Please help in Folland Analysis Proposition 2.11 I don't really understand proposition 2.11 in Folland. So please help me to explain, as well as give some hints to prove it.
The part makes me confuse is $f=g$ $\mu-a.e$ ,does it means f equals to g ? what about the $\mu- a.e$ , I am confused about the notation here.
 A: 
The following implications are valid if and only if the measure $\mu$ is complete:
a.) If $f$ is measurable and $f = g$ $\mu$-a.e., then $g$ is measurable.
b.) If $f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., then $f$ is measurable.

Proof a.):
Suppose $\mu$ is complete and $E$ is the exceptional set where $f(x)\neq g(x)$. Suppose $A$ is measurable. Then $g^{-1}(A) = ( g^{-1}(A) \cap E) \cup ( g^{-1}(A) \cap E^c)$. The set  $g^{-1}(A) \cap E$ is measurable since it is contained in $E$ which has measure zero. We have $g^{-1}(A) \cap E^c= f^{-1}(A) \cap E^c$ (since if $x\in g^{-1}(A)\cap E^c$ then $g(x)\in A$ and $x\in E^c$, so $f(x) = g(x)$ and we have $x\in f^{-1}(A)\cap E^c$ similar for the other direction)  , hence it is measurable and so $g^{-1}(A)$ is measurable, and so $g$ is measurable and so Part (a) holds.
Now suppose Part (a) holds. Let $N \subset E$, where $E$ has measure zero. Let $f=1_{E}$ and $g = 1_{N}$. Then $f=g$ a.e. and so $g$ is measurable. Hence $g^{-1}(\{1\}) = N$ is measurable. Hence $\mu$ is complete.
Proof b.):
We are given $f_n$ to be measurable for $n\in\mathbb{N}$, and $f_n\rightarrow f$ a.e. From proposition 2.7 we can let $$\hat{f} = \lim_{n\rightarrow \infty}\sup f_n$$ since $f_n$ is stated to be measurable, then $\hat{f}$ is also measurable. Also, since $f_n\rightarrow f$ a.e, we then have $\hat{f} = f$ a.e, so by part (a) $f$ is measurable.
Conversely, suppose $\mu$ is not complete. Then there exists a measurable set $E$ such that $\mu(E) = 0$ and a set $F\subset E$ such that $F$ is not measurable. Then for (a), note that $1_F$ is not measurable and $1_F = 0$. Similarly, for (b) define $f_n = 0$ for all $n$ and $f = 1_F$
