Defining a function on measure $0$ set to make it measurable I am confused about something. I am doing some self learning. Currently reading Folland's Real analysis Modern techniques and their applications book and I noticed the following proposition: The measure $\mu$ is complete iff 
1) $f$ is measurable and $f=g$ -a.e., then $g$ is measurable. and 
2) if $f_n$ are measurable and $f_n \to f$ u a.e. $x$, then $f$ is measurable.
On the other hand, in a proof on another section, he finds a function $f$ which is equal to the limit of a sequence of measurable functions at almost every $x$. Then he proceeds to define $f=0$ on the measure $0$ set, and then says f is now measurable. But in that theorem no mention of completeness was made.
I don't understand why in one case we need completeness and in the other no completeness is mentioned.
Perhaps I am misinterpreting what $f=g$ a.e. $x$ means? Does it mean automatically that $f=g$ on a measurable set $E$, with the complement of $E$ having measure $0$? Or do it just refer to a subset of a measurable set? Finally if you can explain how this relates to the fact that for measurable sequence $f_n$, the set {$x: \lim f_n(x)$ exists} is measurable. Please help clarify. Thank you
 A: $f=g$ a.e. can be translated into: $$\text{a measurable set }N\text{ exists such that }\{x\mid f(x)\neq g(x)\}\subseteq N\text{ and }\mu(N)=0\tag1$$
If $f=\lim f_n$ and every $f_n$ is measurable then also $f$ is measurable. 
In the weaker case $f=\lim f_n$ a.e. according to $(1)$ we have $f1_{N^{\complement}}=\lim f_n1_{N^{\complement}}$ so can conclude that $f1_{N^{\complement}}$ is measurable. But what about $f=f1_{N^{\complement}}+f1_N$? It will be measurable iff $f1_N$ is measurable. Observe that the preimage of every set wrt $f1_N$ will have the form $A$ or $A\cup N^{\complement}$ where $A\subseteq N$. If $\mu$ is a complete measure then on base of $A\subseteq N$ we can conclude that $A$ is a measurable set and consequently that $f1_N$ is a measurable function. Then also $f$ is measurable. If $\mu$ is not a complete measure then there is no guarantee that $f1_N$ is measurable. That is the reason for taking $f1_{N^{\complement}}$ instead of the original $f$. 
So if we know that $f=\lim f_n$ a.e. and we are after a measurable function $g$ that satisfies $g=\lim f_n$ a.e. then $g=f1_{N^{\complement}}$ can always be used, but $g=f$ cannot always be used, since it might be that $f$ is not measurable. In the special case where $\mu$ is a complete measure this obstacle is taken away and $g=f$ can also be used. 

If $\mu$ is a complete measure then every subset of such a measurable set $N$ with $\mu(N)=0$ will be measurable itself, so $(1)$ can be refrased in that situation by:$$\mu(\{x\mid f(x)\neq g(x)\})=0\tag2$$
