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?