Egorov's Theorem: Royden / Fitzpatrick

Question

Royden / Fitzpatrick (4ed) state Egorov's theorem as:

Assume $E$ has finite measure. Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise on $E$ to the real-valued function $f$. Then for each $\epsilon > 0$, there is a closed set $F$ contained in $E$ for which $\{f_n\}\to f$ uniformly on $F$ and $m(E\backslash F) < \epsilon$.

The proof doesn't depend on the use of closed sets (we find a set where convergence is uniform and then take its closed inner approximation to form a closed set), so why do we have $F$ closed? In particular, what is an example of where the the closed property is useful?

Answer 1

Mainly because we can. It is a stronger result and is useful for it to be closed, so why not?

Answer 2

Doing measure theory on the real line is concrete and intuitive, but it can make "open sets" and "measurable sets" seem more closely related than they need to be. We have to always remember that a topological space and a measure space are distinct notions (even though they can be related in the style of Borel). Given an arbitrary measure space, we can define the topology however we want -- we can even do not define any topology at all. (See Is every topological space is measurable?)

Egorov's theorem is a result in measure theory that has nothing to do with topology. For general measure spaces, the statement is simply that (Zygmund & Wheeden, p 167) there exists a measurable set A contained in E ...

Sometimes, when a topological structure is also defined and specific regularity relationships exist, we can do inner closed approxiamtion of measurable sets. (See Inner approximation by closed sets if $E$ is measurable)

The Egorov Theorem, combined with this regularity relationship, give us the statement you quote.