TITLE: If $f_n \geq 0$ and $f_n \to f$ in measure, then $\int f \leq \liminf \int f_n$
QUESTION [9 upvotes]: I am reading Folland's, Real Analysis and I am stuck at the following exercise (2.33).

If $f_n \geq 0$ and $f_n \to f$ in measure, then $\int f \leq \liminf \int f_n$

It smells like Fatou's, however I couldn't see anything. Thanks in advance!

REPLY [16 votes]: By the definition of $\liminf$, there is a subsequence $f_{n_k}$ of $f_n$ such that:
$$
\lim_{k \to \infty} \int f_{n_k} = \liminf_{n \to \infty} \int f_n
$$
$f_{n_k}$ also converges in measure to $f$. To see why, notice that the sequence $\mu\left(\left\{ x: \left| f_n(x) - f(x) \right|\geq \epsilon \right\}\right)$ converges to $0$ as $n \to \infty$. Thus, its subsequence $\mu\left(\left\{ x: \left| f_{n_k}(x) - f(x) \right|\geq \epsilon \right\}\right)$ also converges to $0$ as $k \to \infty$. This is a basic result in sequence convergence.
By theorem 2.30 in Folland's book, there is a subsequence of $f_{n_k}$ that converges pointwise a.e. to $f$. Call it $f_{n_{k_j}}$. Notice that it also satisfies:
$$
\lim_{j \to \infty} \int f_{n_{k_j}} = \liminf_{n \to \infty} \int f_n
$$
Apply Fatou's lemma to $\{f_{n_{k_j}}\}$, and put it all together to find that:
$$
\int f = \int \lim_{j \to \infty} f_{n_{k_j}} \le \lim_{j \to \infty} \int f_{n_{k_j}} = \liminf_{n \to \infty} \int f_n
$$