Measure preserving transform and convergent random variables I have been trying to learn about Ergodic Theorem for a while and now I have a problem I can't solve. 
Assume $T$ is a measure preserving transform and $X_n\rightarrow X$ everywhere. Also, assume that $E(\sup_n |X_n|)<\infty$. Then I need to show $\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}{X_k(T^k(\omega))}$ converges with probability 1.
It's easy to show $X$ is integrable, so I was trying to use ergodic theorem (Birkhoff's theorem) to prove this but I couldn't get better results.
Any suggestion is really appreciated.
 A: This is a kind of "uniform ergodic theorem" and extends naturally the case $X_k=X$ for each $k$. 
Notice that $X_k=X_k-X+X$ and by Birkhoff's ergodic theorem, 
$$\frac 1n\sum_{k=0}^{n-1}X\circ T^k\to \mathbb E[X\mid\mathcal I]\quad\mbox{ a.s.},$$
where $\mathcal I$ denotes the $\sigma$-algebra of invariant sets, that is, $\mathcal I=\{A, T^{-1}(A)=A\}$.
If we manage to show that 
$$\frac 1n\sum_{k=0}^{n-1}(X_k-X)\circ T^k\to 0\quad\mbox{ a.s.},$$
we will be done. Replacing the $X_k$'s by $|X_k-X|=:Y_k$ we are reduced to show that 
$$\frac 1n\sum_{k=0}^{n-1}Y_k\circ T^k\to 0\quad\mbox{ a.s.}$$
under the assumption $Y_k\to 0$ a.s. and $\sup_kY_k$ is integrable. 
Consider a sequence $(\delta_l)_{l\geqslant 0}$ of positive real numbers such that for each $l$, $\mathbb E[\chi_A\sup_nY_n]\lt 2^{-l}$ if $\mu(A)\lt \delta_l$. We then use Egoroff's theorem: for each $l$, there is a measurable set $A_l$ such that 
$$\lim_{k\to+\infty}\sup_{\omega\in A_l}Y_k=0\quad\mbox{ and }\mu(\Omega\setminus A_l)\lt\delta_l.$$
Define $M_n:=\frac 1n\sum_{k=0}^{n-1}Y_k\circ T^k=M'_{n,l}+M''_{n,l}$ with $M'_{n,l}:=\frac 1n\sum_{k=0}^{n-1}(\chi_{A_l}Y_k)\circ T^k$ and $M''_{n,l}:=\frac 1n\sum_{k=0}^{n-1}(\chi_{\Omega\setminus A_l}Y_k)\circ T^k$. 
Since for each $k$, $\chi_{A_l}Y_k\leqslant \sup_{\omega\in A_l}Y_k(\omega)$, we have 
$$\tag{1} 0\leqslant M'_{n,l}\leqslant \frac 1n\sum_{k=0}^{n-1}\sup_{\omega\in A_l}Y_k(\omega).$$
Notice that 
$$0\leqslant \mathbb E[\sup_nM''_{n,l}]\leqslant \mathbb E[\sup_n\max_{0\leqslant k\leqslant n-1}\chi_{\Omega\setminus A_l}Y_k]\leqslant \delta_l\leqslant 2^{-l},$$
hence by the Borel-Cantelli lemma,
$$\tag{2}\lim_{l\to\infty}\sup_{n\geqslant 1}M''_{n,l}=0\mbox{ a.s.}$$
Combining (1) and (2), it follows 
$$0\leqslant M_n\leqslant \frac 1n\sum_{k=0}^{n-1}\sup_{\omega\in A_l}Y_k(\omega)+
\sup_mM''_{m,l},$$
hence taking the $\limsup_{n\to+\infty}$, we obtain for each $l$,
$$0\leqslant \limsup_{n\to \infty}M_n\leqslant \sup_mM''_{m,l}.$$
Now let $l$ going to infinity to conclude.
A: I realize what I did in the other answer was maybe overcomplicated. Define 
$D_l:=\sup_{k\geqslant l}|X_k-X|$ (which is integrable for each $l$), and notice that by Birkhoff's ergodic theorem, 
$$\frac 1n\sum_{j=0}^{n-1}D_l\circ T^j\to\mathbb E[D_l\mid\mathcal I]\quad \mbox{a.e.}$$
with the same notations as in the other answer. Since 
$$\frac 1n\sum_{j=0}^{n-1}|X_k-X|\circ T^j=\frac 1n\sum_{j=0}^{l-1}|X_j-X|\circ T^j+\frac 1n\sum_{j=l}^{n-1}\sup_{k\geqslant l}|X_k-X|\circ T^j\\
\leqslant \frac 1n\sum_{j=0}^{l-1}|X_j-X|\circ T^j+\frac 1n\sum_{j=0}^{n-1}\sup_{k\geqslant l}|X_k-X|\circ T^j+ \frac 1n\sum_{j=0}^{l-1}\sup_{k\geqslant l}|X_k-X|\circ T^j,$$
we obtain that for almost every $\omega$ and any $l$, 
$$\limsup_{n\to +\infty}\frac 1n\sum_{j=0}^{n-1}|X_k-X|\circ T^j(\omega)\leqslant 
\mathbb E[D_l\mid\mathcal I](\omega).$$
By monotone convergence, $\lim_{l\to +\infty}\mathbb E[D_l\mid\mathcal I](\omega)=0$ almost everywhere and we are done.
