I am wondering this since the Borel algebra is generated by all open sets, or by all closed sets, or by all half-open intervals. My questions:
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However, is it also generated by all $G_\delta$ sets? Similar question for by $F_\sigma$ sets, or by the finitely many combinations, e.g., $G_{\delta\sigma\delta\sigma}$, which means countable intersections of $G_{\delta\sigma\delta}$ sets and so on, etc.
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Trying to answer my Q1 for $G_\delta$ set: Firstly, an open set is also (generated by) a $G_\delta$ set. Now, if I have a $G_\delta$ set, then it is also formed by (intersections of) open sets. So the answer to my question is yes for $G_\delta$ set?
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If Q2 is correct, can this be continued to $G_{\delta\sigma\delta\sigma…\delta}$ and still result in the answer "yes"?
Any suggestion is appreciated. Thanks!
Best Answer
As noted, any open set is by definition a $G_{\delta}$ set and a closed set is a $F_{\sigma}$ set. Hence, the Borel $\sigma-$algebra, $\mathscr{B}$ is of course contained in the $\sigma$-algebra generated by the $G_{\delta}$ sets, say $\mathscr{C}$ i.e. $\mathscr{B} \subset \mathscr{C}$. But note that any $G_{\delta}$ set is contained in $\mathscr{B}$ and thus $\mathscr{C} \subset \mathscr{B}$ hence $\mathscr{C} = \mathscr{B}$.
Same argument works for your second question