Lusin's theorem:
let $f$ be a real-valued measurable function on $E$ with finite measure.
$\forall \epsilon>0, \exists$ closed $F\subseteq E$ with $m(E\setminus F)<\epsilon $ such that $f$ is continuous on F
I am trying to drop the required finiteness of the measure of $E$;
My idea is to express $E$ as the union of a countable disjoint family of bounded measurable sets $(E_k)_{k=1}^\infty$, therefore we can find a countable family of closed sets $(F_k)_{k=1}^\infty$ with $m(E_k\setminus F_k)<\epsilon/2^{k+2}$ such that $f$ is continuous on $F_k$. Then we take $F=\cup_{k=1}^\infty F_k$. Thus $f$ is f is continuous on $F$ with $m(E\setminus F) \le m(\cup _{k=1}^\infty (E_k/F_k))\le \sum_k m(E_k\setminus F_k)\le\epsilon/4<\epsilon $. But, since the countable union of closed sets may not be closed, the idea here is flawed.
Is there any suggestion?
Best Answer
I got a idea:
by the measurability of $F$, we can take a closed $F^*\subseteq F$ such that $m(F\setminus F^*)<\epsilon/2$. Thus $m(E\setminus F^*)\le m((E\setminus F)\cup(F\setminus F^*))\le m(E\setminus F)+m(F\setminus F^*)<\epsilon$. And, $f$ is continuous on $F^* $