Exercise 1.4.31 (Egorov's theorem II)

Pick an arbitrary $𝜖>0$. Whatever our set $E$ will be, we can already include the null set $E_0=\{x\in X:\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}_n(x)\ne f(x)\}$ in it, and assume that ${\left(f\right)}_{n\in \mathbb{N}}$ converges pointwise to $f$ everywhere. By definition, for all $x\in X∕E_0$ we have

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\frac{1}{m}>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N:\left|f_n(x)-f(x)\right|\le \frac{1}{m}.$$

We would like to improve this to

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\frac{1}{m}>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X∕E:\left|f_n(x)-f(x)\right|\le \frac{1}{m}.$$

In other words, we would like to single out the part $E$ of $X$ on which the difference between $f_n$ and $f$ exceeds the given threshold under the condition that these parts are small enough (in measure theoretic terms). Consider the sets

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\frac{1}{m}>0:E_{N,m}:=\left\{x\in X∕E_0|\exists <!--\; FUNCTION\; APPLICATION\; -->n\ge N:\left|f_n(x)-f(x)\right|\ge \frac{1}{m}\right\}.$$

Notice that we can apply the downward convergence to ${(E_{N,m})}_{N,m\in \mathbb{N}}$ in $N$ by Exercise 1.4.23 since

• ${(E_{N,m})}_{N,m\in \mathbb{N}}$ are decreasing in $N$, i.e., $E_{N,m}\subseteq E_{N-1,m}$ for any $N\in \mathbb{N}$.
• Their intersection gives us $\forall <!--\; FUNCTION\; APPLICATION\; -->\frac{1}{m}>0:{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{N=0}^{\infty }\left\{x\in X∕E_0|\exists <!--\; FUNCTION\; APPLICATION\; -->n\ge N:\left|f_n(x)-f(x)\right|\ge \frac{1}{m}\right\}=\varnothing .$
• All of the sets have finite measure $\mu (E_{N,m})\le \mu (X)<\infty$.

We then have

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\frac{1}{m}>0:\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mu (E_{N,m})=0.$$

In particular, by axiom of choice for each $m\in \mathbb{N}$ we can find a $N_m\in \mathbb{N}$ such that for all $N\ge N_m$ we have $\mu (E_{N_m,m})\le 𝜖∕2^m$. Now set

$$E:=E_0\cup \left(\bigcup _{m=1}^{\infty }{E}_{N_m,m}\right).$$

Then $E$ is measurable, and by countable additivity $m(E)\le 𝜖$. By construction we have the uniform convergence ${\left(f\right)}_{n\in \mathbb{N}}\to f$ on $X∕E$.