Let $G$ a locally compact abelian group and let $\mu_G$ an Haar measure of $G$. Suppose that $f\in L^1(\mu_G)$ is such that for all measurable $A$ of finite $\mu_G$-measure it happens that $f*\chi_A=0$ $\mu_G$-a.e.. I want to prove that $f=0$ $\mu_G$-a.e..
In $(\mathbb{R}^n,+)$ I would proceed taking $B_r$ as the ball centered in the origin of radius $r>0$ and then using Lebesgue differentiation theorem to get that: $$0=\frac{1}{|B_r|}f*\chi_{B_r}\rightarrow f, r\rightarrow0 \ a.e.$$
However, for example, in $(\mathbb{Z},+)$ we don't have at our disposal Lebesgue differentiation theorem but here the result can be neatly proved convolving with indicator function of $\{0\}$.
So, how can I prove the result in general? Should I try to decompose the group as a product $G=G_1\times G_2$ where $G_1$ and $\hat{G_2}$ are discrete (if such a decomposition can be performed at all) and try to use a technique that mix the previous two strategies (if such a mixed strategy can be implemented at all), or is there another simpler (maybe obvious) path?
By taking Fourier transforms, we get $\hat{f} \hat{ \chi_A } = 0$ for all $A$ of finite measure. In particular, $\int \hat{f} \hat{\chi_A} d\mu = 0$. That is, in the $L^2(G^*)$ inner product, $\hat{f}$ is orthogonal to all the $\hat{\chi_A}$.
Now, since the span of the $\chi_A$ are a dense set in $L^2(G)$, it follows that their Fourier transforms are dense in $L^2(G^*)$.
Thus, $\hat{f}$ is orthogonal to every function in $L^2(G^*)$. In particular, it is orthogonal to all the functions of the form $1_K \hat{f}$, where $K$ is a compact subset of $G^*$. This implies that $\hat{f} = 0$.
Thus, $f = 0$, by $L^1$ Fourier inversion.
(I didn't really check any details, there may be subtleties I'm missing.)