Proof of Fatou's Lemma

For $k\in\mathbb{N}$ define $g_k:X\to[0,\infty]$ by $g_k(x)=\inf_{m\ge k}f_m(x)$. Each $g_k$ is measurable in the sense of Lebesgue Integral of a Nonnegative Measurable Function: for $a\in\mathbb{R}$,

$$\{g_k\ge a\}=\bigcap_{m\ge k}\{f_m\ge a\},\qquad\text{with }\{f_m\ge a\}=\bigcap_{j\in\mathbb{N}}\{f_m>a-1/j\}\in\mathcal{F},$$

so $\{g_k\ge a\}\in\mathcal{F}$ by closure of the $\sigma$-algebra under countable intersections, and $\{g_k>a\}=\bigcup_{j\in\mathbb{N}}\{g_k\ge a+1/j\}\in\mathcal{F}$.

The sequence $(g_k)_k$ is nondecreasing pointwise (the infimum is over a smaller index set as $k$ grows), and by definition $\sup_k g_k=\liminf_m f_m$ pointwise. By Monotone Convergence Theorem, $\liminf_m f_m$ is measurable and

$$\int_X\Bigl(\liminf_m f_m\Bigr)\,d\mu=\sup_k\int_X g_k\,d\mu.$$

For every $m\ge k$ we have $g_k\le f_m$ pointwise, so by monotonicity of the nonnegative integral (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) $\int_X g_k\,d\mu\le\int_X f_m\,d\mu$; taking the infimum over $m\ge k$,

$$\int_X g_k\,d\mu\ \le\ \inf_{m\ge k}\int_X f_m\,d\mu.$$

Taking the supremum over $k$ and using the definition of $\liminf$ for sequences in $[0,\infty]$ given in the statement,

$$\int_X\Bigl(\liminf_m f_m\Bigr)\,d\mu=\sup_k\int_X g_k\,d\mu\ \le\ \sup_k\ \inf_{m\ge k}\int_X f_m\,d\mu=\liminf_m\int_X f_m\,d\mu.\qquad\blacksquare$$