I'm working through problem 4.1.2 of Einsidler and Ward's Ergodic Theory with a view towards number theory", which asks to show that the weak limit of ergodic measures need not be ergodic using the following scheme:
Let $x=(x_1,x_2,\dots)\in 2^\mathbb{N}$ (with $2=\{0,1\}$) be a binary sequence such that every block of $k$ digits appears in it with asymptotic frequency $2^{-k}$. Such an $x$ exists by the pointwise ergodic theorem.
Define $y_n\in 2^\mathbb{Z}$ for each $n\in\mathbb{N}$ by the conditions:
- $y_n|_{[0,2n-1]}=(x_1,x_2,\dots,x_n,0,0,\dots,0)$ ($n$ $0$s).
- $\sigma^{2n}(y_n)=y_n$
where $\sigma:2^\mathbb{Z}\to 2^\mathbb{Z}$ is the two-sided left shift. It is clear that $y_n\to y$ in the product topology, where $y$ is defined by $$ y(k)= \begin{cases} x_{k+1} & k\ge 0\\ 0 & k<0 \end{cases} $$ since $y_n$ and $y$ agree on $[-n,n]$.
We construct probability measures supported on the orbit of $y_n$ under $\sigma$: $$\mu_n:=\frac{1}{2n}\sum_{k=0}^{2n-1}\delta_{\sigma^k y_n}$$ where $\delta_z$ denotes the Dirac delta concentrated at $z$. Further, let $\nu:=\frac{1}{2n}\sum_{k=0}^{2n-1}\delta_{\sigma^k y}$.
I'd like to show that $\mu_n\to \nu$ in the weak* topology. Letting $f\in C(2^\mathbb{Z})$, we have that:
$$\mu_n(f)-\nu(f)=\frac{1}{2n}\sum_{k=0}^{2n-1}f(\sigma^ky_n)-f(\sigma^k y)$$ which looks like it should tend to $0$ as $n\to \infty$, but I'm stuck in giving a proof. I can bound $d(\sigma^k y_n,\sigma^k y)\le \frac{1}{n+1-k}$ where $d$ is the standard metric on $2^\mathbb{Z}$ (defined by $d(x,y)=\inf \{1/(k+1): x(i)=y(i)\ \forall |i|\le k\}$), but couldn't get much further.