real analysis - Lebesgue integral of non-negative measurable sequence of functions (not monotone)

Suppose that $f, (f_n)$ are nonnegative measurable functions, that $f_n \to f$ pointwise, and that $f_n \leq f$ for all n. Prove that:

$\int f = \lim_{n \to \infty} \int f_n$

My attempt

One direction seems fairly obvious.
Since $f_n \leq f$ for all n, then:
$\int f_n \leq \int f$ for all n.

So we should have:
$\lim_{n \to \infty} \int f_n \leq \int f$

In the other direction, use Fatou’s Lemma to see that:

$\int f \leq \lim_{n \to \infty} \inf \int f_n$

However, it’s not actually clear that $\lim_{n \to \infty} \int f_n$ is well-defined, so it doesn’t necessarily make sense to get there from the $\lim_{n \to \infty} \inf$.

As a concept, my idea would then (or maybe instead?) create a subsequence from $(f_n)$ that is monotone and then invoke MCT? But I am not sure how to go about this.

Answer

Fatou's Lemma gives
\begin{align*}
\int f\leq\liminf\int f_{n}.
\end{align*}
From

\begin{align*}
f_{n}\leq f,
\end{align*}
we get
\begin{align*}
\int f_{n}\leq\int f,
\end{align*}
and hence
\begin{align*}
\limsup\int f_{n}\leq\int f.

\end{align*}
We conclude that
\begin{align*}
\int f\leq\liminf\int f_{n}\leq\limsup\int f_{n}\leq\int f,
\end{align*}
so the limit exists and
\begin{align*}
\lim\int f_{n}=\int f.
\end{align*}