Fatou’s Lemma

Let $f$ be an integrable function on a measure space; let $g_n$ be a sequence of functions on the same space.

• If $g_n$ are lower bounded by $f$, i.e., $g_n\ge f$ for all $n$, then $\int \mathrm{liminf}{g}_n\mathrm{d}\mu \le \mathrm{liminf}\int g_n\mathrm{d}\mu$;
• If $g_n$ are upper bounded by $f$, i.e., $g_n\le f$ for all $n$, then $\int \mathrm{limsup}{g}_n\mathrm{d}\mu \ge \mathrm{limsup}\int g_n\mathrm{d}\mu$.

Proof

We only prove the first statement. Fixing $n\in \mathbb{N}$, for any $m\ge n$, we have

$$\underset{k\ge n}{\inf }{g}_k-f\le g_m-f.$$

Taking integrals gives

$$\int \underset{k\ge n}{\inf }{g}_k-f\mathrm{d}\mu \le \int g_m-f\mathrm{d}\mu .$$

Taking infimum over $m\ge n$ gives

$$\int \underset{k\ge n}{\inf }{g}_k-f\mathrm{d}\mu \le \underset{m\ge n}{\inf }\int g_m-f\mathrm{d}\mu .$$

Since the integrand on the LHS is nonnegative, by MCT, taking limit over $n$ gives

$$\underset{n}{\lim }\int \underset{k\ge n}{\inf }{g}_k-f\mathrm{d}\mu =\int \underset{n}{\mathrm{liminf}}{g}_n-f\mathrm{d}\mu \le \underset{n}{\mathrm{liminf}}\int g_n-f\mathrm{d}\mu .$$

By the linearity of integral, we have

$$\int \mathrm{liminf}{g}_n\mathrm{d}\mu \le \mathrm{liminf}\int g_n\mathrm{d}\mu .$$

Counter Example

Let $X_n=1_{[n,n+1]}$ or $X_n=1_{[0,1/n]}\cdot n$. Then ${\mathrm{liminf}}_{n\to \infty }{X}_n=0$ but $\mathbb{E}[X_n]=1$ for all $n$.