I have a function $f\in L^1(\mathbb{R},\mathbb{R})$ such that $$\int_{[0,x]}f(t)\:\mathrm{d}t=0,\qquad\forall x>0\qquad\text{and}\qquad \int_{[x,0]}f(t)\:\mathrm{d}t=0,\qquad\forall x<0.$$
I want to prove that if $g$ is a continuous function with compact support, then $f\star g=0$.
For that, I know that it exists $M>0$ such that $\text{supp}(g)\subset[-M,M]$. Then, $$|f\star g(x)|=\left|\int_{-M}^M f(x-t)g(t)\:\mathrm{d}t\right|\leq \int_{-M}^M |f(x-t)g(t)|\:\mathrm{d}t\leq \left(\sup_{-M\leq t\leq M}|g(t)|\right)\int_{-M}^M |f(x-t)|\:\mathrm{d}t.$$ That is, if is enough to show that for all $x\in\mathbb{R}$, $\int_{-M}^M |f(x-t)|\:\mathrm{d}t=0$. That seems really plausible to me. However it can't seem to prove it.
I also want to use that $f\star g=0$ to show that $f=0$ almost everywhere. For that I thought about using a nice $g$. If I could use $g=\chi_I$ for any interval $I$, I could use Lebesgue derivation theorem. But such function is not continuous.