# Non-integrable ergodic theory

Can anyone help me out with proofs/counterexamples? I'm working on an operator-valued multiplicative ergodic theorem and need what may(?) be a well-known fact. This fact (if true) would help me get rid of an annoying asymmetry in the conclusion of a theorem.

I'm assuming that $T$ is an invertible ergodic transformation of a probability space $(X,\mathcal B,\mu)$ and that $f$ is a measurable (but not necessarily integrable) function on $X$.

Is it true that $f(T^nx)/n \to 0$ a.e. if and only if $f(T^{-n}x)/n\to 0$ a.e.?

(1) In probability language if you define $X_n=f(T^nx)$ this is a stationary sequence of random variables. Borel-Cantelli 1 shows that if $\mathbb E |X_0|<\infty$ (i.e. $f\in L^1$) then $X_n/n\to 0$ as $n\to\pm\infty$: The probability that $|X_n|/n > 1/k$ is $\mathbb P(|X_0| > n/k)$. The sum of this series is over-estimated by $k\mathbb E|X_0| < \infty$. Hence almost surely $|X_n|/n < 1/k$ for all large $n$. Since this is true for all $k$ you get $X_n/n\to 0$.

If the $X_n$ are i.i.d. random variables, then the converse holds by Borel-Cantelli 2. This shows that for i.i.d. random variables, the boxed question has an affirmative answer.

(2) In the case where the $X_n$ are not i.i.d. I believe there are examples where $X_n/n\to 0$ almost surely even though $\mathbb E|X_0|=\infty$.

In the case that $f\in L^1$, both sides of the implication in the main question are true. The unresolved case is $f\not\in L^1$.

-
Don't you have then $f(T^n x)/n \to 0$ a.e. even if $f \notin L^1$, by a simple truncation argument? –  Mark Apr 28 '11 at 10:01
As Anthony said, if the sequence of random variables $f(T^n \cdot)$ are independent, then by Borel Cantelli Lemma, $f(T^n \cdot)/n$ converges to $0$ a.e.\ if and only if $f \in L^1$. –  camomille Apr 28 '11 at 10:25
Concretely if $X_n$ takes the value $2^k$ with probability $2^{-k}$ for $k\ge 1$ then you expect to see your first $2^k$ at $t_k\approx 2^k$ so that $X_{t_k}/t_k\approx 1$. –  Anthony Quas Apr 28 '11 at 12:56


I found this question very interesting and gave it much thought this week. I believe I have a proof now. I think it would be interesting to see what generalizations one can get from this argument. Below I use probabilistic notation, which I'm more used to.

Let $\{X_n\}_{n\in \Z}$ be a stationary stochastic process, taking values in $\R$. Let $L=\limsup_{n\to -\infty} \frac{X_n}{|n|}$ and $R=\limsup_{n\to \infty} \frac{X_n}{|n|}$.

Theorem: $L=R$, almost surely.

Proof: It is enough to prove the theorem for ergodic processes. We may also assume, WLOG, that all the values are nonnegative integers.

Let $A$ be the event that for some $n<0$ we have $X_n\ge |n|$ and let $B$ be the event that for some $n>0$ we have $X_n\ge |n|$.

Lemma: $\P(A) \le 2 \P(B) .$

Proof of lemma: For a given realization of $X_n$, let $I$ be all indices $i$ for which $T^i X$ is in $A$ and let $J$ be all the indices for which $T^i X$ is in $B$. We claim that the density of $J$ is no more than twice the density of $I$, which then implies the conclusion.

$I$ can be written as $\cup_{n \in \Z} \{n,\ldots,n+X_n\}$, while $J=\cup_{n \in \Z} \{n-X_n,\ldots,n\}$. In particular, $J$ is contained in $\bar{J}=\cup_{n \in \Z} \{n-X_n,\ldots,n+X_n\}$. But if we write $I$ as a union of disjoint intervals, then in $\bar{J}$ each of these intervals is extended to the left by at most the length of the interval. Hence, the density of $\bar{J}\supset J$ is at most twice that of $I$. $\blacksquare$

Of course, by symmetry, we also have $\P(B) \le 2 \P(A)$. Let $A_K$ be the event that for some $n<0$ we have $X_n \ge \max(|n|,K)$ and define $B_K$ analogously. By applying the lemma to the process $\{Y_n\}_{n\in \Z}$ defined by $Y_n=X_n$ if $X_n\ge K$ and $Y_n=0$ otherwise, we get that $\P(A_K) \le 2 \P(B_K)$ and vice verse.

In particular, $\lim_{K\to \infty} \P(A_K) =0$ if and only if $\lim_{K\to \infty} \P(B_K) = 0$. These limits necessary exists, since these are monotone decreasing events.

For ergodic processes, we have that $L$ and $R$ are a.s. constant. Now, if $L<1$ a.s. then we have $\P(A_K)\to 0$. In the other direction, if $\P(A_K)\to 0$ then a.s. $L\le 1$. Similar implications hold for $R$ and $B_K$. Hence, we get that if $L<1$ then $R\le 1$ and vice verse. By applying these to a rescaled process $X_n / \alpha$, we get that for any $\alpha$ we have $L<\alpha$ implies $R\le \alpha$ (and vice verse), so we must have $L=R$. $\blacksquare$

-
Thanks for the nice answer... –  Anthony Quas May 16 '11 at 8:04

This is not an answer to the question, but this is too long for a comment.

I give what looks like an example where $f$ is not $L^1$ but where $f(T^n\cdot)/n \to 0$ a.e. I will state it in a probabilistic way (sorry about that).

Let $\mu$ be a probability measure on $\{1,2,\dots\}$. I assume that $m=\int k\mu(dk)$ is finite. I define a new probability measure by $\hat{\mu}(dk)=k/m.\mu(dk)$. Let us consider a Markov chain $((A_n,B_n))$ on $\Omega=\{(a,b) : a \ge 1, 1 \le b \le a\} \subset N \times N$ with the following transitions probabilities :

1) From $(a,b)$ with $b>1$ one goes to $(a,b-1)$.

2) From $(a,1)$ one goes to $(c,c)$ where $c$ is chosen according to $\mu$.

Now let $\nu$ be a probability measure on $\Omega$ defined by $\nu(\{(a,b)\})=\mu(\{a\})/m$. The above Markov chain admits $\nu$ as a stationary distribution and I consider it under this distribution. Note that $A_n$ is distributed according to $\hat{\mu}$. Now set $X_n=g(A_n)$ where $g$ is in $L^1(\mu)$ but not in $L^1(\hat{\mu})$. I think that $(X_n)_n$ is a counterexample (consider the Markov Chain seen at the times at which it jumps from $(a,\cdot)$ to $(c,c)$, then the first coordinates make a sequence of i.i.d.r.v. distributed according to $\mu$).

-
Thanks, with two backslashes it works. –  camomille Apr 28 '11 at 11:08
Nice example. It's similar to the one I had in mind, but here you can directly see the conclusion from the Borel-Cantelli arguments in my posting. –  Anthony Quas Apr 28 '11 at 12:49