# Real Analysis Folland Proposition 1.5

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

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

Also,

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

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?

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