1
$\begingroup$

Fatou's Lemma is

Let $(\Omega, \Sigma, \mu)$ be a measure space and $X \in \Sigma$. Let $f_n; f_n : X \to [0, +\infty]$ be a sequence of $(\Sigma, \mathcal{B}_{\mathbb{R}_{\geq 0}})$-measurable functions. Define $f(x) = \lim\inf_{n\to\infty} f_n$. Then $f$ is $(\Sigma, \mathcal{B}_{\mathbb{R}_{\geq 0}})$-measurable and

$\begin{align} \int_X f d\mu \leq \liminf_{n\to\infty} \int_X f_n d\mu \end{align}$

  • Is this true if you generalize the codomain of $f$ and $f_n$ to be all of $\mathbb{R}$ and $(\Sigma, \mathcal{B}_{\mathbb{R}})$-measurable?

  • Is this true if you replace $\liminf$ with $\limsup$?

Provide counterexamples please if possible. No this is not a homework question.

$\endgroup$
3
  • 1
    $\begingroup$ For second question, search for reverse Fatou's lemma $\endgroup$ – user608030 Oct 25 '18 at 4:53
  • $\begingroup$ Zachary Selk your answer was extremely useful. Thank you. $\endgroup$ – kakashi10192020 Oct 25 '18 at 4:55
  • 1
    $\begingroup$ The first question is equivalent to the second. Just replace the functions in the counterexample to second question by their negatives. $\endgroup$ – Kavi Rama Murthy Oct 25 '18 at 5:34
3
$\begingroup$

The answers to both questions are no. Here's a counterexample for both cases.

Let $X = \mathbb{R}$. Let $f_{n}(x)=-1$ if $x \in [n, n+1)$, and $0$ otherwise. Then $$ 0=\int f d\mu > \liminf_{n} \int_{n} f_{n} d\mu = \limsup_{n} \int f_{n} d\mu =-1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.