Suppose $\mu$ is a finite measure on the Borel subsets of $\mathbb R$ such that
$f(x)=\int_{\mathbb R}f(x+t)\mu(dt)$ a.e.
where $f$ is real-valued, bounded, and integrable. Compute $\mu(\{0\})$.
Of course I know that the answer is 1, by letting $f(x)=\mathbb 1_{\{0\}}$, assuming that the equality $f(x)=\int_{\mathbb R}f(x+t)\mu(dt)$ also holds at $x=0$.
However, the equality holds 'almost' everywhere, not everywhere, so we have to prove that the equality holds at $x=0$. I think that it suffices to prove that $\mu(\{0\})\neq 0$ since the equality holds a.e., so the set of points where the equality don't hold has zero measure. Does anyone have ideas?
It's possible that we don't have $f(0) = \int f(t) \, \mu(dt)$ but since the equality is valid almost everywhere we have $$ 1 = \chi_{\{x\}}(x) = \int \chi_{\{x\}}(x+t) \, \mu(dt) = \int \chi_{\{0\}}(t) \, \mu(dt) = \mu(\{0\}) $$ for almost every $x \in \mathbb R$. Here we have used that $\chi_{\{x\}}(x+t) = \chi_{\{0\}}(t)$.
As this equality is valid for almost every $x \in \mathbb R$ it must be valid for some $x$. Thus, $\mu(\{0\}) = 1$.
Also see Suppose $\mu$ is a finite measure on the Borel sets of $R$ such that $f(x) = \int_R f(x + t) \mu(dt)$ a.e., show $\mu(\{0\}) = 1$.