How to prove that If for $g\in L_{p}$ $p\in(1,2]$ and for all $\phi \in C_{0}^{\infty}$ we have $\int_{\mathbb{R}} g \phi d\mu =0 $ then $g=0$ a.e where $\mu $ is the Lebesgue measure.
I was thinking about introducing some $\eta_{n}$ which are tending to 1 as $n \to \infty $
Let $q=\frac p {p-1}$. A well known result, proved using Lusin's Theorem says that $C_0^{\infty}$ functions are dense in $L^{q}$. If $h \in L^{q}$ choose $\{h_n\} \subset C_0^{\infty}$ such that $\|h_n-h\|_q \to 0$. It then follows (by Holder's inequality) that $\int gh=\lim \int g h_n$. Hence $\int gh=0$ for all $h \in L^{q}$ and this implies $\int_a^{b}g(x)\, dx=0$ whenever $a<b$. This in turn implies $g=0$ almost everywhere.
Remark: $p\leq 2$ is not necessary.