Let \(\ts{ f_k } _{k=1}^{\infty }\) be a sequence of extended real-valued Lebesgue measurable functions.
Prove that \(\sup_k f_k\) is a Lebesgue measurable function.
Prove that if \(\lim_{k \to \infty } f_k(x)\) exists for every \(x \in \RR^n\) then \(\lim_{k\to \infty } f_k\) is also a measurable function.
Solution
1. (a) \(\sup_k f_k\) is measurable.
for every \(a \in \RR\), \[ \Big\{\sup_k f_k \le a\Big\} = \bigcap_k \big\{f_k \le a\big\}, \] a countable intersection of measurable sets (each \(f_k\) is measurable), hence measurable.
2. (b) Reduce \(\lim_k f_k\) to operations on measurable functions.
where the limit exists, \[ \lim_{k\to\infty} f_k = \liminf_{k\to\infty} f_k = \sup_{m} \inf_{k \ge m} f_k, \] so it suffices to show \(\inf_k f_k\) and \(\sup_k f_k\) are measurable.
3. \(\inf_k f_k\) is measurable.
\(\{\inf_k f_k \ge a\} = \cap_k\{f_k \ge a\}\), a countable intersection of measurable sets.
4. Conclude (b).
by 1 and 3, \(\sup_m\inf_{k\ge m} f_k\) is measurable (sup and inf of measurable functions are measurable), and it equals \(\lim_k f_k\) on the set where the limit exists — all of \(\RR^n\) by hypothesis.
5. Q.E.D.