Unclear moments of the proof of Lusin's Theorem 
When the author considers the set $F'$ some moments are unclear.
1) Is it true that $A_{\varepsilon/3}$ contains $\bigcup\limits_{n\geq N}E_n$ ? If yes, how to prove it?
2) Why the author takes the condition $\sum \limits_{n\geq N}2^{-n}<\varepsilon/3$? What is the consequence of this inequality?
Would be very grateful if anyone can answer my questions.
 A: 1) No, not necessarily, and this is not needed. 2) To get the desired $m(E - F_{\epsilon}) \leq \epsilon$ (see below).
Found the book. Both this proof and the one I know are based on a) Egorov's theorem (Theorem 4.4 there) and b) the fact that for any measurable set $F$ and any $\delta > 0$ there is a closed measurable subset $G$ of $F$ such that $m(F - G) < \delta$ (part of Theorem 3.4 there).
In the proof above: $m(E - A_{\epsilon/3}) \leq \epsilon/3$; $A_{\epsilon/3} - F' \subseteq \bigcup_{n \geq N}E_n$ and therefore (here is the answer to your second question!) $m(A_{\epsilon/3} - F') \leq \epsilon/3$; finally, $m(F' - F_{\epsilon}) \leq \epsilon/3$. Together these give $m(E - F_{\epsilon}) \leq \epsilon$.
An alternative way is to prove the theorem for step functions first (in full strength, using just the Theorem 3.4), and, for general case, to use Egorov's theorem (which gives $F \subseteq E$ with $m(E - F) \leq \epsilon/2$ and $f_n \to f$ uniformly on $F$) and the case proven already (which gives closed $F_n \subseteq F$ with $m(F - F_n) \leq \epsilon / 2^{n + 1}$ and $f_n$ continuous on $F_n$; finally we're done with $F_{\epsilon} = \bigcap_{n = 1}^{\infty} F_n$).
