Thread: Measurable Functions

Let be a sequence of real-valued measurable functions on . If converges to a function almost everywhere on , prove that there exists a sequence of measurable subsets of such that

and converges uniformly to on each .

How do I even start this question?

Thanks in advance.

If converges almost everywhere to , then there is a zero set of points on which it does not converge. A zero set has certain properties. For one thing, it has measure zero. So, if , then you are guaranteed that the measure will be zero. So, let's construct . Since the zero set is a subset of , by the well ordering principle of real numbers, there exists a greatest lower bound of . Can you find a really small interval that will include , but not ? Will that interval be measurable? How do you know? Will converge to uniformly on that interval? Again, how do you know?

Is a_1 a point in the set Z or just part of the partition of [0,1]?

It is a point in the set Z.

Alright I managed to solve it. It's actually applying Egorov's Theorem to each E_n and intersect [a,b]\E_n for all n to get m([a,b]\(\cup E_n)) = 0 and f_n converges uniformly to f on each E_n.

Thanks.