# Fatou's Lemma proof from Royden 4th

I'm having a hard time understanding the proof presented by Royden, Fitzpatrick in Real Analysis 4th edition. The proof is described as follows...

Fatou's Lemma Let $$\{f_n\}$$ be a sequence of nonnegative measurable functions on $$E$$. $$$$\textit{If}~\{f_n\}\rightarrow f~\textit{pointwise a.e. on}~E,~\textit{then}~\int_E{f}\leq \liminf\int_E{f_n}.$$$$

Proof In view of addititivity over domains of integration, by possibly excising from $$E$$ a set of measure zero, we assume the pointwise convergence is on all of $$E$$. The function $$f$$ is nonnegative and measurable since it is the pointwise limit of a sequence of such functions. To verify the inequality in Fatou's Lemma it is necessary and sufficient to show that if $$h$$ is any bounded measurable function of finite support for which $$0\leq h \leq f$$ on $$E$$, then $$$$\int_E h \leq \liminf \int_E f_n.$$$$ Let $$h$$ be such a function. Choose $$M\geq 0$$ for which $$\lvert h \rvert\leq M$$ on $$E$$. Define $$E_0 = \left\{x \in E~\middle|~h(x)\neq 0 \right\}$$. Then $$m(E_0) < \infty$$. Let $$n$$ be a natural number. Define a function $$h_n$$ on $$E$$ by $$$$h_n = \min\left\{h,f_n\right\}~\text{on}~E.$$$$ Observe that the function $$h_n$$ is measurable, that $$$$0\leq h_n \leq M~\text{on}~E_0~\text{and}~h_n \equiv 0~\text{on}~E\sim E_0.$$$$ Furthermore, for each $$x$$ in $$E$$, since $$h(x) \leq f(x)$$ and $$\left\{f_n(x)\right\} \rightarrow f(x)$$, $$\left\{h_n(x)\right\} \rightarrow h(x)$$. We infer from the Bounded Convergence Theorem applied to the uniformly bounded sequence of restrictions to $$h_n$$ to the set of finite measure $$E_0$$, and the vanishing of each $$h_n$$ on $$E \sim E_0$$, that $$$$\lim_{n\rightarrow \infty}\int_E{h_n} = \lim_{n\rightarrow \infty}\int_{E_0}{h_n} = \int_{E_0}{h} = \int_{E}{h}.$$$$ However, for each $$n$$, $$h_n \leq f_n$$ on $$E$$ and therefore, by the definition of the integral of $$f_n$$ over $$E$$, $$\int_E{h_n} \leq \int_E{f_n}$$. Thus, $$$$\int_E{h} = \lim_{n\rightarrow \infty} \int_{E}{h_n} \leq \liminf \int_E f_n.$$$$

So I understand everything up until the final conclusion. It seems that the conclusion I reach from the inequality $$\int_E h_n \leq \int_E f_n$$ is that $$$$\int_E{h} = \lim_{n\rightarrow \infty} \int_{E}{h_n} \leq \lim_{n\rightarrow \infty} \int_E f_n.$$$$

Where does the $$\liminf$$ come from?

• Suppose you have $a_n \le b_n$ and $a_n \to a$. All you can say is that $a \le \liminf_n b_n$. It may be the case that $b_n$ does not have a limit. – copper.hat Oct 18 '18 at 19:55
• @copper.hat Interesting, that makes a lot of sense. So the $\liminf$ statement is actually weaker. Just as a quick followup, we know that $\liminf b_n$ converges (possibly to infinity) by monotone convergence right? – jodag Oct 18 '18 at 19:59
• Lets say we restricted Fatou's lemma such that the sequence $\left\{\int_E f_n\right\}$ converges to some extended real value. Then would it be fair to replace the $\liminf$ with $\lim$ in Fatou's Lemma? I guess what I'm trying to ask is, is the $\liminf$ only there to account for the case where $\lim_{n\rightarrow \infty} \int_E f_n$ doesn't exist? – jodag Oct 18 '18 at 20:06
• Well, given any sequence $b_n$ we can compute the (possibly extended value) $\liminf_n b_n$, it has nothing to so with monotone convergence as such. – copper.hat Oct 18 '18 at 20:07
• @copper.hat I realized that "monotone convergence" was ambiguous. I was referring to the convergence of monotone sequences which I can see should have been explicit especially in the context of Lebesgue integration! – jodag Oct 18 '18 at 20:21