Note that one must be careful about what ae. means. When we say that $f=g$ ae., it means that there is a measurable set $E$ of measure zero such that
$f(x)=g(x)$ for $x \notin E$.
Suppose $\mu$ is complete. Let $E$ be the exceptional set where $f(x) \ne g(x)$.
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 $g^{-1}(A) \cap E^c= f^{-1}(A) \cap E^c$, 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 $f=g$ ae. and so $g$ is measurable. Hence
$g^{-1}(\{1\}) = N$ is measurable. Hence $\mu$ is complete.
Part (b) is similar. Note that if $h_n \to h$ with the $h_n$ measurable, then
$h$ is measurable. So, this really can be reduced to Part (a) fairly easily.
Your proof is essentially correct. The proof of the last part, that is (b. $\Rightarrow$ $\mu$ is complete), is very similar to the proof that (a. $\Rightarrow$ $\mu$ is complete).
Here is the proof in details.
Proposition 2.11 (Exercise 10) - 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:
($\mu$ is complete $\Rightarrow$ a.)
Suppose $\mu$ is complete. Suppose $f$ is measurable and $f = g$ $\mu$-a.e..
Since $f = g$ $\mu$-a.e., there exists a measurable set $E$ such that $\mu(E) = 0$ and, for all $x\notin E$, $f(x) = g(x)$.
Given any Borel set $B\subset \mathbb{R}$, we have
$$g^{-1}(B)= (g^{-1}(B)\cap E) \cup (g^{-1}(B) \cap E^c)=(g^{-1}(B)\cap E)\cup(f^{-1}(B)\setminus E)$$
since $f$ is measurable we have that $f^{-1}(B)$ is measurable, since $E$ is measurable, $f^{-1}(B)\setminus E$ is measurable. Since $\mu$ is complete and $\mu(E)=0$, we have $g^{-1}(B)\cap E \subset E$ is measurable. Thus $g^{-1}(B)$ is measurable.
(a. $\Rightarrow$ $\mu$ is complete)
Suppose a. holds. Let $E$ be a measurable set such that $\mu(E)=0$ and let $A$ be any subset of $E$.
Take $f=0$ (the null function) and $g=\chi_A$ (the indicator function of $A$). We have that $f$ is measurable and $f=g$ $\mu$-a.e., so by a.), we have that $g$ is measurable. Since $g^{-1}(\{1\})=A$, we have that $A$ is measurable. So $\mu$ is complete.
($\mu$ is complete $\Rightarrow$ b.)
Suppose $\mu$ is complete. Suppose $f_n$ is measurable for $n\in\mathbb{N}$, and $f_n\rightarrow f$ a.e..
Let $$\hat{f} = \lim_{n\rightarrow \infty}\sup f_n$$
Since $f_n$ is measurable, by proposition 2.7 we have that $\hat{f}$ is measurable. Thus given the fact that $f_n\rightarrow f$ e.e., we have $\hat{f} = f$ a.e., so, since $\mu$ is complete, a.) holds and we can conclude that $f$ is measurable.
(b. $\Rightarrow$ $\mu$ is complete)
Suppose b. holds. Let $E$ be a measurable set such that $\mu(E)=0$ and let $A$ be any subset of $E$.
Take $f_n=0$, for all $n\in \mathbb{N}$ and $f=\chi_A$ (the indicator function of $A$). We have that$f_n$ is measurable for $n\in \mathbb{N}$ and $f_n\rightarrow f$ $\mu$-a.e., so by b.), we have that $f$ is measurable. Since $f^{-1}(\{1\})=A$, we have that $A$ is measurable. So $\mu$ is complete.
Best Answer
Fix $\varepsilon>0$. For each $N \geq 1$, apply the original version restricted to the set $G_N:=\{g \geq 1/N\}$ (which clearly has finite measure if $\int g \, d\mu<\infty$), in order to get a set $E_N \subset G_N$ with $\mu(E_N)<\frac{\varepsilon}{2^N}$ such that $f_n \to f$ uniformly on $G_N \!\setminus\! E_N$. Now let $E:=\bigcup_{N=1}^\infty E_N$.
Obviously $\mu(E)<\varepsilon$. It's not too difficult to show that $f_n \to f$ uniformly on $X \!\setminus\! E$.
[If you're having trouble with this last bit: First, as an exercise, show that $f_n(x)=\mathbf{1}_{[-n,n]}(x)e^{-|x|}$ converges to $f(x)=e^{-|x|}$ uniformly on $\mathbb{R}$. Then generalise the reasoning.]