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Anthony Quas
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non-integrable subadditive ergodic theorem

Dear MO_World,

I have (another) question about relaxing the assumptions in the sub-additive ergodic theorem. Apologies if this is something I should know already...

There are a number of statements of the Kingman sub-additive ergodic theorem and its extensions. Here is a fairly typical version:

Let $T\colon X\to X$ be a measure-preserving transformation of a probability space. Let $(f_n)$ be a sequence of measurable functions from $X$ to $[-\infty,\infty)$ satisfying $f_{n+m}(x)\le f_n(x)+f_m(T^nx)$ for every $m,n\in\mathbb N$ and $x\in X$. Suppose further that $f_1^+$ is integrable. Then $f_n(x)/n$ is convergent for almost every $x$.

There are also statements about convergence in the mean in certain situations, but I focus on pointwise convergence.

My general question is: When can you remove the integrability assumption?

In the following example the conclusion holds, even though the hypothesis fails. Consider a distribution on the positive reals so that if $(X_n)$ is a sequence of iid random variables with the given distribution, then $\mathbb E(\min(X_1,X_2,\ldots,X_n))=\infty$ for all $n$. (As an example, consider a random variable that takes the value $2^{k^3}$ with probability $2^{-k}$ for each $k\ge 1$. Let $M_n$ be $\min(X_1,X_2,\ldots,X_n)$. Then $\mathbb P(M_n=2^{k^3})\approx 2^{-nk}$ so that $\mathbb E M_n=\infty$.

Let $\Omega=\{2^{k^3}\colon k\ge 1\}^\mathbb Z$ equipped with the Bernoulli probability measure arising from the distribution above. Now define the sequence $f_n(\omega)$ as follows: $f_{2n}(\omega)=0$; $f_{2n+1}(\omega)=\min_{0\le j\le 2n+1}\omega j$. This sequence of functions is clearly sub-additive. Also $0\le f_n(\omega)/n\le \omega_0/n$, so that $f_n/n\to 0$ everywhere.

My specific question is as follows:

Suppose that $(f_n)$ is a sub-additive sequence of functions taking values in $[-\infty,\infty)$. If we assume that $\limsup f_n(x)/n `<` \infty$ a.e., then does it follow that $f_n(x)/n$ converges almost surely? [ The above example shows that we cannot expect convergence in norm, even if the $f_n$ are non-negative ]
Anthony Quas
  • 23.2k
  • 5
  • 63
  • 98