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If $f_n (\omega) = \sum_{i=1}^n f_1 (T^i \omega)$ and $T$ is an ergodic action with respect to the measure $\mu$ then it is know as Birkhoff's theorem that

$$ \lim_{n \rightarrow \infty} \frac{f_n}{n} = \int_{\Omega} f_1(\omega) d\mu. $$

I was wondering what happens if one studies $h_n (\omega) = \sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega)$ is it still true that

$$ \lim_{n \rightarrow \infty} \frac{h_n}{n} = \int_{\Omega} f(\omega)g(\omega) d\mu? $$

If we assume that $g(T^{2i}\omega)$ form a family of i i d random variables.

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  • $\begingroup$ I think the integral you have on the right is not the correct limit. The integral of $f\circ T^{i}.g\circ T^{2i}$ is $\int_{\Omega}f(\omega)g\left(T^i\omega\right)d\mu$ by the $T$-invariance of the measure. Assuming that all of the functions are bounded so that we are allowed to swap limit and integral, the integral of the limit $\lim_{n \rightarrow \infty} \frac{h_n}{n} $ then should be $$\lim_{n\to\infty}\int_{\Omega}f(\omega)g_n(\omega)d\mu=\int_{\Omega}f(\omega)d\mu.\int_{\Omega}g(\omega)d\mu$$ since the Birkhoff averages $g_n$ of $g$ tend to $\int_{\Omega}g(\omega)d\mu$ $\mu$-a.e. $\endgroup$ – KhashF Feb 6 '19 at 20:17
  • $\begingroup$ Having that said, I think it is better to ask whether $\lim_{n \rightarrow \infty} \frac{h_n}{n}=\int_{\Omega}f(\omega)d\mu.\int_{\Omega}g(\omega)d\mu$ $\mu$-a.e.? This is true if you make your assumptions about $T$ stronger, e.g. suppose $T$ is weak mixing. Because then, $T\times T^2$ would be an ergodic transformation of the product space $\left(\Omega\times\Omega,\mu\otimes\mu\right)$ and it suffices to apply the Birkhoff Ergodic Theorem to the function $(\omega,\omega')\mapsto f(\omega).g(\omega')$ on this space. $\endgroup$ – KhashF Feb 6 '19 at 20:32
  • $\begingroup$ I@KhashF : I had thought of something like what you are suggesting. However, with your outer-product function, do you really get $h_n=\sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega)$? Or do you get $\sum_{i=1}^n f (T^i \omega)g(T^{2i} \omega')$ instead? $\endgroup$ – Iosif Pinelis Feb 6 '19 at 23:15
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$\newcommand{\om}{\omega} \newcommand{\Om}{\Omega} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb{Z}} \newcommand{\de}{\delta}$ Let $\Om:=\{-1,1\}^{\N_0}$ and $\mu:=(\frac12\de_{-1}+\frac12\de_1)^{\otimes\N_0}$, where $\N_0:=\{0,1,\dots\}$ and $\de_a$ denotes the Dirac point measure at $a$. For $\om\in\Om$, suppose that $f(\om)=g(\om)=\om_0$, where $\om_j:=\om(j)$ for $j\in\N_0$. For $\om\in\Om$ and $j\in\N_0$, let $(T\om)_j:=\om_{j+1}$, so that $X_j(\om):=f(T^j\om)=\om_j$.

Then $X_0,X_1,\dots$ are independent Rademacher random variables, with $P(X_j=\pm1)=1/2$. Let $S_n:=\sum_1^n X_i X_{2i}$. Then the question becomes the following: is it true that \begin{equation} A_n:=\tfrac1n\,S_n\to EX_0^2 \end{equation} almost surely (or at least in probability), as $n\to\infty$?

We have \begin{equation} E S_n^2=\sum_1^n EX_i^2 X_{2i}^2+2\sum_{1\le i<j\le n}EX_i X_{2i}X_j\, EX_{2j}=n, \end{equation} so that $EA_n^2=1/n\to0$ and hence $A_n\to0$ in probability. On the other hand, $EX_0^2=1\ne0$.
So, the answer to your question is no in general.

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These averages converge almost surely, although as the previous comment shows it is not clear how to identify the limit. The almost sure convergence is a theorem of Bourgain from the following reference.

"Double recurrence and almost sure convergence", J. Reine Angew. Math. 404, pp 140--161, 1990.

It is unknown whether averages of the form $\frac{1}{n} \sum_{k=0}^{n-1} f_1(T^kx)f_2(T^{2k}x)f_3(T^{3k}x)$ converge almost surely.

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