# 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.

• "converges with probability 1 to some random variable with expectation equal to E(X)" is a bit odd. – mookid Mar 6 '14 at 23:49
• Sorry, do you mean it's wrong? or the way I have expressed it is wired? – Cohlan Mar 6 '14 at 23:51
• Isn't it "converges with probability 1 to E(X)" instead? – mookid Mar 6 '14 at 23:54
• I think once we can prove that thing converges with probability 1, the rest would be pretty easy. So maybe I should edit the question. – Cohlan Mar 6 '14 at 23:54
• put the "\" before Sup and lim in this case ;) – mookid Mar 6 '14 at 23:55

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.

• Thanks for this one as well. – Cohlan Mar 8 '14 at 20:52
• You are welcome. – Davide Giraudo Mar 8 '14 at 20:55

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.

• That's great, Thank you very much. – Cohlan Mar 8 '14 at 20:51