1. ## Measurable Functions

Let $\displaystyle (f_n)$ be a sequence of real-valued measurable functions on $\displaystyle [a,b]$. If $\displaystyle (f_n)$ converges to a function $\displaystyle f$ almost everywhere on $\displaystyle [a,b]$, prove that there exists a sequence $\displaystyle (E_n)$ of measurable subsets of $\displaystyle [a,b]$ such that
$\displaystyle m([a,b] \setminus \cup E_n) = 0$
and $\displaystyle (f_n)$ converges uniformly to $\displaystyle f$ on each $\displaystyle E_n$.

How do I even start this question?

2. ## Re: Measurable Functions

If $\displaystyle (f_n)$ converges almost everywhere to $\displaystyle f$, 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 $\displaystyle $a,b$ \setminus \cup E_n = Z\text{ where }Z\text{ is your zero set}$, then you are guaranteed that the measure will be zero. So, let's construct $\displaystyle E_1$. Since the zero set is a subset of $\displaystyle $a,b$$, by the well ordering principle of real numbers, there exists a greatest lower bound of $\displaystyle Z$. Can you find a really small interval $\displaystyle E_1$ that will include $\displaystyle a$, but not $\displaystyle a_1$? Will that interval be measurable? How do you know? Will $\displaystyle (f_n)$ converge to $\displaystyle f$ uniformly on that interval? Again, how do you know?

3. ## Re: Measurable Functions

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

4. ## Re: Measurable Functions

It is a point in the set Z.

5. ## Re: Measurable Functions

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.