How to understand that a function is measurable for a.e. x?

Question

Suppose that $\{ f_{n} \}$ is a sequence of measurable functions that are all bounded by $M$, are supported on a set $E$ of finite measure, and $f_{n}(x) \to f(x)$ a.e. $x$ as $n \to \infty$. Then $f$ is measurable, bounded, supported on $E$ for a.e. $x$, and $$ \int |f_{n} - f| \to 0\quad \text{as} \ n \to \infty. $$

What's the exact meaning of $f$ is measurable on $E$ for a.e. $x$?

In my understanding, we use the term "almost everywhere" for point-wise property such as $$ f(x) = g(x) \quad \text{a.e.}\ x \in E, $$ or $$ \lim_{n \to \infty} f_n(x) = f(x) \quad \text{a.e.} $$ However the statement that a function is measurable is specified for a certain set and is not a point-wise property.

Does that mean $f$ is measurable on $E - E_0$ for some set $E_0$ of measure zero? I found that somewhat weird.

Answer 1

Yes. If you're working with a complete measure, then all negligible sets (i.e. measure 0) are measurable, so being measurable a.e. is equivalent to be measurable. If you're not in a complete measure, then it's possible for measurable functions to converge pointwise to a non-measurable function, so you need the a.e. restrictions.

Answer 2

Suppose the convergence occurs on a set $E_0\subset E$ of full measure. Then $f$ is determined on $E_0$. However it could be the case that the measure is not complete, so that there is a subset $E_1\subset E_0$ which is not measurable. Evidently we may take $f(x) = 1(x\in E_0)$ for $x\in E_0$.