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 $E$. \begin{equation} \textit{If}~\{f_n\}\rightarrow f~\textit{pointwise a.e. on}~E,~\textit{then}~\int_E{f}\leq \liminf\int_E{f_n}. \end{equation}

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 \begin{equation} \int_E h \leq \liminf \int_E f_n. \end{equation} 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 \begin{equation} h_n = \min\left\{h,f_n\right\}~\text{on}~E. \end{equation} Observe that the function $h_n$ is measurable, that \begin{equation} 0\leq h_n \leq M~\text{on}~E_0~\text{and}~h_n \equiv 0~\text{on}~E\sim E_0. \end{equation} 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 \begin{equation} \lim_{n\rightarrow \infty}\int_E{h_n} = \lim_{n\rightarrow \infty}\int_{E_0}{h_n} = \int_{E_0}{h} = \int_{E}{h}. \end{equation} 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, \begin{equation} \int_E{h} = \lim_{n\rightarrow \infty} \int_{E}{h_n} \leq \liminf \int_E f_n. \end{equation}

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 \begin{equation} \int_E{h} = \lim_{n\rightarrow \infty} \int_{E}{h_n} \leq \lim_{n\rightarrow \infty} \int_E f_n. \end{equation}

Where does the $\liminf$ come from?

  • 2
    $\begingroup$ 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. $\endgroup$ – copper.hat Oct 18 '18 at 19:55
  • $\begingroup$ @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? $\endgroup$ – jodag Oct 18 '18 at 19:59
  • $\begingroup$ 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? $\endgroup$ – jodag Oct 18 '18 at 20:06
  • $\begingroup$ 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. $\endgroup$ – copper.hat Oct 18 '18 at 20:07
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    $\begingroup$ @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! $\endgroup$ – jodag Oct 18 '18 at 20:21

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