The following is a proof that Folland gives on Egoroff's theorem:
My question:
Why do we know that $E_n(k)$ decreases as $n$ increases? Isn't there a possibility that $f_{m+1}(x)$ gets "further away" from $f(x)$ than $f_m(x)$ so that $E_{m}(k) \subset E_{m+1}(k)$. Or maybe they are even non-comparable -- for the values of $x$ such that $\{x:|f_{m+1}(x)-f(x)|\geq 1/k\}$ are not the same as $\{x:|f_{m}(x)-f(x)|\geq 1/k\}$? -- I think the fact that we assume $f_n \rightarrow f$ everywhere has something to do with this. But that raises a separate question -- we drastically altered the problem by doing this. Do we "foresee" that we will be taking the measure of $E_n(k)$ later?
You are thinking much too hard. Pay attention to the whole definition of $E_n(k)$. $E_{n+1}(k)$ is smaller because it is a union of strictly fewer of the same sets.
If we fix $k$ and let $A_{m} = \{x : |f_m(x) - f(x)| \ge k^{-1}\}$, then $E_n(k) = \bigcup_{m=n}^\infty A_{m}$. So $E_n = A_n \cup A_{n+1} \cup \dots = A_n \cup E_{n+1}$. In particular $E_{n+1} \subset E_n$.
If $f_{m+1}$ is "further away" from $f$ then that would make $A_{m+1}$ larger than $A_m$. But that's irrelevant to the question at hand about the $E_n$.