# Mathematics Prelims

## July 20, 2008

### Fatou’s Lemma, Part I

Filed under: Analysis,Measure Theory — cjohnson @ 4:38 pm

If $E \subset \mathbb{R}$ is a Lebesgue measurable set and $(f_n : E \to \mathbb{R})_{n \in \mathbb{N}}$ is a sequence of non-negative measurable functions, then

$\displaystyle \liminf_{n \to \infty} \int_E f_n dm \geq \int_e \liminf_{n \to \infty} f_n dm$

Proof: Let $g_n = \inf_{k \geq n} f_k$ and $f = \liminf_{n \to \infty} f_n = \lim_{n \to \infty} g_n$.  Let $\phi$ be a simple function with $\phi \leq f$.   Assume $f > 0$ on $E$, as the case where $f = 0$ is trivial.  Now define $\overline{\phi}$ as follows.

$\overline{\phi}(x) = \left\{ \begin{array}{lr} \phi(x) - \epsilon &: \phi(x) > 0 \\ 0 &: \phi(x) = 0 \end{array} \right.$

Where $\epsilon > 0$ is a value that ensures $\overline{\phi} \geq 0$.  Now note that $\overline{\phi} < f$ and $g_n \nearrow f$, so there must exist an $N \in \mathbb{N}$ such that $g_n \geq \overline{\phi}$ for all $n > N$.

Let $A_k = \{ x \in E : g_k(x) \geq \overline{\phi}(x) \}$ and notice that $A_k \subseteq A_{k+1}$.  Also, $\bigcup_{k=1}^\infty A_k = E$.  For $k \geq n$ we have the following.

$\displaystyle \int_{A_n \cap E} \overline{\phi} \, dm \leq \int_{A_n \cap E} g_n \, dm$

$\displaystyle \qquad \leq \int_{A_n \cap E} f_k \, dm$

$\displaystyle \qquad \leq \int_E f_k \, dm$

So we have

$\displaystyle \int_{A_n \cap E} \overline{\phi} \, dm \leq \liminf_{k \to \infty} \int_E f_k \, dm$

Now notice that $\overline{\phi}$ is a simple function, and so we can write it as

$\displaystyle \overline{\phi} = \sum_{i=1}^\ell c_i 1_{B_i}$

Now we have

$\displaystyle \int_{A_n \cap E} \overline{\phi} \, dm = \sum_{i=1}^\ell c_i m(A_n \cap E \cap B_i)$

Letting $n \to \infty$ this gives us

$\displaystyle \int_E \overline{\phi} \, dm = \sum_{i=1}^\ell c_i m(E \cap B_i)$

This leaves us with

$\displaystyle \int_E \overline{\phi} \, dm \leq \liminf_{k \to \infty} \int_E f_k \, dm$

If $m(\{ x \in E : \phi(x) > 0 \}) < \infty$, then we have

$\int_E \overline{\phi} \, dm = \int_E \phi \, dm - \epsilon \, m(\{ x \in E : \phi (x) > 0 \})$

This gives the desired result if we let $\epsilon \to 0$ as $\phi$ is an arbitrary simple function with $\phi \leq f$.

Note: This is an awesomely useful lemma, but I hate this proof.