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So the definition of a measurable function is as follows:

Let $f: (X, \mathcal{A}) \rightarrow\overline{\mathbb{R}}$ be a function on the measurable space $(X, \mathcal{A})$. Then $f$ is said to be measurable if $f^{-1}((a, \infty]) \in \mathcal{A}$ i.e. if the pre-image of that interval belongs to the $\sigma$-algebra $\mathcal{A}$

So my question is: do we say $f$ is Borel measurable precisely when $\mathcal{A}$ is the Borel $\sigma$-algebra on $X$

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  • $\begingroup$ Yes, but note that you also need $X$ to be a topological space in order for the Borel sigma algebra (the sigma generated by the open sets) to be defined. $\endgroup$ – jkn Apr 18 '15 at 14:13
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Yes this is the right way to define it, here's the source (definition)

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