# Fatou's Lemma

## The Basic Idea

Given a sequence of functions $\{f_n\}$ which converge pointwise to some limit function $f$, it is not always true that $$\int \lim_{n\to\infty}f_n = \lim_{n\to\infty}\int f_n.$$ (Take this sequence for example.) Fatou's Lemma, the Monotone Convergence Theorem (MCT), and the Dominated Convergence Theorem (DCT) are three major results in the theory of Lebesgue integration which answer the question "When do $\displaystyle{ \lim_{n\to\infty} }$ and $\int$ commute?" The MCT and DCT tell us that if  you place certain restrictions on both the $f_n$ and $f$, then you can interchange the limit and integral. On the other hand, Fatou's Lemma says, "Here's the best you can do if you don't put any restrictions on the functions."

Below, we'll give the formal statement of Fatou's Lemma as well as the proof. Then we'll look at an exercise from Rudin's Real and Complex Analysis (a.k.a "Big Rudin") which illustrates that the inequality in Fatou's Lemma can be a strict inequality.

## From English to Math

### Proof

For each $k\in\mathbb{N}$, let $g_k=\displaystyle{\inf_{n\geq k}f_n}$ and define $$h=\lim_{k\to\infty}g_k=\lim_{k\to\infty}\inf_{n\geq k}f_n=\liminf_{n\to\infty}f_n.$$

1st observation: $\int g_k \leq \int f_n$ for all $n\geq k$. This follows easily from the fact that for a fixed $x\in X$, $\displaystyle{\inf_{n\geq k}\{f_n(x)\}}\leq f_n(x)$ whenever $n\geq k$ (by definition of infimum). Hence $\int \displaystyle{\inf_{n\geq k} f_n} \leq \int f_n$ for all $n\geq k$, as claimed. This allows us to write  \begin{align}  \int g_k\leq \inf_{n\geq k}\int f_n. \qquad \qquad (1)  \end{align}

2nd observation: $\{g_k\}$ is an increasing sequence and $\displaystyle{\lim_{k\to\infty} g_k}=h$ pointwise. Thus, by the Monotone Convergence Theorem,  \begin{align*}  \int\liminf_{n\to\infty} f_n =\int h = \lim_{k\to\infty} \int g_k \leq \lim_{k\to\infty} \inf_{n\geq k}\int f_n = \liminf_{n\to\infty} \int f_n  \end{align*}  where the inequality in the middle follows from (1).

Finally, $\liminf f_n$ is measurable as we've proved before in the footnotes here.

### Exercise from Big Rudin

The following is taken from chapter 1 of Rudin's Real and Complex Analysis(Rudin, RCA, #1.8) Let $E\subset \mathbb{R}$ be Lebesgue measurable, and for $n\geq 0$ define $$f_n=\begin{cases} \chi_E &\text{if n is even};\\ 1-\chi_E &\text{if n is odd.} \end{cases}$$ What is the relevance of this example to Fatou's Lemma?  For simplicity, let's just consider what happens when $X=[0,2]\subset \mathbb{R}$ and we let $E=(1,2]\subset X$. Then we get the following sequence of functions $$f_n=\begin{cases} \chi_{(1,2]} &\text{if n is even};\\ \chi_{[0,1]} &\text{if n is odd}. \end{cases}$$ The first few of these functions look like this:

Notice that as $n$ increases, the graphs switch back and forth. For any given $n$, $$\int_{[0,2]}f_n=1$$ but $\liminf_nf_n=0$. (Recall that $\liminf_n f_n$ is the infimum of all subsequential limits of $\{f_n\}$). This shows us that $$0=\int_{[0,2]} \liminf_{n\to\infty} f_n < \liminf_{n\to\infty}\int_{[0,2]} f_n=1$$ proving that a strict inequality in Fatou's Lemma is possible.

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