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 $\mu$.



Understanding a proof that the simplex of shift invariant probability measures on $\{0,1\}^\mathbb{Z}$ is Poulsen?
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.


