Let say that an infinite subsets $A$ of $\mathbb{N}$ is "nice w.r.to ergodic limits", if it can replace $\mathbb{N}$ in the individual ergodic theorem, that is, if it is such that the following statement is true:

For any probability space $(X,\Sigma,\mu),$ for any measure-preserving transformation $T$ on $X,$ for any $f\in L^1(X,\mu)$ the ergodic means along $A,$

$$M(f,T,A,t,x):=\big|\{j\in A\, : j\leq t\}\big|^{\,-1}\sum_{j\in A,\, j\leq t}f(T^{j}x)$$

converge for a.e. $x\in X$ to the conditional expectation w.r.to the $T$ invariant $\sigma$-algebra, $\mathbb{E}(f|\Sigma_T),$ as $t \to +\infty.$

So $\mathbb{N}$ itself is nice in this sense, by Birkhoff's theorem; if $A$ is nice and $m$ is a positive integer, the set of translates $A+m$ is nice (the set of convergence with $T$ along $A+m$ coinciding with the $T^{\, m}$ pre-image of the set of convergence with $T$ along $A$). Also, a disjoint union of two nice sets is nice.

Is there any other structure on the family of these sets? What about e.g. the union of two of them? (at a glance it seems to me that something more can be said for the analogous cases of other ergodic theorems, e.g. for the $L^p$ convergence.

Looking at this very related question, and its answer, make me think that the situation may be non-trivial and interesting enough, so that it should have been studied.

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    $\begingroup$ If $A$ is nice, and the symmetric difference of $A,B$ is finite, then $B$ is nice. $\endgroup$ – Kevin O'Bryant Oct 23 '10 at 15:43
  • $\begingroup$ And slightly more generally, if $|A\Delta B\cap [0,x)|=o(|A\cap [0,x)|)$ as $x\to\infty .$ $\endgroup$ – Pietro Majer Oct 23 '10 at 17:03
  • $\begingroup$ It is not true that $mA$ is nice, unless you restrict your space to non-atomic ones... $\endgroup$ – Ori Gurel-Gurevich Oct 23 '10 at 17:28
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    $\begingroup$ @Ori: sorry, I can't see why... What's the problem with atoms? $\endgroup$ – Pietro Majer Oct 23 '10 at 21:10
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    $\begingroup$ if the measure is uniform on {0,1} then the transformation T swapping 0 and 1 is ergodic, but $T^2$ isn't, so according to your definition $2\mathbb{N} isn't nice. Or maybe I misunderstood something? $\endgroup$ – Ori Gurel-Gurevich Oct 24 '10 at 3:49

These are called good universal sets. Bourgain (1987) proved that sequences of the form $p(n)$, $n \in {\bf N}$, $p$ a non constant polynomial, are good.

He also proved (1988) that the set of primes is a good universal set for $L^p$ functions, $p> {(1+\sqrt{3})\over 2}$. This was later improved to $p>1$ by Wierdl, see its article for a short introduction to the problem http://www.springerlink.com/content/e027w4211n7784h1/fulltext.pdf . There is now an extensive litterature on the problem, following Bourgain's articles.

In another direction, note that a transformation T is mixing iff the ergodic theorem for T holds for all subsequences (see e.g. the book of Krengel, "ergodic theorems").


There is a large survey article by Rosenblatt and Wierdl which discusses questions of this sort (and their answers): "Pointwise Ergodic Theorems via Harmonic Analysis" which can be found in the book "Ergodic Theory and its Connections with Harmonic Analysis". Unfortunately, the article (and the book) is from 1993 and I am unaware of more recent surveys (though I'd be happy to hear of such). Anyway, it contains a discussion of "good" and "bad" sequences (or subsets of $\mathbb{N}$, if you wish) in various situations ($L^2$ convergence, a.e. convergence, etc.).


There is a term: "bad universal" sequences in ergodic theory.


Maybe it is the opposite of what you are asking about, or of what you want to ask about?


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