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.
Best Answer
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: $$\int\limits_{E} \lim\limits_{n \to \infty} inf f_n \leq \lim\limits_{n \to \infty} inf \int\limits_{E} f_n \leq \lim\limits_{n \to \infty} \sup \int\limits_{E} f_n \leq \int\limits_{E} \lim\limits_{n \to \infty} \sup f_n$$