Measurable Functions

**Markeur** · Sep 25th 2011, 10:00 AM

Let $(f_n)$ be a sequence of real-valued measurable functions on $[a,b]$ . If $(f_n)$ converges to a function $f$ almost everywhere on $[a,b]$ , prove that there exists a sequence $(E_n)$ of measurable subsets of $[a,b]$ such that

$m([a,b] \setminus \cup E_n) = 0$

and $(f_n)$ converges uniformly to $f$ on each $E_n$ .

How do I even start this question?

Thanks in advance.

**SlipEternal** · Sep 25th 2011, 02:25 PM

If $(f_n)$ converges almost everywhere to $f$ , 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 $\[a,b\] \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 $E_1$ . Since the zero set is a subset of $a,b$ , by the well ordering principle of real numbers, there exists a greatest lower bound of $Z$ . Can you find a really small interval $E_1$ that will include $a$ , but not $a_1$ ? Will that interval be measurable? How do you know? Will $(f_n)$ converge to $f$ uniformly on that interval? Again, how do you know?

**Markeur** · Sep 25th 2011, 09:38 PM

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

**SlipEternal** · Sep 26th 2011, 02:45 AM

It is a point in the set Z.

**Markeur** · Sep 27th 2011, 08:50 AM

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.

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