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Katok/ Hasselblatt, "Modern Theory of Dynamical Systems", p. 156:

Every probability distribution $p=(p_0,\ldots,p_{N-1})$ where $0\leq p_i\leq 1$ for $i=0,\ldots,N-1$ and $\sum_{i=0}^{N-1}p_i=1$, determines the product measure $\mu_p$ on the space $$ \Omega_n=\{\omega=(\ldots,\omega_{-1},\omega_0,\omega_1,\ldots)|\omega_i\in\{0,1,\ldots,n-1\}\text{ for }i\in\mathbb{Z}\}. $$ Namely, for any cylinder $$ C_{\alpha_1,\ldots,\alpha_k}^{n_1,\ldots,n_k}=\{\omega\in\Omega_n|\omega_{n_i}=\alpha_i, i=1,\ldots,k\} $$ we set $$ \mu_p(C_{\alpha_1,\ldots,\alpha_k}^{n_1,\ldots,n_k})=\prod_{i=1}^kp_i $$ and then extend $\mu_p$ in the standard fashion to the $\sigma$-algebra of all Borel sets.

I have some problems to understand the bold marked text, because my understanding was the following:

The set consisting of all the cylinders, I denote it by $C$, is invariant under intersections, which means that $\mu_p$ can be extended uniquely to $\sigma(C)$, the smallest $\sigma$-algebra containing the cylinders which is here the product $\sigma$-algebra (and sometimes called cylinder $\sigma$-algebra).

But in general, $\sigma(C)\subseteq\mathcal{B}(\tau)$, where $\tau$ is the topology on $\Omega_n$ with base consisting of the cylinders and $\mathcal{B}(\tau)$ being the Borel $\sigma$-algebra.

So my question is: How to extend $\mu_p$ to all Borel sets as the quote says, i.e. how to extend $\mu_p$ to $\mathcal{B}(\tau)$?

(If this is what the quote means...)


The only explanation I have is that maybe in this context $\sigma(C)=\mathcal{B}(\tau)$, however I cannot prove that $\mathcal{B}(\tau)\subseteq\sigma(C)$.

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  • $\begingroup$ What are the factor spaces, you should explain your notation. $\endgroup$ – Michael Greinecker Jun 16 '18 at 16:42
  • $\begingroup$ I guess you mean $\Omega_n=\{0,1,\ldots,n-1\}^{\mathbb{Z}}$. $\endgroup$ – Rhjg Jun 16 '18 at 16:43
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    $\begingroup$ In that case, the product $\sigma$-algebra is the Borel $\sigma$-algebra. $\endgroup$ – Michael Greinecker Jun 16 '18 at 16:46
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    $\begingroup$ Yes, that's how I would show it. There is a countable basis consisting of open rectangls, which are all measurable. $\endgroup$ – Michael Greinecker Jun 16 '18 at 16:51
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    $\begingroup$ Exactly. You can now answer your own question. $\endgroup$ – Michael Greinecker Jun 16 '18 at 16:56
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Here, we have $\sigma(C)=\mathcal{B}(\tau)$ since the space is separable.

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