Meaning of the expression "$\mu$ a.e."

In the monotone convergence theorem:

Let $g_n\geq 0$ be a sequence of measurable functions, such that $g_n \uparrow g\;\; \mu \text{ a.e.},$ i.e. $g_n(\omega) \leq g_{n+1}(\omega)\;\; \mu \text{ a.e.},$ and $\displaystyle \lim_{n\to\infty} g_n(\omega) = g(\omega)\;\; \mu \text{ a.e.},$ then $\int g_n d\mu \uparrow \int g d\mu.$

I wrote the theorem just for context. And I understand that "$\text{a.e.}$" is a way of excluding sets or subsets that are not measurable, and that the measure is $\mu.$

But I don't see the intent or meaning of the expression $$\mu\text{ a.e.}$$

1 Answer

It means almost everywhere with respect to the measure $\mu$: for short, $\mu$-almost-everywhere. Id est, that the set of points where the condition fails is contained in a measurable set $S$ such that $\mu(S)=0$.