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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 $\Sigma$?

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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

Every continuous $f\colon \mathbb{R} \to \mathbb{R} \cup \{-\infty,+\infty\}$ is measurable with respect to the $\sigma$-algebra $\Sigma$ if and only if $\Sigma\supset \mathscr{B}(\mathbb{R})$.

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If the $\sigma$-algebra contains the Borel $\sigma$-algebra, such as with the $\sigma$-algebra of Lebesgue measurable sets, $\Sigma(m^{*})$, then yes, any continuous function is measurable. But it is not true for any arbitrary $\sigma$-algebra on $\mathbb{R}$. For example, you could take the $\sigma$-algebra $\Sigma = \{\mathbb{R}, \emptyset \}$ on $\mathbb{R}$. Under this $\sigma$-algebra, the function $f(x) = x$ is not measurable, even though it is a continuous function.

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