Suppose there exists $\displaystyle y\in V$ such that $\displaystyle f(y)>0$ then since $\displaystyle f$ is continous and $\displaystyle V$ open there exists an $\displaystyle r>0$ such that for all $\displaystyle x\in B_r(y) \subset V$ we have $\displaystyle f(x)>0$ and since $\displaystyle B_r(y)$ is Jordan measurable $\displaystyle \int_{B_r(y)} f >0$ a contradiction.