The proposition and its proof is given below:
Proposition 9 Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise a.e. on $E$ to the function $f$. Then $f$ is measurable.
Proof Let $E_0$ be a subset of $E$ for which $m(E_0) = 0$ and $\{f_n\}$ converges to $f$ pointwise on $E \sim E_0$. Since $m(E_0) = 0$, it follows from Proposition 5 that $f$ is measurable if and only if its restriction to $E \sim E_0$ is measurable. Therefore, by possibly replacing $E$ by $E \sim E_0$, we may assume the sequence converges pointwise on all of $E$.
Fix a number $c$. We must show that $\{x \in E |f(x) < c\}$ is measurable. Observe that for a point $x \in E$, since $\lim_{n \to \infty} f_n(x) = f(x)$, $$ f(x) < c $$ if and only if there are natural numbers $n$ and $k$ for which $f_j(x) < c - 1/n$ for all $j \geq k$.
But for any natural numbers $n$ and $j$, since the function $f_j$ is measurable, the set $\{x \in E | f_j(x) < c - 1/n\}$ is measurable. Therefore, for any J$k$, the intersection of the countably collection of measurable sets $$ \bigcap_{j=k}^{\infty}\{x \in E | f_j(x) < c - 1/n\} $$ also is measurable. Consequently, since the union of a countable collection of measurable sets is measurable, $$ \{x \in E | f(x) < c\} = \bigcup_{1\leq k,n < \infty} \left[ \bigcap_{j=k}^{\infty}\{x \in E | f_j(x) < c - 1/n\} \right] $$ is measurable. $\square$
My questions are:
why $f(x) < c$ iff there are natural numbers $n$ and $k$ for which $f_{j}(x) < c-{1/n}$ for all $j \geqslant k$? what is the importance of the $k$ and why we subtract from $c$ the value $1/n$?
why we took the intersection and start it specifically from $j=k$? and why we take the union after that?
Can anyone explain these points for me please?
$\textbf{For 1:}$
$\Leftarrow :$ Suppose that there exist natural numbers $n$ and $k$ such that for $j\ge k$ we have that $f_j(x)<c-\frac{1}{n}$. Well then we must have that $\lim_{j>k}f_j(x)=f(x)<c- \frac{1}{n}$ and this tells us that $f(x)<c$.
$\Rightarrow$: Now suppose that for any integer $n$ and any integer $k$ that there exists an integer $j>k$ so that $f_j(x)>c -\frac{1}{n}$. Since this is true for every integer $n$ then we must have that $\lim_j f_j(x) = f(x) \ge c$ and so $f(x) \not < c$. By the contra-positive we see that the $\Rightarrow$ direction is true.
$\textbf{For 2:}$
The statement that $$ f(x) < c \; \; \text{iff} \; \; \text{there exists integers $n$ and $k$ so if $j>k$ then} \; f_j(x)<c-\frac{1}{n} \tag{1}$$ is the same thing as saying $$ \{x \in E | f(x) < c\} = \bigcup_{1\leq k,n < \infty} \left[ \bigcap_{j=k}^{\infty}\{x \in E | f_j(x) < c - 1/n\} \right] \tag{2} $$ To see this, suppose that $x$ is such that (1) is true. Well then there exists integers $n$ and $k$ so that if $j>k$ then $f_j(x)<c-\frac{1}{n}$. But this means that $x \in \bigcap\limits_{j \ge k}\{x \in E | f_j(x) < c - 1/n\}$ and therefore (1) implies (2).
Now suppose that (2) is true. Then by the definition of the union and intersection, there exists a pair of integers $n$ and $k$ so that for every integer $j>k$ we have that $f_j(x) <c - \frac{1}{n}$. This is exactly statement (1) so (2) implies (1).