Let $X=\{0,1\}^{\mathbb{Z}}$ be the space of all bi-infinite sequences $x=(x_n)_{n\in\mathbb{Z}}$ of zeros and ones. On $X$, define the Left-Shift $\sigma\colon X\to X, (\sigma(x))_n=x_{n+1}$.

Now, consider the probability space $(X,B(X),\delta_x)$, where $B(X)$ is the Borel-$\sigma$-algebra and $\delta_x$ is the Dirac-measure for some fixed $x\in X$.

Now, for $B\in B(X)$, I am considering the measures $$ \mu_n(B):=\frac{1}{n}\sum_{i=0}^{n-1}\delta_x(\sigma^{-i}(B)) $$

and would like to know to which measure the sequence $(\mu_n)_n\in\mathbb{N}$ converges as $n\to\infty$ (that it converges follows from the compactness of $M(X)$, the set of all probability measures on $B(X)$, in the weak*-topology). Should be the Dirac measures.

  • $\begingroup$ 2 questions: Why must the limit measure exist? And. have you tried to identify the limit in any particular case, such as in the sequence of all $0$s, so $x_i=0$ for all $i$? $\endgroup$ Mar 27 at 12:55
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    $\begingroup$ I am not so sure that this actually converges in general. Compactness only gives you convergence of some subsequence (or even sub-net). Also, you should specify the topology with respect to which you consider the convergence. $\endgroup$
    – PhoemueX
    Mar 27 at 19:51

There are sequences of binary digits that are not Cesàro summable, that is, for which the sequence of partial averages $a_n=\sum_{i=1}^n x_i/n$ does not converge. (You can construct examples by concatenating ever-longer blocks of consecutive $0$s and $1$s. If the block boundaries are $n=3^k$, for instance, the quantities $a_n$ oscillate between about $1/3$ and $2/3$, as the block ending at $3^k$ is twice as long as all the previous blocks put together.)

Starting with such a sequence $x$ your measures $\mu_n$ do not converge in the weak* topology. Take, for instance the projection map $f:X\to\mathbb R$ that sends the sequence $u=(u_n)$ to its $0$-th component $u_0$. It is continuous and bounded, and clearly $\langle \mu_n, f \rangle=a_n$. If the $\mu_n$ had a weak* limit $\mu$, then $a_n$ would converge to $\langle \mu, f\rangle$.

In response to comments: It is possible that you are confusing having limit points with has a limit, that is, converges. Yes, your set of measures $\mu_n$ is relatively compact, and hence has convergent subsequences, but it need not be the case that those subsequences all converge to the same limit.

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    $\begingroup$ Yes. Slightly more generally: if $x$ is periodic, with period $p$, then your limit measure splits its mass evenly between the $p$ shifts of $x$. Or, if $x$ differs from a periodic sequence in finitely many entries, then the limit measure splits its mass between the periodic translates, and assigns mass $0$ to the original $x$ and its translates. Your question was/is interesting, but you did not clearly indicate what you knew, tried, or what the context of the question was. $\endgroup$ Mar 30 at 18:08
  • $\begingroup$ Yes. Since $x$ has period $p$, the set of all translates of $x$ actually consists of $p$ distinct points, etc etc. $\endgroup$ Mar 31 at 14:14
  • $\begingroup$ I'm sorry: I don't understand your last comment. "Enough" for what? $\endgroup$ Mar 31 at 17:34
  • $\begingroup$ Yes, I think you are right. $\endgroup$ Mar 31 at 18:48

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