$f_n$ is a sequence of measurable functions on (0,1) such that $f_n$ converges to $f$ point wise. Prove that there exists a countable collection of measurable sets $\{A_n: n=0,1,2,3…\}$ such that $f_n\rightarrow f$ on $A_n$ and $m(\cup A_n)=1$.
I should use the Egorov theorem here, that is $for\ each\ \epsilon>0\ there\ exists\ A_\epsilon\ such\ that\ m((0,1)\cap A_\epsilon^c )<\epsilon\ and\ f_n\rightarrow f\ on\ A_\epsilon.$
My question is how do I utilize Egorov theorem to construct such a sets.
here $m$ is the measure.
Best Answer
For each $k \in \Bbb N_{\ge 0}$, there is a measurable set $A_k$with $m(A_k^c) < \frac{1}{2^k}$ such that $f_n \to f$ on $A_k$. If $S = \cap_{k = 0}^\infty A_k^c$ then $m(S) \le m(A_k^c) < \frac{1}{2^k}$ for every $k$. Hence $m(S) = 0$. Since $\cup A_k = S^c$, then $m(\cup A_k) = 1 - m(S) = 1$.