5
$\begingroup$

Let $g$ be a non-negative integrable function over $E$ and suppose $\{f_n\}$ is a sequence of measurable functions on $E$ such that for each $n$, $|f_n| \leq g$ a.e. on $E$. Show that $$ \int \liminf f_n \leq \liminf \int f_n \leq \limsup \int f_n \leq \int \limsup f_n.$$

I know that this problem is an application of the Lebesgue dominated convergence theorem.

Any idea of how to go about it thanks, I am really having a hard time with this problem.

$\endgroup$
0

3 Answers 3

7
$\begingroup$

By possibly excising a set of measure 0 we can assume that $|f_n| \leq g$ holds on $E$.\ Let $$g_n =\inf\limits_{k\geq n} f_k \leq f_n \ then \ g_n \rightarrow \lim\limits_{n \to \infty} \inf f_n$$ Note that from $-g \leq |f_n| \leq g$ for all $n$ it also follows that $-g \leq g_n \leq g$ for all $n$ and thus $|g_n| \leq g$\ Using LDCT it follows that $$\int\limits_{E} \lim\limits_{n \to \infty} \inf f_n = \lim\limits_{n \to \infty} \int\limits_{E} g_n = \lim\limits_{n \to \infty} inf \int\limits_{E} g_n \leq \lim\limits_{n \to \infty} \int\limits_{E} f_n$$\ Also $$\int\limits_{E} \lim\limits_{n \to \infty} inf f_n \leq \lim\limits_{n \to \infty} \int\limits_{E} f_n \leq \int\limits_{E} \lim\limits_{n \to \infty} \sup f_n \ \ (*)$$ Similarly,Let $h_n =\sup\limits_{k \geq n} f_k \geq f_n$ then $h_n \rightarrow \lim\limits_{n \to \infty} \sup f_n$ and note that $|h_n| \leq g$\ Again, using LDCT we get $$\int\limits_{E} \lim\limits_{n \to \infty} \sup f_n = \lim\limits_{n \to \infty} \int\limits_{E} h_n = \lim\limits_{n \to \infty} \sup \int\limits_{E} g_n \geq \lim\limits_{n \to \infty} \int\limits_{E} f_n$$ Also $$\int\limits_{E} \lim\limits_{n \to \infty} \sup f_n \geq \lim\limits_{n \to \infty} \int\limits_{E} f_n \geq \int\limits_{E} \lim\limits_{n \to \infty} \inf f_n \ \ (**)$$ From (*) and (**) we have: $$\int\limits_{E} \lim\limits_{n \to \infty} inf f_n \leq \lim\limits_{n \to \infty} inf \int\limits_{E} f_n \leq \lim\limits_{n \to \infty} \sup \int\limits_{E} f_n \leq \int\limits_{E} \lim\limits_{n \to \infty} \sup f_n$$

$\endgroup$
2
$\begingroup$

Hint: Apply Fatou's Lemma to the sequences $g + f_n, g- f_n.$

$\endgroup$
6
  • $\begingroup$ Would you be able to elaborate a bit further? Once Fatou's Lemma is applied to $g\pm f_n$, we get $\int_E\liminf_{n\to\infty} (g\pm f_n)\leq \liminf_{n\to\infty} \int_E (g\pm f_n)$ and $\limsup_{n\to\infty}\int_E (g+f_n)\leq \int_E\limsup_{n\to\infty} (g+f_n)$, but how to proceed from there? $\endgroup$ Aug 2, 2016 at 16:39
  • $\begingroup$ You just need to verify that if $a_n$ is a sequence in $\mathbb R,$ then $\liminf (c+a_n) = c +\liminf a_n,$ where $c$ is a constant. $\endgroup$
    – zhw.
    Aug 2, 2016 at 17:18
  • $\begingroup$ $\int \liminf (g\pm f_n)=\int( g\pm \liminf f_n)=\int g\pm\int \liminf f_n$ It is correct? $\endgroup$
    – eraldcoil
    Dec 8, 2018 at 4:30
  • $\begingroup$ @eraldcoil Yes. $\endgroup$
    – zhw.
    Dec 8, 2018 at 16:27
  • $\begingroup$ A doubt! $(f_n)$ is measurable but $g$ is only integrable. Why g is measurable? We need$ (g\pm f_n)$ to be measurable in order to use Fatou's lemma ... Integrable implies measurable??? $\endgroup$
    – eraldcoil
    Dec 11, 2018 at 13:30
0
$\begingroup$

We can assume that $|f_n| \leq g$.

let $g_n= $inf $f_k < f_n $. $g_n \rightarrow$ lim inf $f_n$. Since $-g < f_n < g$ for all $n$, it follows that $-g < g_n < g$ thus $|g_n| \leq g$. Now by the lebesgue dominated convergence theorem we get $\int_{E}$ lim inf $f_n =$ lim $\int g_n =$ lim inf $\int_{E} g_n < $ lim inf $\int_{E} f_n$. by the monotonicity of the integral thus we have $\int_{E}$ lim inf $f_n \leq $ lim inf $\int_E f_n \leq$ lim sup $\int_E f_n$

$\endgroup$

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.