Let $A\subseteq \mathbb{R}^n$ be a Jordan region ($A$ is nonempty) and let $f : A\to \mathbb{R}, f(a)\geq 0$ for all $a\in A.$ Suppose $f$ is integrable over $A$ and that $\displaystyle\int_A f(x)dx = 0.$ Prove that there exists a $B\subseteq A$ so that $Vol(B) = \int_B 1 dx = 0$ and $f(x) = 0\,\forall x \in A\backslash B.$
I think I should go about this by contradiction. Suppose that for every $B\subseteq A,$ either $Vol(B)\neq 0$ or $\exists x \in A\backslash B$ so that $f(x) > 0.$ Clearly, for any point $a \in A,B= \{a\}\subseteq A$ and $Vol(B) = 0,$ so we need to show that $\forall B\subseteq A, \exists x \in A\backslash B$ is false. Observe that in order for $Vol(B)$ to be $0$, we must have that $Int(B) = 0$ because otherwise we can find a rectangle $R$ inside $B$ so that $Vol(R) < Vol(B) = 0,$ a contradiction. Observe that since $A\cap (B\backslash A) = \emptyset, 0 = \int_A f(x) dx = \int_B f(x)dx + \int_{A\backslash B} f(x)dx\Rightarrow \int_B f(x)dx = -\int_{A\backslash B} f(x)dx.$ Since $f(x) \geq 0$ on $A,$ we have that $\int_{A\backslash B} f(x)dx \geq \int_{A\backslash B} 0 dx = 0\Rightarrow 0\leq \int_B f(x)dx = -\int_{A\backslash B} f(x)dx \leq 0\Rightarrow \int_B f(x) dx = 0.$ Suppose that $Vol(B) = \int_B 1 dx\neq 0.$ Then by definition, $Vol(B) > 0$. Hence $\int_B 1dx > 0.$
We need to somehow derive a contradiction from this. And we need to show that $\exists B \subseteq A$ so that $\forall x \in A\backslash B, f(x) = 0.$ I think it might be useful to consider when $Int(A) = \emptyset$ and when $Int(A)\neq \emptyset.$
Take $B = \{x \in A : f(x) > 0\}$. We can write $B = \bigcup\limits_{i = 1}^\infty \{x \in A : f(x) > 1/i\}$. Then $Vol(B) = \lim\limits_{i \to \infty} Vol(\{x \in A : f(x) > 1/i\})$. But $Vol(\{x \in A : f(x) > 1/i\}) = 0$, since if the volume were $r > 0$, the integral over $A$ would be at least $r/i > 0$. So $Vol(B) = 0$. And clearly, if $x \in A \setminus B$, $f(x) = 0$.