Integral of product of g with a $C_{0}$ functions is always 0. Does g have to be Zero?

Question

So g is in $L^{1}(U)$ (not necessary in $L^{2}$!) for some open $U$ and i know, that for all continouus functions with compact support on $U$: $f$ the integral $\int_{U}fg$ is zero. Can i conclude, that $g$ is zero $a.e$? If so, any references, if not any counterexample?

Answer 1

If $U\subset \mathbb{R}^n$, then the claim is true. For any open set of U, say V, there exists a sequence of continuous functions $\{h_n\}_{n=1}^{\infty}$ such that all $h_n$'s are compactly supported and $0\leq h_n \leq h_{n+1} \to \chi_{V}$(indicator function of $V$).Indeed let $\{K_n\}_{n=1}^{\infty}$ be a sequence of increasing compact sets with $\bigcup_{n=1}^{\infty} K_n = V$. Now set $h_n(x) = d(x, V^c)/(d(x, K_n) + d(x, V^c)).$ Then by the dominated convergence theorem this leads to $$\lim_{n\to \infty} \int_U h_n g = \int_V g,$$ and thus $\int_V g = 0$. From this we can show that the integral of $g$ on any Borel set of $U$ is zero. Therefore $g =0$ a.e.