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