Egorov's Theorem and finite measure restriction

Here's the statement of Egorov's Theorem from my book:

Quote:

Assume set E has finite (Leb) measure. Let {fn} 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 {fn} converges to f uniformly on F and measure(E~F) < EPSILON

If E has infinite measure, it possible for the conclusion of this Theorem to fail.

measure theory - Egorov's Theorem - Counterexample in Infinite Case - Mathematics Stack Exchange

real analysis - Littlewood's Principles and Counterexamples - Mathematics Stack Exchange

The above discussions give one such example when E = R (real line) and Fn = X[n, n+1] (characteristic function), fn converges pointwise to f = 0.

So, what happens in this case? I'm not quite sure how it works.

The way I understand it, if measure(E~F) < EPSILON is to hold, F needs to be "big", inadvertently covering some interval [n, n+1] where Fn = 1. In other words, there exists some x in F where fn(x) = 1 (for all n) so {fn} cannot converge to f = 0 uniformly on F (for EPSILON < 1 ?)

Any suggestions? Thanks