Proof of Fatou's Lemma

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[Fatou's Lemma.]  If $|X_n|\geq0$ a.e. on $\Lambda$, then $$\int_\Lambda\underset{n\rightarrow\infty}{\liminf}X_n\,d\mathscr{P}\leq\underset{n\rightarrow\infty}{\liminf}\int_\Lambda X_n\,d\mathscr{P}.$$Furthermore, if for all $n$, $|X_n|\leq Y$ a.e. on $\Lambda$ with $\mathscr{E}(Y)<\infty$, the above remains true as well as $$\int_\Lambda\underset{n\rightarrow\infty}{\limsup}X_n\,d\mathscr{P}\geq\underset{n\rightarrow\infty}{\limsup}\int_\Lambda X_n\,d\mathscr{P}.$$In the second statement, it would be false if the condition involving $Y$ is omitted.

$\bullet$ Proof.

For the first inequality, we use Monotone Convergence Theorem as following.

[Theorem] Monotone Convergence Theorem

If $X_n\geq0$ and $X_n\uparrow X$ a.e. on $\Lambda$, then $$\underset{n\rightarrow\infty}{\lim}\int_\Lambda X_n\,d\mathscr{P}=\int_\Lambda X\,d\mathscr{P}=\int_\Lambda\underset{n\rightarrow\infty}{\lim}X_n\,d\mathscr{P}.$$See Convergence Theorems

Back to the proof.  Let $Y_m=\underset{n\geq m}{\inf}X_n$.  Since $X_n\geq0$, we have $$0\leq Y_1\leq Y_2\cdots\uparrow.$$and then $$\underset{m\rightarrow\infty}{\lim}Y_m=\underset{m\geq1}{\sup}Y_m=\underset{m\geq1}{\sup}\underset{n\geq m}{\inf}X_n=\underset{n\rightarrow\infty}{\liminf}X_n.$$Thus, by Monotone Convergence Theorem, we have $Y_m\uparrow$ and $Y_m\rightarrow\underset{n\rightarrow\infty}{\liminf}X_n\quad\mbox{a.e.}$, thus $$\underset{m\rightarrow\infty}{\lim}\mathscr{E}(Y_m)=\mathscr{E}(\underset{m\rightarrow\infty}{\lim}Y_m)=\mathscr{E}(\underset{n\rightarrow\infty}{\liminf}X_n).$$Since for all $m$, $Y_m\leq X_m$, then $$\mathscr{E}(\underset{n\rightarrow\infty}{\liminf}X_n)=\underset{n\rightarrow\infty}{\lim}\mathscr{E}(Y_n)\leq\underset{n\rightarrow\infty}{\lim}\inf\mathscr{E}(Y_n)\leq\underset{n\rightarrow\infty}{\liminf}\mathscr{E}(X_n).$$

For the second inequality, let $Z_n=Y-X_n$, then $Z_n\geq0$ on $\Lambda$.  Since $\int_\Lambda Yd\mathscr{P}<\infty$, we have $$\begin{array}{rl} \int_\Lambda\underset{n\rightarrow\infty}{\underline{\lim}}Z_nd\mathscr{P}\leq\underset{n\rightarrow\infty}{\underline{\lim}}\int_\Lambda Z_nd\mathscr{P} &   \Rightarrow \int_\Lambda\underset{n\rightarrow\infty}{\underline{\lim}}Y-X_nd\mathscr{P}\leq\underset{n\rightarrow\infty}{\underline{\lim}}\int_\Lambda Y-X_nd\mathscr{P} \\   & \Rightarrow \int_\Lambda-\underset{n\rightarrow\infty}{\underline{\lim}}X_nd\mathscr{P}\leq-\underset{n\rightarrow\infty}{\underline{\lim}}\int_\Lambda X_nd\mathscr{P} \\   & \Rightarrow \int_\Lambda\underset{n\rightarrow\infty}{\overline{\lim}}X_nd\mathscr{P}\geq\underset{n\rightarrow\infty}{\overline{\lim}}\int_\Lambda X_nd\mathscr{P}. \end{array}$$

$\Box$

$\bullet$ Counterexample.  If $\int_\Lambda Y\,d\mathscr{P}<\infty$ does not hold, the conclusion about $\limsup$ fails.  Here is a counterexample.  Define $X_n:\,\mathbb{R}^+\rightarrow\mathbb{R}$ and $$X_n=\begin{cases}n,&\mbox{if }\omega\in(0,\frac{1}{n})\\0,&\mbox{o.w.}\end{cases},\,\mbox{ for }n\geq1.$$Then $$X_n\leq\underset{1\leq k\leq n}{\sup}X_k=\begin{cases}k,&\mbox{if }\omega\in[\frac{1}{k+1},\frac{1}{k})\\0,&\mbox{o.w.}\end{cases},\,\mbox{ for }1\leq k\leq n.$$But, $$\int_\mathbb{R} \underset{1\leq k\leq n}{\sup}X_k\,d\mathscr{P}=\sum_{k=1}^nk\left[\frac{1}{k}-\frac{1}{k+1}\right]=\sum_{k=1}^n\frac{1}{k+1}\rightarrow\infty\mbox{ as }n\rightarrow\infty.$$Thus $\{X_n\}$ is unbounded.

Then consider $$\underset{n\rightarrow\infty}{\overline{\lim}}\left(0,\frac{1}{n}\right)=\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty\left(0,\frac{1}{m}\right)=\bigcap_{n=1}^\infty\left(0,\frac{1}{n}\right)=\emptyset.$$Which implies $\underset{n\rightarrow\infty}{\overline{\lim}}X_n=0$ for all $\omega$.  For all $n$, we know that $\int_\mathbb{R}X_n\,d\mathscr{P}=n\frac{1}{n}=1$, hence $$\int_\mathbb{R}\underset{n\rightarrow\infty}{\overline{\lim}}X_n\,d\mathscr{P}=\int_\mathbb{R}0\,d\mathscr{P}=0<1=\underset{n\rightarrow\infty}{\overline{\lim}}\int_\mathbb{R}X_n\,d\mathscr{P}.$$Summarizing above, if $\{X_n\}$ is unbounded, i.e. $\int_\Lambda Y\,d\mathscr{P}<\infty$ does not hold, then the second statement would be false.

$\Box$