# Further Clarification Needed: Martingale to prove f(x+s) = f(x)

Given $f$ is a bounded continuous function on $\mathbf{R}$ and $\mu$ is a probability measure such that for all $x \in \mathbf{R}$ $$f(x) = \int_\mathbf{R}f(x+y)\mu(dy)$$

Please help to show that $f(x+s) = f(x)$ for all $s$ in the support of $\mu$. (The support of $\mu$ is the smallest set $E$ s.t. $\mu(E)=1$.) The hint is that we can try to use martingales to show $E[f(x+X_0)-f(x)]^2=0$, where $X_0$ has distribution $\mu$.

So far I've constructed a martingale $M_n=f(x+S_n)$. Here $S_n = \sum_{j=1}^nX_j$ and $\{ X_j\}$ are i.i.d. with distribution $\mu$. But I've no idea on how to connect this martingale with the previous hint.

I came cross another post about the same question on the forum, but it ends with the martingale that I stated here. So I'm wondering if anyone can provide some further insights about how to finish the proof. Since this is a hw problem, I've been working on it for three days only to get to this point. I sincerely appreciate any inputs that people will give.

Thanks a ton!

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See here -it's the same question. – saz Dec 18 '12 at 6:57
Is your problem how to show this is a martingale or how to use the fact that this is a martingale? – Did Dec 18 '12 at 7:41
This is a straight question from a take-home exam and I believe that seeking an answer for it on this platform is a violation of the rules of the exam and is unethical. Sorry for being fussy but I think, sometimes we have to rise beyond cheap tricks. – user53661 Dec 19 '12 at 0:34