I am reading measure in the analysis independently, and got stuck on a problem. Hope you guys will help me out.
Let $f_n:[a,b]\to \Bbb R$ be a sequence of measurable functions such that $f_n\to f$. Prove that there exists a sequence $E_k$ of closed sets such that $f_n$ converges uniformly on each $E_k$ and $|{[a,b]}\setminus\sum_{k=1}^\infty E_k|=0$
Thank you.
There exist sets $A_k$ such that $|A_k^{c}| <\frac 1 {2^{k}}$ and $f_n \to f$ uniformly on $A_k$. There exists a closed set $E_k \subset A_k$ such that $|A_k \setminus E_k| <\frac 1 k$. Of course, $f_n \to f$ uniformly on $E_k$. To see that $|[a,b]\setminus \cup E_k|=0$ verify the following set theoretic fact: $(\cup E_k)^{c} \subset \lim \sup (A_k^{c}) \cup \lim\inf (A_k\setminus E_k)$. The fact that $\sum |A_k^{c}| <\infty$ implies that $|\lim \sup (A_k^{c})|=0$. Next, $|\lim \inf (A_k\setminus E_k)| \leq \lim\inf |(A_k\setminus E_k)|=0$. Hence, $|(\cup E_k)^{c}|=0$. [Notations: $\lim \sup S_k$ denotes the set of points that belong to inifinitely many of the sets $S_n$ and $\lim \inf S_k$ denotes the set of points that belong to $S_n$ whenever $n$ is sufficiently large].