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?