Problem (Choquet-Deny theorem): Let $\sigma$ be a probability measure on $\mathbb R$, and let $f : \mathbb R \to \mathbb R$ be a bounded (Borel) measurable function. We will prove the $f$ satisfies the equation $$ f(x) = \int_{\mathbb R} f(x + y)d\sigma(y), ∀x \in \mathbb R$$ if and only if for each $x ∈\mathbb R, f(x + y) = f(x)$ for σ-almost all $y ∈ \mathbb R.$
(a) Let $f$ be a solution to the convolution equation, and let $x ∈ \mathbb R$. Let $Y_n := x + X_1 +\dots + X_n$ where the ${X_k}$ are i.i.d. random variables with distribution σ. Prove that $f(Y_n)$ is an uniformly integrable martingale (with respect to the natural filtration).
b) Prove $\lim f(Y_n)=f(x)$ a.s
hint: Use Martingale convergence and the Hewitt-Savage zero-one law
My attemps
in clause (a) I havn't managed to show it is a martingale, can I get a hint.
I tried showing
$E(f(Y_{n+1})|F_n)=f(Y_n)$ and $for \ A\in F_n:\ \ \ E(f(Y_n)*1_A)=E(f(Y_{n+1})*1_A)$
but in both approach havn't gotten much progress.
In (b) I dont understand why would $\lim f(Y_n)$ be in $\mathcal E$ so i can use Hewitt-Savage zero-one law
Thank you,
A set $A\in\mathcal F_n$ can be written as $(X_1,\dots,X_n)\in B$ for some $B\in\mathcal B(\mathbb R)$. Therefore, one has $$ \mathbb E\left[f\left(Y_{n+1}\right)\mathbf{1}_A\right]= \mathbb E\left[f\left(Y_n+X_{n+1}\right)\mathbf{1}_B(X_1,\dots,X_n)\right] = \mathbb E\left[\int f\left(Y_n+y\right)\mathbf{1}_B(X_1,\dots,X_n)d\sigma(y)\right], $$ where we used Funini's theorem and independence between $X_{n+1}$ and $(X_1,\dots,X_n)$. From the assumption on $f$, we derive that $$ \mathbb E\left[f\left(Y_{n+1}\right)\mathbf{1}_A\right]= \mathbb E\left[ f\left(Y_n \right)\mathbf{1}_A \right] $$ which proves the martingale property.
For the question (b), note that if you exchange the first say $10$ indexes, it will not affect $f(Y_n)$ for $n$ larger than $10$.