Let $U \subset \mathbb{R}^3$ and $ f :U \rightarrow \mathbb{R}$ be a continuous function.
I want to show that if, for bounded and open sets $V \subset U$,
$$\int_{V}f(\textbf{x}) d\textbf{x} = 0$$
then $f = 0$ on $U$.
I want to prove this by contradiction. I want to suppose supposing that $f >0$ somewhere on $U$ without loss of generality. Taking an open subset of $f(U)$, I can say that the preimage of this set is open, giving me a (hopefully) bounded and open subset of $U$ that gives a nonzero integral.
The problem I am having is finding a set whose preimage is both bounded and open.
Making sure that the set itself is open guarantees that its preimage is open, due to $f$ being continuous. However, showing that the preimage is bounded seems harder. Are there any sets I should look at or theorems about continuous functions I can keep in mind? I'm aware that there are similar questions, but they involve the measurability of sets instead of bounded and open ones.
If $f(x_0)>0$, then there is an ball $B$ surrounding $x_0$ where $f(x)$ is near $f(x_0)$, say $f(x_0)/2<f(x)<2f(x_0)$. This is due to continuity. Then $\int_B f(x)\,dx$ is at least $f(x_0)/2$ times the volume of $B$ (so nonzero).