# Extension of Lusin's theorem

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?

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 $$F^*$$