# Convergence of measurable functions and the completion of a measure

I am currently studying from Folland's Real Analysis: Modern Techniques and Their Applications. I know the following:

1. If $\mu$ is a complete measure, and $\{ f_{n} \}$ is a sequence of measurable functions such that $f_{n} \to f$ a.e., then $f$ is also measurable.

2. If $(X, \mathcal{M} , \mu)$ is a measure space and $(X ,\bar{\mathcal{M}}, \bar{\mu})$ is its completion, and if $f$ is a $\bar{\mathcal{M}}$-measurable function on $X$, then there is an $\mathcal{M}$-measurable function $g$ such that $f=g$ $\bar{\mu}$- almost everywhere.

Now, at the beginning of Folland's proof of the dominated convergence theorem, he argues (very briefly) that if $f_{n} \to f$ a.e., and $\{ f_{n} \}$ is a sequence in $L^{1}$, then "$f$ is measurable (perhaps after redefinition on a null set)", which he says follows from (1) and (2) above.

What does he mean by this? And wouldn't we then be able to apply this logic to any old measurable sequence $\{ h_{n} \}$ such that $h_{n} \to h$ a.e., and claim that "$h$ is measurable (perhaps after redefinition on a null set)" regardless of whether the measure we use is complete? Any help is appreciated.

• @saz: I don't see the problem: it means that there is a measurable set $E$, with $\mu(E) = 0$, such that for all $x \in E^c$, we have $f_n(x) \to f(x)$. – Nate Eldredge Oct 20 '15 at 19:01
• @NateEldredge Yeah, you are right... I'll remove (this part of) my comment. – saz Oct 20 '15 at 19:05
• This question is related: math.stackexchange.com/q/1434943 – saz Oct 20 '15 at 19:06

Folland's definition of "a.e." is stated a bit imprecisely: "If a statement about points $x\in X$ is true except for $x$ in some null set, we say that it is true almost everywhere". (Notation: $(X,\mathcal M,\mu)$ is a measure space; "null set" means an element $E$ of $\mathcal M$ with $\mu(E)=0$.) If one allows the reading "with the possible exception of" for "except for" then $$f_n\to f\hbox{ a.e.}$$ means $$\hbox{there exists }N\in\mathcal M \hbox{ with }\mu(N)=0\hbox{ such that }\{x\in X:f_n(x)\not\to f(x)\}\subset N.$$ Things being so, it may happen that each $f_n$ is $\mathcal M$-measurable but $f$ is not. This can be fixed by redefining $f$ appropriately on $N$ (e.g., $f(x):=\limsup_n f_n(x)$ there). [Added: It is clear from subsequent uses of the term in his text (e.g., Prop. 2.11 on page 47 of the 2nd edition) that Folland intends the "possible exception" reading of the definition of "a.e."]
If the measure space is complete, then the above containment $$\{x\in X:f_n(x)\not\to f(x)\}\subset N$$ forces $N_0:=\{x\in X:f_n(x)\not\to f(x)\}\in\mathcal M$ and $\mu(N_0)=0$.