# 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

**Fatou's Lemma:** Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n:X\to[0,\infty]\}$ a sequence of nonnegative measurable functions. Then the function $\displaystyle{ \liminf_{n\to\infty} f_n}$ is measureable and $$\int_X \liminf_{n\to\infty} f_n \;d\mu \;\; \leq \;\; \liminf_{n\to\infty} \int_X f_n\;d\mu .$$

### 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.