I don't understand the last sentence of Proposition 1.5.

In short, $X_j$ is not necessarily in $\mathcal{E}_j$, therefore, we cannot use Proposition 1.4.

In more detail, ${\bigotimes}_1^n \mathcal{B}_{X_j}$ is, by definition, a $\sigma$-algebra on $X$ generated by

$$\{\pi_j^{-1}(E_j) : E_j\in\mathcal{B}_{X_j}, j=1,...,n \}.$$

Here, $\mathcal{B}_{X_j}$ is the Borel $\sigma$-algebra on a separable metric space $X_j$ and $X=\prod_1^nX_j$ equipped with the product metric defined as

$$\rho ((x_1,x_2,...,x_n),(y_1,y_2,...,y_n))=\textrm{max}(\rho_1(x_1,y_1),\rho_2(x_2,y_2),...,\rho_n(x_n,y_n)) .$$


$$\pi_j^{-1}(E_j)=\{x\in X : x_j\in E_j \}. $$

We showed that $\mathcal{B}_{X_j}$ is generated by $\mathcal{E}_j$ and $\mathcal{B}_X$ is generated by $\{\prod_1^n E_j : E_j\in \mathcal{E}_j \}$. The definition of $\mathcal{E}_j$ is given in Proposition 1.5.

Now if $X_j\in\mathcal{E}_j$, then, by Proposition 1.4, ${\bigotimes}_1^n \mathcal{B}_{X_j}$ is generated by $\{\prod_1^n E_j : E_j\in \mathcal{E}_j \}$. Consequently, since ${\bigotimes}_1^n \mathcal{B}_{X_j}$ and $\mathcal{B}_X$ are generated by the same set, ${\bigotimes}_1^n \mathcal{B}_{X_j}=\mathcal{B}_X$.

However, $X_j$ is not necessarily in $\mathcal{E}_j$. Or am I missing something?

Thank you in advance.

I attached the image of the page with Propositions 1.4 and 1.5.

enter image description here


2 Answers 2


Just notice that in this case if we define $\tilde{\mathcal{E}}_j = \mathcal{E}_j\cup\{X_j\}$, then the $\sigma$-algebra generated by both sets is the same, since $X_j$ can be written as an infinite union of those open balls.

  • $\begingroup$ Thank you very much! It was so simple! $\endgroup$
    – Yerbolat
    Apr 17, 2018 at 2:06
  • $\begingroup$ I was also confused by the last sentence of the proof of proposition 1.5. But notice that the conclusion follows from THE FIRST part of proposition 1.4 (using Lemma 1.1). In fact the conclusion also follows by proposition 1.3 (using Lemma 1.1) $\endgroup$
    – Kimarokko
    Dec 17, 2018 at 18:57

$X_j$ itself is an open set so it’s a countable union of the generating set $\mathcal{E}_j$.


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