du,dv)=0\qquad\forall \,x,y>0. \end{equation}

Here, $\mu$ is the two-dimensional Lebesgue measure. Is it then true that $f= 0$ [$\mu$-a.e]? If not, is there any counterexample?

Any hints to approach the problem would be appreciated.

Answer 1

Write $f=f_1-f_2$ with $f_1\ge0$ and $f_2\ge0$. Define measures $\mu_i$ by $$ \mu_i(E)=\int_E f_i \,d\mu. $$ From the assumption, it immediately follows that $\mu_1(R)=\mu_2(R)$ for any rectangle $R=(a,b)\times(c,d)$. A uniqueness theorem on the extension of measures implies that $\mu_1=\mu_2$ on Borel sets. Since they are both absolutely continuous wrt Lebesgue measure, they agree on the Lebesgue sigma-algebra. It is now easy to show that $f_1=f_2$ a.e., and therefore $f=0$ a.e.