# Determine the limit measure

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.

• 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 $i$? Mar 27 at 12:55
• 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. 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.
• 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. Mar 30 at 18:08
• Yes. Since $x$ has period $p$, the set of all translates of $x$ actually consists of $p$ distinct points, etc etc. Mar 31 at 14:14