0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The answer to your question beginning "Is it true..." is no. $\endgroup$ – ncmathsadist Jun 3 '18 at 23:21
2
$\begingroup$

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 $\mathbb R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.