Fatou's Lemma

Fatou's lemma is a fundamental result in measure theory that deals with the relationship between limits and integrals of sequences of non-negative measurable functions. See the wikipedia page for further info.

Statement of Fatou's Lemma

Let $(f_n)$ be a sequence of non-negative measurable functions on a measure space $(X,\mu )$. Then:

$\int {\mathrm{liminf}}_{n\to \infty }{f}_{n}d\mu \le {\mathrm{liminf}}_{n\to \infty }\int f_{n}d\mu$

In words, this means that the integral of the limit inferior of a sequence of functions is less than or equal to the limit inferior of their integrals.

Preliminaries

To attack this proof, we first need to understand some basic results Recall that ${\mathrm{liminf}}_{n\to \infty }{f}_n(x)$ is defined as: ${\mathrm{liminf}}_{n\to \infty }{f}_n(x)={\sup }_{n\ge 1}{\inf }_{k\ge n}{f}_k(x)$

First, let's understand what we want liminf to capture. Informally, the liminf should be the "lowest value that the sequence eventually stays above."

Let's break this down:

1. For a fixed $n$, consider ${\inf }_{k\ge n}{f}_k(x)$ This represents the lowest value the sequence takes from position $n$ onwards.

2. Now, why do we take ${\sup }_{n\ge 1}$ of these infimums? Let's see with an example:

Consider the sequence: $1,0,2,0,3,0,4,0,5,0,...$

Let's compute ${\inf }_{k\ge n}{f}_k(x)$ for different $n$:

• For $n=1$: $\inf \{1,0,2,0,3,0,...\}=0$
• For $n=2$: $\inf \{0,2,0,3,0,...\}=0$
• For $n=3$: $\inf \{2,0,3,0,...\}=0$ And so on...

Notice that no matter how far out we go, there will always be zeros, so each infimum is 0.

In contrast, consider: $0,0,1,1,1,...$

Here:

• For $n=1$: $\inf \{0,0,1,1,1,...\}=0$
• For $n=2$: $\inf \{0,1,1,1,...\}=0$
• For $n=3$: $\inf \{1,1,1,...\}=1$
• For $n=4$: $\inf \{1,1,...\}=1$

By taking ${\sup }_{n\ge 1}$ of these infimums, we capture the value that the sequence eventually stays above (1 in this case).

The key insights are:

1. ${\inf }_{k\ge n}{f}_k(x)$ gives us a lower bound for all terms from position $n$ onwards
2. As $n$ increases, these lower bounds might increase (they can't decrease, as we're looking at a subset of the previous terms)
3. Taking the supremum of all these lower bounds gives us the highest lower bound that's valid "eventually"

This is exactly what we want liminf to be: the highest value that eventually serves as a lower bound for the sequence.

First, let's recall that for any sequence $(f_n)$: ${\mathrm{liminf}}_{n\to \infty }{f}_n(x)={\lim }_{n\to \infty }\inf \{f_k(x):k\ge n\}$

Proof:

Let's denote $a_n=\inf \{f_k(x):k\ge n\}$ for fixed $x$. We need to prove:

${\mathrm{liminf}}_{n\to \infty }{f}_n(x)={\lim }_{n\to \infty }{a}_n$

We use the definition of

$$\underset{n\to \infty }{\mathrm{liminf}}{f}_n(x)=\underset{n\ge 1}{\sup }\underset{k\ge n}{\inf }{f}_k(x)$$

Now, let's prove both inequalities:

4. First, let's show $(a_n)$ is increasing: For $m>n$: $a_m=\inf \{f_k(x):k\ge m\}\ge \inf \{f_k(x):k\ge n\}=a_n$ This is because we're taking the infimum over a smaller set.

5. Since $(a_n)$ is increasing, ${\lim }_{n\to \infty }{a}_n$ exists by monotone convergence theorem and: ${\lim }_{n\to \infty }{a}_n={\sup }_{n\ge 1}{a}_n$

6. But $a_n={\inf }_{k\ge n}{f}_k(x)$, so: ${\lim }_{n\to \infty }{a}_n={\sup }_{n\ge 1}{\inf }_{k\ge n}{f}_k(x)$

7. And this is exactly the definition of ${\mathrm{liminf}}_{n\to \infty }{f}_n(x)$

Therefore: ${\mathrm{liminf}}_{n\to \infty }{f}_n(x)={\sup }_{n\ge 1}{\inf }_{k\ge n}{f}_k(x)={\lim }_{n\to \infty }\inf \{f_k(x):k\ge n\}$

The key insight here is that $(a_n)$ being increasing means its limit exists and equals its supremum. This increasing property comes from taking infimums over progressively smaller sets as $n$ increases.

Proof

1. We first start from the lemma above.

2. For each $n$, define $g_n(x)=\inf \{f_k(x):k\ge n\}$ Note that $g_n(x)\le f_n(x)$ for all $x$ and $n$

3. The sequence $(g_n)$ is increasing: For any $m>n$, $g_m(x)=\inf \{f_k(x):k\ge m\}\ge \inf \{f_k(x):k\ge n\}=g_n(x)$

4. By definition: ${\mathrm{liminf}}_{n\to \infty }{f}_n(x)={\lim }_{n\to \infty }{g}_n(x)$

5. Since $(g_n)$ is increasing, by the Monotone Convergence Theorem: $\int {\mathrm{liminf}}_{n\to \infty }{f}_{n}d\mu =\int {\lim }_{n\to \infty }{g}_{n}d\mu ={\lim }_{n\to \infty }\int g_{n}d\mu$

6. For each $n$: $\int g_{n}d\mu \le \int f_{n}d\mu$ (because $g_n(x)\le f_n(x)$ for all $x$, and the integral preserves inequalities)

7. Therefore: ${\lim }_{n\to \infty }\int g_{n}d\mu \le {\mathrm{liminf}}_{n\to \infty }\int f_{n}d\mu$

8. Combining steps 5 and 7: $\int {\mathrm{liminf}}_{n\to \infty }{f}_{n}d\mu ={\lim }_{n\to \infty }\int g_{n}d\mu \le {\mathrm{liminf}}_{n\to \infty }\int f_{n}d\mu$

This completes the proof.

Key Intuition

The lemma essentially states that when taking limits of integrals, you can't "lose" area in the limit. The integral of the limit inferior provides a lower bound for the limit inferior of the integrals.