Prove $f=0$ when $ \int_{A} f(x)dx=0$ for every Jordan Measurable set $A$. Let $f: \mathbb R^n \to \mathbb R$ be a continuous function.
Suppose $ \int_{A} f(x)dx=0$ for every Jordan Measurable set $A$.
Prove $f=0$.

I considered using the Theory of Functions of a Real Variable or attempting to prove by contradiction. Or would the proof similar to this question (Showing that $f = 0 $ a.e. if for any measurable set $E$, $\int_E f = 0$)? I'm unsure which is the correct method and how to make this proof concise without over-complicating it.
Any help/solutions appreciated!
Suppose that $f$ is not zero everywhere, so that we can choose an $x\in\mathbb R^n$ with $f(x)\not=0$. Because $f$ is continuous, we can choose an $\epsilon$ so that $f$ is non-zero in $U_\epsilon(x)$. Than $f$ is either completely positive or completely negative in $U_\epsilon(x)$. Also $U_\epsilon(x)$ is Jordan-measurable with positive measure. But the integral of a positive/negative function over a set of positive measure is positive/negative. Contradiction.
Suppose that $f$ is not zero everywhere, so that we can choose an $x\in\mathbb R^n$ with $f(x)\not=0$. Because $f$ is continuous, we can choose an $\epsilon$ so that $f$ is non-zero in $U_\epsilon(x)$. Than $f$ is either completely positive or completely negative in $U_\epsilon(x)$. Also $U_\epsilon(x)$ is Jordan-measurable with positive measure. But the integral of a positive/negative function over a set of positive measure is positive/negative. Contradiction.