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Problem of the Week #1 - June 4th, 2012

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Chris L T521

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Jan 26, 2012
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Hello everyone! Welcome to the inaugural POTW for Graduate Students. My purpose for setting this up is to get some of our more advanced members to participate in our POTWs (I didn't want them to feel like they were left out or anything like that (Smile)).

As with the POTWs for the Secondary/High School and University students, Jameson and I will post a problem each Monday around 12:00 AM Eastern Standard Time (EST), and you'll have till Saturday at 11:59 PM EST to submit your solutions. With that said, let's get this started!! (Smile)

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Problem: Let $g$ be a Lebesgue integrable function on a measurable set $E\subset\mathbb{R}$ and suppose that $\{f_n\}$ is a sequence of measurable functions such that $|f_n(x)|\leq g(x)$ $m$-a.e. on $E$. Show that

\[\int_E \liminf_{n\to\infty}f_n\,dm \leq \liminf_{n\to\infty}\int_E f_n\,dm \leq \limsup_{n\to\infty}\int_E f_n\,dm \leq \int_E \limsup_{n\to\infty}f_n\,dm.\]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one tried this problem. :-/


Here's my solution.


Proof: Let $g$ be an integrable function, and let $\{f_n\}$ be a sequence of measurable functions with $|f_n|\leq g$ $m$-a.e. on $E$. Then $\{f_n+g\}$ is a sequence of nonnegative functions on $E$. Thus, by Fatou's Lemma, we have


\[\int_E \liminf_{n\to\infty} f_n\,dm + \int_E g\,dm \leq \int_E\liminf_{n\to\infty}(f_n+g)\,dm \leq \liminf_{n\to\infty}\int_E (f_n+g)\,dm \leq \liminf_{n\to\infty} \int_E f_n\,dm + \int_E g\,dm.\]


Thus, $\displaystyle\int_E\liminf_{n\to\infty} f_n\,dm \leq \liminf_{n\to\infty}\int_E f_n\,dm$.


Similarly, $\{g-f_n\}$ is a sequence of nonnegative measurable functions on $E$. Therefore,


\[\int_E g\,dm + \int_E\liminf_{n\to\infty}(-f_n)\,dm \leq \int_E\liminf_{n\to\infty}(g-f_n)\,dm \leq \liminf_{n\to\infty}\int_E(g-f_n)\,dm \leq \int_E g\,dm + \liminf_{n\to\infty}\int_E(-f_n)\,dm.\]


Recalling that $\displaystyle \liminf_{n\to\infty}(-f_n) = -\limsup_{n\to\infty} f_n$, we see that $\displaystyle\limsup_{n\to\infty}\int_E f_n\,dm\leq \int_E \limsup_{n\to\infty} f_n\,dm$. We also have (by "definition" of liminf & limsup) that $\displaystyle \liminf_{n\to\infty}\int_E f_n\,dm \leq \limsup_{n\to\infty}\int_E f_n\,dm$. Therefore, we see that


\[\int_E \liminf_{n\to\infty}f_n\,dm\leq \liminf_{n\to\infty} \int_E f_n\,dm \leq \limsup_{n\to\infty}\int_E f_n\,dm \leq \int_E \limsup_{n\to\infty} f_n\,dm\]


and this completes the proof. Q.E.D.
 
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