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Let $(f_n)\subset\epsilon$. If there is an integrable function g such that $f_n \le g$ for every n, then $\mu(\limsup f_n)\ge \limsup \mu f_n$ . I want to prove it using Fatou's Lemma. But I have a problem what should I do after getting $\mu (\liminf(g-f_n))\le \liminf\mu (g-f_n)$. Please help me.

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Hint: For any real number $a$ and any sequence of real numbers $b_n$, we have $\liminf(a-b_n)=a-\limsup b_n$. – Nate Eldredge May 24 '12 at 17:12
@NateEldredge Thank you for your hint, but I wonder is liminf($a_n-b_n$)=liminf $a_n$-liminf $b_n$ right? – user30795 May 24 '12 at 17:25
No, it isn't. (Can you find a counterexample? Try a sequence $b_n$ which doesn't converge, so its liminf and limsup are different.) – Nate Eldredge May 24 '12 at 17:29
What is $\epsilon$? – Davide Giraudo May 24 '12 at 19:24
@DavideGiraudo: I assumed that $\epsilon$ is the collection of measurable functions, as it is the only reasonable explanation that I could think of. Considering, that this requirement is one of the two assumptions in the Fatou's lemma, other which is stated right after the first verse. – T. Eskin May 24 '12 at 19:29
up vote 2 down vote accepted

I'm assuming that with $\mu$ you mean the integral operator.

Since $f_{n}\leq g$ then $g-f_{n}\geq 0$ for all $n$, and each such is in addition measurable. In particular since $g$ is integrable then $\mu(g)<\infty$. Hence we can apply Fatou (**) to the sequence $(g-f_{n})_{n=1}^{\infty}$ and we get \begin{align*} \mu(g)-\mu(\limsup_{n\to\infty} f_{n}) &=\mu(g-\limsup_{n\to\infty} f_{n})=\mu(\liminf_{n\to\infty} (g-f_{n}))\overset{**}{\leq} \liminf_{n\to\infty} \mu(g-f_{n}) \\ &=\liminf_{n\to\infty} (\mu(g)-\mu(f_{n}))=\mu(g)-\limsup_{n\to\infty}\mu(f_{n}). \end{align*} By subtracting $\mu(g)$ from both sides and by multiplying with $-1$ we obtain the desired result \begin{align*} \limsup_{n\to\infty}\mu(f_{n})\leq \mu(\limsup_{n\to\infty}f_{n}). \end{align*}

Extra Lemma that we used: Note that above we used the fact (which seemed to puzzle you) that $\liminf (-a_{n})=-\limsup(a_{n})$. This follows by observing that for any $A\subset \mathbb{R}$ we have $\inf(-A)=-\sup A$. Here's a sketch of how to prove it. We can assume that either of them (hence both) are finite: otherwise the equality is trivial (e.g. if $-\sup A=-\infty$, then $\sup A=\infty$ so we can find a sequence in $A$ not bounded from above. Put minus signs infront, you get sequence not bounded from below in $-A$, whence $\inf(-A)=-\infty$. Similarly the other case). First of all if $-a\in -A$ then $a\in A$ and $\inf A\leq a\leq \sup A$, so moreover $-\inf A\geq -a \geq -\sup A$. Thus $-\sup A$ is a lower bound for $-A$. We show that it is in addition the biggest lower bound (i.e. infimum). Let $\varepsilon>0$. There now exists (by supremum criteria) $a_{\varepsilon}\in A$ so that $\sup A -\varepsilon \leq a_{\varepsilon}$, whence $-a_{\varepsilon}\leq -\sup A + \varepsilon$. Since $-a_{\varepsilon}\in -A$, then what we have shown is that for every $\varepsilon>0$ there exists $-a_{\varepsilon}\in -A$ with $-a_{\varepsilon}\leq -\sup A + \varepsilon$, which by infimum criteria means precisely that $-\sup A$ is the biggest lower bound of $-A$, i.e. $-\sup A=\inf(-A)$, as required.

Hence using the above result we get exactly the equality that we wanted to use \begin{align*} \liminf (-a_{n})=\lim_{n\to\infty}(\inf_{k\geq n}-a_{k})=\lim_{n\to\infty}(-\sup_{k\geq n}a_{k})=-\lim_{n\to\infty}(\sup_{k\geq n}a_{k})=-\limsup(a_{n}). \end{align*}

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There are some problems here: 1) $f_{n} \leq g$ does not imply the $f_{n}$ are integrable – madprob Jun 28 '12 at 23:58
@madprob: You're right, thanks. And even though I wrote it there, I actually didn't use it in the proof. Indeed $\infty-\infty$ can never occur there as $\mu(g)<\infty$. – T. Eskin Jun 29 '12 at 5:56

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