If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e.

Question

I can show this for $f\in L^1(\mathbb{R})$ by creating a sequence of continuous functions of compact support $\varphi_n\to \frac{f}{|f|+1}$ so $\int\frac{f^2}{|f|+1}=0$ by Dominated Convergence so we must have $f=0$ in $L^1$. However, I'm unsure how to extend this for $f\in L^1_{loc}$, the space of locally integrable functions. Any help with this would be greatly appreciated. Thanks!

Answer 1

Consider $f \phi$, where $\phi \geq 0$ is smooth and compactly supported in a ball $B(0,r)$. Take $\phi=1$ inside some smaller ball $B(0,r/2)$. Then $f\phi$ is in $L^1$ and satisfies the same hypotheses as $f$, so $f\phi=0$ a.e., and hence $f=0$ a.e. in $B(0,r/2)$. Now take a sequence of $r_n = n$ approaching $\infty$ to find that $f=0$ a.e. in $B(0,n/2)$ for all $n$, i.e. $f=0$ a.e.