I have a doubt in the proof of lusin theorem. Before i ask the question, let me state down the proof first.

Lusin Theorem : If f is a measurable function on A where m(A)< , then given any , there exist a closed set such that and f is continuous on .

PROOF :

First, let be a simple function on A. Given , for each k there exists a closed set such that

Now C= is closed, and f is continuous on C since C is equal to the disjoint union of closed sets and f is constant on each .

Now let f be any measurable function. Since , we may assume f is non-negative. Since f is measurable, for a sequence { } of simple functions.

By the first part of the proof, given , there exists for each a closed set such that and is continuous on .

Let.

Then is closed and

= = < = .

By egoroff theorem, there is a measurable subset such that and uniformly on C.

Now each is continuous on so that , being the uniform limit of the , is continuous on . If is closed we are through. If not, take closed set such that .

MY QUESTION is : Can be empty set ? Why?