Royden Real Analysis, Chapter $3$ Proposition $9$ This is from Royden Real Analysis Chapter $3$ Proposition $9$ Page $60$
Let $f_{n}$ be a sequence of measurable functions on $E$ that converges pointwise almost everywhere on $E$ to the function $f$ . Then $f$ is measurable.
In the proof author have introduced natural numbers $n$ and $k$ for which $f_{j}(x)<c-1/n$ for all $j\geq k$
The proof is completed by stating the following. 
Since the union of a countable collection of measurable sets is measurable
$\left \{ x \in E | f(x)< c \right \}$ = $\bigcup_{1\leq k,n<\infty}$$[\bigcap_{j=k}^{\infty}\left \{ x\in E\left | f_{j}(x)<c-1/n \right \} \right]$
is measurable. Hence $f$ is measurable. 
I do not understand the purpose of introducing $n$ in this proof. Is it possible to avoid introducing $n$ and just use $f_{j}(x)<c$ and take union over $k$ alone. 
 A: The trouble is that even if $f_j(x) < c$ for every $j$, we still might have $f(x) = c$. To ensure that $f(x) < c$, we need $f_j(x)$ to be less than $c$ by some amount which does not depend on $j$. That is the role that $\frac{1}{n}$ plays in the proof.
For instance, consider the sequence $f_n(x) = -\frac{1}{n}$, which converges to the function $f(x) = 0$. Take $c = 0$. Then $\{x : f(x) < c\} = \varnothing$, but $$\bigcup_{k = 1}^\infty \bigcap_{j = k}^\infty \{x : f_j(x) < c\} = \mathbb{R}.$$
A: The point is that $$\bigcup_{k=1}^\infty \bigcap_{j=k}^\infty\{x\mid f_j(x) < c\}$$
would capture all points where the limit is less than $c$, but also perhaps some where the limit is $c$.
A: Observe that $S=\bigcup_{1\leq k,n<\infty}[\bigcap_{j=k}^{\infty}\left \{ x\in E\left | f_{j}(x)<c-1/n \right \} \right]$ means precisely the following:
$x\in S$ whenever there are $k,n\in \mathbb N$ such that for all $j\geq k,\ $ $f_j(x)<c-1/n$ whereas 
$S'=\bigcup_{1\leq k<\infty}[\bigcap_{j=k}^{\infty}\left \{ x\in E\left | f_{j}(x)<c \right \} \right]$ means 
$x\in S'$ whenever there is a $k\in \mathbb N$ such that for all $j\geq k,\ f_j(x)<c$ which does not exclude the possiblity that $f(x)=c$.
