# Equivalence of different definitions of measurable functions

I have found two definitions of measurable functions (or measurable maps) in different texts, and I'm struggling to see how they are equivalent.

The first is: A function $$f:X\rightarrow \mathbb{R}$$ is measurable with respect to a $$\sigma$$-algebra $$\mathcal{A}$$ of $$X$$ and the Borel $$\sigma$$-algebra $$\mathcal{B}$$ of $$\mathbb{R}$$ if, for each $$B \in \mathcal{B}$$, there exists $$A \in \mathcal{A}$$ such that $$f^{-1}(B) = A$$

The second definition looks very different to me, and I don't understand how they are equivalent. It says:

An extended real-valued function $$f:X\rightarrow [-\infty,\infty]$$ is measurable if its domain is measurable, and for each real number $$\alpha$$, the set $$\{x:f(x)>\alpha\}$$ is measurable.

Are these equivalent? I'd like to use the 2nd definition to prove that if $$g(x)$$ is measurable and $$g(x) \neq 0$$ for all $$x$$, then $$\frac{1}{g(x)}$$ is also measurable, but I'm not sure how to relate that to the first definition.

• Notice that $\{x:f(x)>\alpha\}=f^{-1}(\alpha,\infty).$ Borel $\sigma$ algebra is generated by opens. Commented Apr 19, 2020 at 20:08
• $f^{-1}$ comutes with intersections and unions, even if they are infinite. This should suffice Commented Apr 19, 2020 at 20:08

Prove using the first definition of measurability that a function $$f: A \to B$$ is measurable w/r/t $$\sigma$$-algebras $$\mathcal{A,B}$$ resp. iff $$f^{-1}(S) \in \mathcal{A}$$ for all $$S$$ in a set $$\mathcal{S} \subseteq \mathcal{B}$$ s.t. $$\mathcal{S}$$ generates $$\mathcal{B}$$ (meaning the $$\mathcal{B}$$ is the smallest $$\sigma$$-algebra containing $$\mathcal{S}$$).
Then use that together with the note that $$(\alpha,\infty]$$ generates the Borel $$\sigma$$-algebra on $$\overline{\mathbb{R}}$$, the extended reals.