If $f$ is continuous and $B\in\mathcal B$ is a Borel set, what is and isn't $f^{-1}(B)$?
What are open sets in relation to borel sets? If $\mathcal O=\{\text{open sets}\}$ surely $\mathcal B\not\subset \mathcal O$ because there are closed sets in $\mathcal B$. But are all open sets Borel sets?
Is there a "classification" that states what are the pre-images/images (although I think it's not as simple with images) of open sets/Borel sets/Lebesgue measurable sets by a continuous/smooth/measurable function (any combination of those)? If yes, a link would be great.
Thanks
Best Answer
Yes, open sets are Borel sets by definition. $\mathcal{B}$ is the smallest $\sigma$-algebra that contains all open sets.
If $f:X \to Y$ is continuous, then $\mathcal{S} := \{A \subseteq Y : f^{-1}[A] \in \mathcal{B}_X\}$ is a $\sigma$-algebra on $Y$ that contains all open sets because $f$ is continuous, so $f^{-1}[O]$ is open in $X$ when $O$ is open in $Y$, so $f^{-1}[O]$ is in particular in $\mathcal{B}_X$, so $O \in \mathcal{S}$.
By minimality of the Borel sets, as defined in the first line, $\mathcal{B}_Y \subseteq \mathcal{S}$ which means exactly that the inverse image of a Borel set in $Y$ is a Borel set in $X$.