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.
Best Answer
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.