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 $\mu$.