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.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

Blog at

%d bloggers like this: