# Thread: Jordan Content Question

1. ## Jordan Content Question

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.

3. Originally Posted by Jose27
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.
But what if f(y) can be both positive or negative?

4. If it's negative use the exact same argument with the inequalities reversed, and note that continuity is crucial since it lets us pick a ball around y where f is positive (or negative).