Question

This is a question on the proof of this fact in chapter 3 of the book "Functional Analysis: Surveys and Recent Results II". There at the end the proof is outlined as follows:

Let $\mu$ be a shift invariant measure on $X=\{0,1\}^\mathbb{Z}$. For each subset $\Lambda\subset\mathbb{Z}$ define $C_\Lambda$ to be the subset of continuous functions on $X$, $C_\Lambda\subset C(X)$ which depend only on coordinates indexed by $\Lambda$. Let $\mu_\Lambda$ be the restriction of $\mu$ to $C_\Lambda$.

Now let $\Lambda_n=[1,n]$ and define $\nu_n=\otimes_{k\in\mathbb{Z}}\mu_{shift^{kn}(\Lambda_n)}$. Now define $$\mu_n=\frac{1}{n}\sum_{s=0}^{n-1}shift^{s}_*(\nu_n).$$

Then $\nu_n$ and $\mu_n$ are ergodic. Also $\mu_n=\nu_n=\mu$ on $C_{\Lambda_n}$, hence $\mu_n\to\mu$.

I do not understand precisely this last statement, that $\mu_n=\mu$ on $C_{\Lambda_n}$.

I don't know how to show it even for $n=2$: Since $C_{\Lambda_2}=C_{[1]}\otimes C_{[2]}$, any element of $C_{\Lambda_2}$ is a linear combination of functions $f(\omega_1)g(\omega_2)$. So if $\mu_2=\nu_2$ one must have $$\int f(\omega_1)g(\omega_2) d\mu_{\Lambda_2}=\int f(\omega_1)d\mu_{[1]}\int g(\omega_2)d\mu_{[2]},$$ which I don't see why should hold for any $\mu$.

Answer 1

The language of the proof given in the book you refer to is a little different from the language I'm accustomed to, but I'll give what I believe is the exact same argument using a slightly different language, and hopefully do it in such a manner that the issue you point out doesn't arise. (Since I'm not quite sure how to explain it away using a language that's less familiar to me.)

Let $X = \{0,1\}^\mathbb{Z}$ with $\sigma\colon X\to X$ the shift map, and let $\mu$ be any $\sigma$-invariant probability measure on $X$. Given $n\in \mathbb{N}$, let $\nu_n$ be the Bernoulli measure for $\sigma^n$ that best approximates $\mu$, and let $\mu_n$ be the invariant measure generated by $\nu_n$.

More precisely, $\nu_n$ is defined as follows. Let $Y = \{0,1,\dots,2^n-1\}^\mathbb{Z}$, with $\tau\colon Y\to Y$ the shift map, and define a homeomorphism $\pi\colon Y \to X$ by identifying symbols in the alphabet of $Y$ with $n$-words in $X$: if $\phi\colon \{0,1,\dots,2^n-1\}\to \{0,1\}^n$ is a bijection, then we put $\pi(y) = \dots \phi(y_{-1}).\phi(y_0)\phi(y_1)\dots$, where juxtaposition denotes concatenation. Note that $\pi$ conjugates $\tau$ to $\sigma^n$ via $\pi\circ \tau = \sigma^n\circ \pi$.

Define a measure $\mu^*$ on $Y$ by $\mu^*(E) = \mu(\pi E)$. Now define a $\tau$-invariant measure $\nu$ on $Y$ by putting $\nu([y_1\dots y_k]) = \prod_{j=1}^k \mu^*([y_j])$. This is the Bernoulli measure that best approximates $\mu^*$. Define $\nu_n$ on $X$ by $\nu_n(E) = \nu(\pi^{-1}E)$. Then $\nu_n$ is a Bernoulli measure for $\sigma^n$ with the property that $\nu_n([x_1\dots x_n]) = \mu([x_1\dots x_n])$ for every $n$-cylinder, but $\nu_n$ is not $\sigma$-invariant.

To rectify this, let $\mu_n = \frac 1n \sum_{k=0}^{n-1} \sigma_*^k \nu_n$. Then $\mu_n$ is $\sigma$-invariant, and moreover, since for every $n$-cylinder $C$ the $\sigma$-invariance of $\mu$ gives $(\sigma_*^k \nu_n)(C) = (\sigma_*^k \mu)(C) = \mu(C)$, we have $\mu_n(C) = \mu(C)$.

I believe that this last set of equalities (the fact that $\nu_n$, $\mu_n$, and $\mu$ agree on $n$-cylinders) is the statement you wanted explained. It seems to me that agreement on $C_{\Lambda_n}$ (in the language of the book) corresponds to agreement on $n$-cylinders (in the language here). In any case, the measure $\mu_n$ is ergodic and $\sigma$-invariant, and approaches $\mu$ as $n\to\infty$, which shows that the space of $\sigma$-invariant measures is the Poulsen simplex.