Defect version of Fatou's Lemma (repost) Someone recently posted the following question, but it was deleted.  Since it was an interesting question and I didn't know the answer, I'm reposting it here.
The question: is the following limit theorem true?

Suppose $(X,\mathcal{A},\mu)$ is a measure space, $f \in L^{1}(X)$, and $\{f_{n}\}_{n \in \mathbb{N}} \subseteq L^{1}(X)$ with $f_{n} \geq 0$ a.e. for all $n \in \mathbb{N}$ and $f_{n} \to f$ pointwise a.e.  If $\{f_{n}\}_{n \in \mathbb{N}}$ is bounded in $L^{1}(X)$, then $f_{n} \to f$ in $L^{1}(X)$ if and only if $\int_{X} f_{n} \, d \mu \to \int_{X} f \, d \mu$.

The answer turns out to be yes.  This isn't completely obvious, and it's interesting since I'm aware of at least one very closely related result in the literature.  (Namely, if $\{\mu_{n}\}_{n \in \mathbb{N}}$ are finite signed Borel measures on a compact metric space $X$ and $\mu_{n} \overset{*}{\rightharpoonup} \mu$, then $\|\mu_{n} - \mu\| \overset{*}{\rightharpoonup} 0$ if and only if $\|\mu_{n}\|(X) \to \|\mu\|(X)$.)
 A: The result you are asking for is just a special case of a more general result, known as Scheffe's lemma:

Suppose $(X,\mathcal{A},\mu)$ is a measure space. Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in L^1 (X)$ and $f_n \to f$ pointwise a.e. in $X$. Then $f_{n} \to f$ in $L^{1}(X)$ iff $\lim_{n \to \infty} \int_X |f_n| d\mu = \int_X |f| d\mu$.

It can be proved in a rather simple way by using Fatou's lemma. Let us prove it:
Proof: (=>) It is trivial, since, from Minkowski's inequality, we have
$$\left | \int_X|f_n|d\mu-\int_X|f|d\mu \right |\leqslant \int_X|\,|f_n| -|f|\,| d\mu\leqslant \int_X|f_n -f| d\mu$$
So, if $f_{n} \to f$ in $L^{1}(X)$ then $\lim_{n \to \infty} \int_X |f_n| d\mu = \int_X |f| d\mu$.
(<=) Note that $|f_n -f|\leqslant |f_n| +|f|$. So, for each $n$, the function $|f_n| +|f| - |f_n -f|$ is non-negative and using Fatou's Lemma, we have
\begin{align} 
2 \int_X|f|d\mu 
&=\int_X \lim\inf(|f_n| +|f| - |f_n -f|)d\mu \leqslant \lim\inf \int_X (|f_n| +|f| - |f_n -f|)d\mu = \\ 
&=\lim\inf \left (\int_X|f_n|d\mu +\int_X|f|d\mu - \int_X|f_n -f|d\mu \right) = \\
&= \left(\lim\inf\int_X|f_n|d\mu\right) +\int_X|f|d\mu - \left(\lim\sup\int_X|f_n -f|d\mu\right)  = \\ 
&=2\int_X|f|d\mu - \left(\lim\sup\int_X|f_n -f|d\mu\right) 
\end{align}
So we have
$$2 \int_X|f|d\mu \leqslant 2\int_X|f|d\mu - \left(\lim\sup\int_X|f_n -f|d\mu\right) $$
Since $f \in L^1 (X)$ , we know that $\int_X|f|d\mu<+\infty$, and so we get
$$\lim\sup\int_X|f_n -f|d\mu \leqslant  0$$
So we can conclude that
$$\lim\int_X|f_n -f|d\mu =  0$$
So, $f_{n} \to f$ in $L^{1}(X)$.
Remark: Note that the hypothesis that $\{f_{n}\}_{n \in \mathbb{N}}$ is bounded in $L^{1}(X)$ is not needed. In fact, it is a consequence of the other hypothesis of the result.
A: The "only if" direction is immediate so I'll only treat the "if" direction.
A sequence in $L^{1}(X)$ converges if and only if it converges locally in measure, it's uniformly integrable, and it's tight.  This follows from the Vitali Convergence Theorem, the Arzela-Ascoli Theorem analogue for $L^{1}$.
Local convergence in measure: $\{f_{n}\}_{n \in \mathbb{N}}$ converges locally in measure (i.e. it converges in measure on finite measure sets) since it's pointwise convergent.
Tightness: $\{f_{n}\}_{n \in \mathbb{N}}$ is tight if and only if, for each $\epsilon > 0$, there is a $N_{\epsilon} \in \mathbb{N}$ and a measurable $A_{\epsilon}$ with $\mu(A_{\epsilon}) < \infty$ such that
\begin{equation*}
\int_{X \setminus A_{\epsilon}} f_{n}(x) \, \mu(dx) < \epsilon.
\end{equation*}
To see this is so, recall that $\{f\}$ is tight so there is an $A_{\epsilon}$ with $\mu(A_{\epsilon}) < \infty$ such that
\begin{equation*}
\int_{X \setminus A_{\epsilon}} f(x) \, \mu(dx) < \epsilon/2.
\end{equation*}
Fatou's Lemma thus gives
\begin{equation*}
\int_{X} f(x) \, \mu(dx) - \epsilon/2 \leq \int_{A_{\epsilon}} f(x) \, \mu(dx) \leq \liminf_{n \to \infty} \int_{X \setminus A_{\epsilon}} f_{n}(x) \, \mu(dx) 
\end{equation*}
and so
\begin{equation*}
\liminf_{n \to \infty} \int_{A_{\epsilon}} f_{n}(x) \, \mu(dx) \leq \int_{X} f(x) \, \mu(dx) - \limsup_{n \to \infty} \int_{X \setminus A_{\epsilon}} f_{n}(x) \, \mu(dx) \leq \epsilon/2.
\end{equation*}
Therefore, there is a $N_{\epsilon} \in \mathbb{N}$ with the desired property.
Uniform integrability:  We proceed by contradiction.  $\{f_{n}\}_{n \in \mathbb{N}}$ is not tight if and only if there is a sequence of measurable sets $\{A_{j}\}_{j \in \mathbb{N}}$, a subsequence $(n_{j})_{j \in \mathbb{N}} \subseteq \mathbb{N}$, and a $\nu > 0$ such that
\begin{equation*}
\inf \left\{ \int_{A_{j}} f_{n_{j}}(x) \, \mu(dx) \, \mid \, j \in \mathbb{N} \right\} \geq \nu, \quad \lim_{j \to \infty} \mu(A_{j}) = 0.
\end{equation*}
Passing to a further subsequence, we see that we can assume that $\mu(A_{j}) \leq 2^{-j}$ for all $j \in \mathbb{N}$, hence $\sum_{j = 1}^{\infty} \mu(A_{j}) < \infty$.  Therefore, by the Borel-Cantelli Lemma, $\mu(\limsup_{j \to \infty} A_{j}) = 0$ (def. of lim-sup of sets can be found here), and an immediate consequence is
\begin{equation*}
\lim_{j \to \infty} f_{n_{j}} \chi_{X \setminus A_{j}} = f \quad \text{a.e.}
\end{equation*}
From this, we see that
\begin{equation*}
\int_{X} f(x) \, \mu(dx) \leq \liminf_{j \to \infty} \int_{X \setminus A_{j}} f_{n_{j}}(x) \, \mu(dx) \leq \int_{X} f(x) \, \mu(dx) - \nu.
\end{equation*}
This contradicts the choice of $\nu > 0$.
(As Ramiro pointed out, the non-negativity of $\{f_{n}\}_{n \in \mathbb{N}}$ isn't necessary.  The proof above is readily adapted to that setting.)
