If $(X, \Sigma, \mu)$ is a complete measure space, and $f$ is a function that is defined almost everywhere, can I use the language that $f$ is measurable? What does it mean for this function that is defined everywhere except on a set of measure $0$ to be measurable? Does it just mean that it has a measurable extension?
The following are equivalence:
$f$ have a measurable extension.
Preimage of any Borel set is measurable.
$f$ is measurable on the measure subspace containing the domain of $f$.
All of them can be proved by remembering that the domain of $f$ is a measurable set. And fill the part outside the domain with $0$ to obtain an extension.