Completness of $\mathscr{L}_p$ spaces, $p$-th mean convergence

Question

Theorem: For $p\geqslant1$, if $\{f_n\}$ is a sequence of functions in $\mathscr{L}_p$ which is a Cauchy sequence in pth mean, then there is an $f$ in $\mathscr{L}_p$ such that $f_n\to f$ in pth mean.

Proof: We again use the device of obtaining a subsequence which will converge almost everywhere to $f$. For any $\epsilon>0$, let $N(\epsilon)$ denote an integer such that $\int|f_r-f_s|^p\:d\mu<\epsilon\:^{p+1}$ for $r,s\geqslant N(\epsilon)$. Put $N_k=N(\epsilon2^{-k}$), and assume that $N_{k+1}>N_k$ for each integer $k$. Then $\mu(E(\epsilon,r,s))<\epsilon$ for $r,s\geqslant N(\epsilon)$, $E(\epsilon,r,s))=\{x:|f_r-f_s|\geqslant\epsilon\}$, $F_k=\bigcup_\limits{I=k}^{\infty}E_i$, we have $\mu(E_k)<2^-k\epsilon$, $\mu(F_k)<2^{1-k}\epsilon$, and if $x$ is not in $F_k$ $|f_{N_{i+1}}-f_{N_{i}}|\geqslant\epsilon 2^{-i}$ for all $i\geqslant k$. Hence the series $\sum_\limits{i=1}^{\infty}(f_{N_{i+1}}-f_{N_{i}})$ converges outside $F=\bigcap_\limits{k=1}^{\infty}F_k$ and $\mu(F_k)=0$. Suppose then that $f_{N_{i}}\to f$ almost everywhere. For a fixed integer $r$, if we put $g_i=|f_{N_{i}}-f_r|^p$, $g=|f-f_r|^p$ we obtain a sequence $g_i$ of non-negative measurable functions with $\lim \inf g_i=\lim g_i=g$ almost everywhere. By theorem 5.7 (Fatou) we have: $\int g\:d\mu\leqslant\lim\inf_{i\to\infty}\int |f_{N_{i}}-f_r|^p\:d\mu<\epsilon$ if $r>N(\epsilon)$. Hence, $g$ is integrable, so that $(f-f_r)\in\mathscr{L}_p$ which implies that $f\in\mathscr{L}_p$. We have also proved that $\int |f-f_r|^p\:d\mu<\epsilon$ if $r>N(\epsilon)$ so that $f_r\to f$ in pth mean. $\blacksquare$

I have seen a much simple proof of the same theorem on class but the teacher insists we should read the book. My problem is that these proofs of the book are so precise that I lose myself on the details. Therefore my questions are:

1-Why is $\mu(F_k)<2^{1-k}\epsilon$? What is the 1 doing there? Why 1?

2-Why $|f_{N_{i+1}}-f_{N_{i}}|\geqslant\epsilon 2^{-i}$ for all $i\geqslant k$? Why use $i\geqslant k$?

3-What does it mean"Hence the series $\sum_\limits{i=1}^{\infty}(f_{N_{i+1}}-f_{N_{i}})$ converges outside $F=\bigcap_\limits{k=1}^{\infty}F_k$ and $\mu(F_k)=0$?

The rest is clear to me. Thanks in advance!

Answer 1

1. This is due to the fact that $$\sum_{i=k}^{\infty}2^{-i }=2^{- k} \sum_{i=0}^{\infty}2^{-i }=2^{1-k}.$$

2. The inequality should be reversed. if $x$ does not belong to $F_k$, then $x\notin E_i$ for each $i\geqslant k$. As a consequence, for such $i$'s, $\left|f_{N_{i+1}}-f_{N_i}\right|\lt \varepsilon 2^{-i}$, which ensures the convergence of the series $ \sum_{i=1}^\infty f_{N_{i+1}}(x)-f_{N_i}(x)$.

3. If $x$ does not belong to $F$, then there is some $k\geqslant 1$ for which $x\notin F_k$, and by the argument in 2., the series $\sum_{i=1}^\infty f_{N_{i+1}}(x)-f_{N_i}(x)$ converges. The part "$\mu\left(F_k\right)=0$" should be replace by $\mu\left(F\right)=0$.