Assume $\mu$ is a finite measure on the Borel subsets of $\mathbb{R}$ such that$$f(x) = \int_{\mathbb{R}} f(x + t)\,\mu(dt), \quad \text{a.e.,}$$whenever $f$ is real-valued, bounded, and integrable. Show that $\mu(\{0\}) = 1$.
Thoughts. Take $f(x) = 1_{|x| < \epsilon}$ and $x = 0$, we get that $\mu((-\epsilon, \epsilon)) = 1$? I am not sure if this is rigorous/how to conclude, could anybody help me out?