Let $A\subset \mathbb{P}(X)$ and let $\sigma(A)$ be the sigma algebra generated by $A$. Then $\sigma(A)$ is defined to be the smallest sigma algebra on $X$ which contains $A$ as a subset.

Is it true that every member of $\sigma(A)$ can be written as the complement and/or countable unions and/or countable intersections of members of $A$?

The reason I'm asking is because I need to understand why every continuous function is Borel-measurable: If $E$ is a borel-measurable set, then $E$ can be written as the complement/countable union/countable intersection of open sets, and so can $f^{-1}(E)$, which implies that $f^{-1}(E)$ is borel-measurable.

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    $\begingroup$ The answer to your question beginning "Is it true..." is no. $\endgroup$ – ncmathsadist Jun 3 '18 at 23:21

You cannot write down all sets in the sigma algebra generated. To prove that continuous functions are measurable consider the class of all Borel sets E in $\mathbb R$ such that $f^{-1}(E)$ is measurable, show that this is a sigma algebra which contains all open sets; hence it contains all Borel sets in $\mathbb R$.


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