Show: If $f_n \ge f$ a. e. then $\int_X \liminf_{n \to \infty}{f_n} \: d\mu \le \liminf_{n \to \infty} \int_X{f_n}d\mu$

Question

Let the functions $f, f_n : X \to \bar{\Bbb R}$ be measurable and $f$ integrable:
I have to proof:

• If $f_n \ge f, \: \mu -$almost everywhere $\forall n \in \Bbb N$ then $$\int_X \liminf_{n \to \infty}{f_n} \: d\mu \le \liminf_{n \to \infty} \int_X{f_n}d\mu$$
• For $f_n \le f, \: \mu -$almost everywhere $\forall n \in \Bbb N$ then $$\int_X \limsup_{n \to \infty}{f_n}d\mu \ge \limsup_{n \to \infty} \int_X{f_n}d\mu$$

I also have to show why one can't give up the condition $f_n \ge f$ in a) respectively $f_n \le f$ in b)

I don't really know how to show all this, and i also don't get why the $f_n$ in case a) are integrable (in case b) it's clear since f is always bigger and integrable) Any ideas or tipps on how to show the two things above? Thanks in advance!

Answer 1

These are all (in their own way) different statements of Fatou's lemma.

If you know Fatou's lemma then (for the first) simply plug $g_n = f_n -f \geq 0$. $g_n$ are measurable (as sum of measurable functions) and the theorem follows (But wait, what is $\liminf f$? And can we just cancel them two terms in both sides of the equation?)

For the second, note that $h_n = -g_n \geq 0$ and we (again) meet the conditions for Fatou's lemma, only this time we have terms like $\limsup -f_n $, but this is easy (If you can connect $\limsup$ with $\liminf$).