I am reading the following paper: http://arxiv.org/pdf/1602.03849v2.pdf
I will explain the general setup below.
Let $x\in X=\mathbb{T}^d$, where $\mathbb{T}^d$ is the d dimensional torus. Let $\rho$ be a Borel prob. measure on SL$_d(\mathbb{Z})$, and set $X_0=x$, and $X_{n+1}=g_{n+1}X_n$ where $(g_n)\subset$ SL$_d(\mathbb{Z})$ is chosen with law $\rho^{\otimes\mathbb{N}}$. Let $\mathbb{P}_x$ be the measure on $X^\mathbb{N}$ associated with the random walk starting at $x$.
For $f$ Borel on $X$, $f\geq 0$, and $x\in X$, define the Markov operator $Pf(x)=\int_{SL_d(\mathbb{Z})}f(gx)d\rho(g).$ We have $\mathbb{E}_x[f(X_n)]=\mathbb{E}_x[P^{n-k}f(X_k)]$ for all $k\leq n$. We let $g$ be a continuous function on $\mathbb{T}^d$. Let $M_n=\sum\limits_{k=0}^{n-1}g(X_{k+1})-Pg(X_k)$. I want to show $M_n$ is a martingale wrt $\mathcal{F}_n=\sigma(X_k \mid k\leq n)$.. It is easy to see this is equivalent to showing that $\mathbb{E}[g(X_{n+1}) \mid \mathcal{F}_n]=Pg(X_n)$.
This is the part which I am having trouble with. Any help would be much appreciated! (I think the Markov property is used somewhere).
Edit: My attempt so far: By the Markov property, it suffices to show $\mathbb{E}[g(X_{n+1}) \mid X_n]=Pg(X_n)$.
Clearly $Pg(X_n)$ is $\sigma(X_n)$ measurable. So it remains to show $\int Pg(X_n)d\mu_{X_n}=\int g(X_{n+1})d\mu_{X_n}$. This part I am having trouble with. Any help would be appreciated!
Notice that $g_{n+1}$ is independent of $\mathcal F_n(=\sigma(g_1,\ldots,g_n))$. Therefore $$ \eqalign{ \Bbb E[G(X_{n+1})\,|\,\mathcal F_n] &=\Bbb E[G(g_{n+1}X_{n})\,|\,\mathcal F_n]\cr &=\int_{SL_d(\Bbb Z)}G(gX_n)\,d\rho(g)\cr &=PG(X_n)}. $$