Fatou's Lemma for Integrals

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\left({f_n}\right)_{n \in \N} \in \mathcal{M}_{\overline{\R}}^+$, $f_n: X \to \overline{\R}$ be a sequence of positive measurable functions.

Let $\displaystyle \liminf_{n \to \infty} f_n: X \to \overline{\R}$ be the pointwise limit inferior of the $f_n$.

Then:


 * $\displaystyle \int \liminf_{n \to \infty} f_n \, \mathrm d\mu \le \liminf_{n \to \infty} \int f_n \, \mathrm d\mu$

where:


 * the integral sign denotes $\mu$-integration; and
 * the right-hand side limit inferior is taken in the extended real numbers $\overline{\R}$.

Proof
We consider the sequence $\inf_{m>n}f_m$. Its is a monotone increasing sequence and:
 * $\inf_{m>n}f_m\leq f_k$

for $k>n$. We have then:
 * $\displaystyle \int\inf_{m>n}f_md\mu\leq \int f_k d\mu$

for $k>n$, and because of that:
 * $\displaystyle \int\inf_{m>n}f_md\mu\leq \inf_{m>n}\int f_m d\mu$.

Now we take the limit $n\rightarrow\infty$ and use the monotone convergence theorem to exchange integral and limit on the left side of the equation and obtain:
 * $\displaystyle \int\liminf_{m>n}f_md\mu\leq \liminf_{m>n}\int f_m d\mu$.

Also see

 * Reverse Fatou's Lemma