A function $f:\Sigma\ni X\to Y$ with a $\sigma$-algebra $\Sigma$ is called measurable if $f^{-1}(A)$ is measurable for all open sets $A\subset Y$.
Now let $f:\mathbb R\to\mathbb R\cup\{\pm\infty\}$ and $\Sigma$ be any $\sigma$-algebra on $\mathbb R$.
Then is every continuous function measurable?
Well clearly, for the Borel-$\sigma$-algebra it's true cause of preimages of open sets are open, but what about any other algebra $\Sigma$?
Best Answer
Generally, no. Consider $f\colon x \mapsto x$. Then $f$ is continuous, but it is measurable only with respect to $\sigma$-algebras $\Sigma \supset \mathscr{B}(\mathbb{R})$.
On the other hand, since every continuous function is measurable with respect to the Borel $\sigma$-algebra $\mathscr{B}(\mathbb{R})$, the continuous functions are a fortiori measurable with respect to all larger $\sigma$-algebras.
So we have the proposition