Almost Uniform convergence Theorem (Egoroff) doubts

Question

Consider $E\in\mathscr{F}$(sigma-algebra), and $E\in\Omega$ defined on a measure space $(\Omega,\mathscr{F},\mu)$. Suppose $\mu(E)<\infty$, and $\{f_n\}$ is a sequence of measurable functions on $E\to\mathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:E\to\mathbb{R}$ which is also finite almost everywhere. Then $f_n\to f$ almost uniformly in $E$.

Proof: By omitting a subset of $E$ of zero measure, we may assume that all the functions $f_n$ and $f$ are finite that $f_n(x)\to f(x)$ for all $x\in E$.

For positive integers $m,n$ let $$A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\}.$$

Then for fixed $m$, the sequence $A^m_1, A^m_2, \dots,$ is an increasing sequence of measurable sets converging to $E$. Since $\mu(E)$ is finite, by theorem 3.2 there is a positive integer $N_m=N_m(m)$ such that $$\mu(E-A^m_n)<\frac{\epsilon}{2^m}$$ for $n \geqslant N_m$. If we put $$F_\epsilon=\bigcup_\limits{m=1}^{\infty}(E-A^m_{Nm}),$$ then $\mu(F_\epsilon)<\epsilon$. Furthermore, given $\delta>0$, we can choose $m$ so that $\frac{1}{m}<\delta$ and then $|f_i(x)-f(x)|<\delta$ for all $i\geqslant N_m$ $x\in (E-F_\epsilon)$, so that $f_n\to f$ uniformly on $(E-F_\epsilon)$$\blacksquare$ .

I do not understand on this theorem proofs the following points:

1. Why is $A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\}$ increasing when $m$ is fixed?

2. How does "$|f_i(x)-f(x)|<\delta$ for all $i\geqslant N_m$ $x\in (E-F_\epsilon)$, so that $f_n\to f$ uniformly" conclude the proof?

3. What is the conclusion?

Thanks in advance!

Answer 1

1. Because the intersection is taking into account less sets, i.e. $$ A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\} \subseteq \bigcap_\limits{i=n+1}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\} = A^m_{n + 1}. $$

2. The result (Egoroff) says, that $f_i$ converges almost uniformly to $f$, i.e. for every $\epsilon > 0$ there exists $F_\epsilon$ with $\mu(F_\epsilon) < \epsilon$ such that convergence on $E - F_\epsilon$ is uniform. But this is just the last sentence of the proof you stated.

3. should be clear from 2.