Question about proving something using reverse Fatou's lemma 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.
 A: 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*}
