Why is $\mu(X) < \infty$ important in Egoroff's theorem?

Question

The following is Folland's proof of Egoroff's theorem.

Without loss of generality we may assume that $f_n \rightarrow f$ everywhere on $X$. For $k, n \in \mathbb{N}$ let $$E_n(k) = \bigcup_{m=n}^\infty \{x : |f_m(x) - f(x)| \geq k^{-1}\}.$$ Then, for fixed $k$, $E_n(k)$ decreases as $n$ increases, and $\bigcap_{n=1}^\infty E_n(k) = \emptyset$, so since $\mu(X) < \infty$ we conclude that $\mu(E_n(k)) \rightarrow 0$ as $n\rightarrow \infty$. Given $\epsilon > 0$ and $k \in \mathbb{N}$, choose $n_k$ so large that $\mu(E_{n_k}(k)) < \epsilon 2^{-k}$ and let $E = \bigcup_{k=1}^\infty E_{n_k}(k)$. Then $\mu(E) < \epsilon$, and we have $|f_n(x) - f(x)| < k^{-1}$ for $n > n_k$ and $x \not\in E$. Thus $f_n \rightarrow f$ uniformly on $E^c$.

Why is it important that $X$ have finite measure and how does it imply $\mu(E_n(k)) \rightarrow \infty$?

Answer 1

We know that if $(A_k)$ is a non-increasing sequence of measurable sets such that $\mu(A_1)<\infty$, then $\lim_{k\to\infty}\mu(A_k)=\mu\left(\bigcap_{k=1}^\infty A_k\right)$, what is called continuity from above.

This continuity from above may not hold if all the $A_k$ have infinite measure, as the example $A_k=\{j,j\geqslant k\}$ as a subset of $\mathbb N$ endowed with the counting measure shows.

Here is a concrete example where the sets $E_n(k)$ can be made explicit. Let $X=\mathbb N$, $\mathcal F=\mathcal P(\mathbb N)$ and $\mu$ is the counting measure. Let $f_n\colon k\mapsto \mathbf{1}_{k=n}$. Then for each fixed $k$, $f_n(k)\to 0$ as $n$ goes to infinity. Moreover, $$ E_n(k)=\{j,j\geqslant n\} $$ and here $\mu(E_n(k))=\infty$ for each $n$ and the proof would collapse at this stage.