I think this is not actually true, but it is a qual problem from Boston College so I am being doubtfull of my counter-example.
Let $X=[0,1]$ and $f_n=1-g_n$ where $g_n$ is the typewriter sequence ($g_n\to 0$ in $L^1$ but not a.e) thus $f_n\to 1$ in $L^1$ but not a.e How is this not a counterexample?
This hypotesis by itself, as your counterexample shows, is not sufficient to guarantee the convergence of $f_n$, much less that $f_n\to 1$. However, if you add some additional hypotesis, the result follows. For example, the following is true:
If $f_n\to f$, $f=1$ a.e.
Proof:
In fact $1\ge f\ge 0$, and thanks to the dominated convergence theorem, we have
$$\mu(X)=\int_X fd\mu=\int_X 1d\mu\\\int_X(1-f)d\mu =0\\ \forall_{t>0}\ \mu(\{1-f\ge t\})=0$$
Thus $f=1$ a.e.
Another proof can be found applying Fatou's lemma. Since $0\leq f_n\leq1$ we know that $0\leq 1-f_n$ and so by Fatous Lemma $$\int_X\liminf{(1-f_n)}\leq\liminf{\int_X(1-f_n)}\stackrel{f_n\to 1(L^1)}{=}0$$ Hence a.e we have $$\liminf{(1-f_n)}=0 \implies\limsup{f_n}=1 \implies f_n\to f$$