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.
@Wolfy, The idea in your attempt proof is correct but it fails basically because when we take a subsequence $\{f_{n_j}\}$ of $\{f_n\}$, we have
$$\liminf \int f_n \leq \liminf \int f_{n_j}$$
The "trick" is start by peeking a subsequence $\{f_{n_i}\}_i$ such that
$$\int f_{n_i} \to \liminf_{n \to \infty} \int f_n $$
in other words:
$$\lim_{i \to \infty}\int f_{n_i} = \liminf_{n \to \infty} \int f_n $$
Here is the proof in details.
Exercise 33 - If $f_n\geq 0$ and $f_n\rightarrow f$ in measure then $\int f\leq \liminf \int f_n$
Proof
Since $\{ \int f_n \}_n$ is just a sequence of complex number, there is subsequence $\{f_{n_i}\}_i$ such that
$$\int f_{n_i} \to \liminf_{n \to \infty} \int f_n $$
Since $f_n\rightarrow f$ in measure, we have that $f_{n_i}\rightarrow f$ in measure. So there is a subsequence $\{f_{n_{i_j}}\}_j$ of $\{f_{n_i}\}_i$ such that $$f_{n_{i_j}} \to f \textrm{ a.e.} \tag{1} $$
Since $\{f_{n_{i_j}}\}_j$ is a subsequence of $\{f_{n_i}\}_i$, we also have
$$\int f_{n_{i_j}} \to \liminf_{n \to \infty} \int f_n \tag{2}$$
So we have from $(1)$ and $(2)$, using Fatou's lemma,
$$ \int f = \int \lim_{j \to \infty } f_{n_{i_j}} = \int \liminf_{j \to \infty } f_{n_{i_j}}\leq \liminf_{j \to \infty } \int f_{n_{i_j}}= \lim_{j \to \infty } \int f_{n_{i_j}}= \liminf_{n \to \infty} \int f_n$$
Best Answer
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.