In a proof of the completeness of $L^\infty$ The following is a proof of the completeness of $L^\infty$ in a lecture note by Hunter:



Here are my questions:


*

*
Can one (literally) replace $1/m$ in the whole proof with $\epsilon$ and replace $N_{j,k,m}$ with $N_{j,k}$?


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[Added:] To be more precise, what would be wrong if one literally does the following change in the proof?
  
  
*
  
*change "for every $m\in\mathbb{N}$" to "for every ${\epsilon>0}$";
  
*change $1/m$ to $\epsilon$;
  
*change $N_{j,k,m}$ to $N_{j,k}$;
  
*change the definition of $N$ to
  $
   N=\bigcup_{j,k\in\mathbb{N}}N_{j,k}
 $
  


*[Added2:] Due to the confusion of the changes above, I would like to explicitly write down the proof I meant after the above change:
If $\{f_k\}_k$ is Cauchy in $L^\infty$ then for every $\color{blue}{\varepsilon > 0}$, there exists an $n(\varepsilon) \in \Bbb{N}$ such that 
\begin{align}  (*)  && && 
\left| f_j(x) - f_k(x) \right| &< \varepsilon \quad \text{for all $j,k \geq n(\varepsilon)$ and $x \in \color{blue}{N_{j,k}^c}$}
\end{align}
where $\color{blue}{N_{j,k}}$ is a null set.  Let 
$$
N = \color{blue}{\bigcup_{j,k \in \Bbb{N}} N_{j,k}}.
$$
Then $N$ is a null set, and for every $x \in N^c$ the sequence $\{f_k(x)\}_{k \in \Bbb{N}}$ is Cauchy in $\Bbb{R}$.  Define measurable $f:X \rightarrow \Bbb{R}$ by 
$$
f(x) = \lim_{k \rightarrow \infty} f_k(x) \quad \text{for $x \in N^c$}  \text{.}  
$$
Taking $k \rightarrow \infty$ in $(*)$, for every $\color{blue}{\varepsilon > 0}$, there exists an $n(\varepsilon) \in \Bbb{N}$ such that $$  \left| f_j(x) - f(x) \right| \leq \varepsilon \quad \text{for $j \geq n(\varepsilon)$ and $x \in N^c$}  \text{.}  $$  It follows that $f$ is essentially bounded and $f_j \rightarrow f$ in $L^\infty$ as $j \rightarrow \infty$.  
 A: It's not quite as simple as that.  The union $$ N = \bigcup_{j,k,m \in \Bbb{N}} N_{j,k,m}  $$ is a countable union, so properties of measures hold.  If we take the most straightforward implementation of your replacement scheme, we need $$  N = \bigcup_{\substack{j,k\in \Bbb{N} \\ 0 < \varepsilon < \varepsilon_0}} N_{j,k,\varepsilon}  $$  which is no longer a countable union.  We can avoid this by @zuggg's method: replace $\varepsilon$ with a decreasing sequence $(\varepsilon_m)_m$.  But then we don't avoid $m$ and we need to make the extra replacement of $N_{j,k,\varepsilon}$ by $N_{j,k,\varepsilon_m}$ and write the justification of that replacement.
Edit:  The question was changed hours after it was answered.  The bulk of the above still holds, but it would apparently be useful to be more detailed in a few areas.  Let's actually write out what you say you  want with dependence on $\varepsilon$ made explicit (instead of lurking implicitly in your replacements):

Suppose $p = \infty$ and $(f_k)_k$ is Cauchy in $L^\infty$.  Then for every $\varepsilon > 0$, there exists an $n(\varepsilon) \in \Bbb{N}$ such that \begin{align}  (*)  && && \left| f_j(x) - f_k(x) \right| &< \varepsilon \quad \text{for all $j,k \geq n(\varepsilon)$ and $x \in N_{j,k,\varepsilon}^c$}
\end{align}
  where $N_{j,k,\varepsilon}^c$ is a null set.  Let $$N_{\varepsilon} = \bigcup_{j,k \in \Bbb{N}} N_{j,k,\varepsilon}  \text{.}  $$  Then $N_\varepsilon$ is a null set, and for every $x \in N_\varepsilon^c$ the sequence $(f_k(x))_{k \in \Bbb{N}}$ is Cauchy in $\Bbb{R}$.  Define measurable $f_\varepsilon:X \rightarrow \Bbb{R}$ by $$f_\varepsilon(x) = \lim_{k \rightarrow \infty} f_k(x) \quad \text{for $x \in N_\varepsilon^c$}  \text{.}  $$
  Taking $k \rightarrow \infty$ in $(*)$, for every $\varepsilon > 0$, there exists an $n(\varepsilon) \in \Bbb{N}$ such that $$  \left| f_j(x) - f_\varepsilon(x) \right| \leq \varepsilon \quad \text{for $j \geq n(\varepsilon)$ and $x \in N_\varepsilon^c$}  \text{.}  $$  It follows that $f_\varepsilon$ is essentially bounded and $f_j \rightarrow f_\varepsilon$ in $L^\infty$ as $j \rightarrow \infty$.  This proves that $L^\infty$ is complete.

The last sentence does not follow.  We have shown that the Cauchy sequence converges to some essentially bounded function for each choice of $\varepsilon$, but we have not shown that $\lim_{\varepsilon \rightarrow 0} f_\varepsilon$ is in $L^\infty$ (particularly, having finite essential supremum).  We don't even have that the various $f_\varepsilon$ have compatible domains.  ($X$ might be the union of an uncountable collection of disjoint null sets indexed by $E$, the collection of ordinals up to the first uncountable ordinal, and each $f_\varepsilon$ might be defined on only the sets indexed by $E \setminus \{e \in E \mid 0 < e < \varepsilon\}$.)  How do we know without adding more to the proof that 


*

*$\lim_{\varepsilon \rightarrow 0} f_\varepsilon$ is even defined a.e., and

*$\lim_{\varepsilon \rightarrow 0} f_\varepsilon \in L^\infty$, that is, the essential supremum of $f_\varepsilon$ does not grow unboundedly, for example, as $1/\varepsilon$, as $\varepsilon \rightarrow 0$?


Well, let $\varepsilon_m \rightarrow 0$ as $m \rightarrow \infty$, ...
On the other hand, the original proof closes off the dependence on $m$ at the union defining $N$, so $N$ and $f$ are independent of $m$.  If you want to make your substitutions and close off the dependence on $\varepsilon$ at the union defining $N$, you must either deal with the uncountable union of null sets -- not impossible because they're nested -- or take a sequence of $\varepsilon$s going to zero and argue about that sequence as suggested by the line "Well, let $\varepsilon_m \rightarrow 0$ as $m \rightarrow \infty$, ..." just before this paragraph.
A: Enlightened by discussion in this related question:
In Rudin's proof of the completeness of $L^\infty$,
I conclude that the proof in "Added2" in OP is correct. More precisely, I rewrite it as follows.
Define for each $j,k\in\mathbb{N}$ a null set $N_{j,k}$ such that
$$
|f_j(x)-f_k(x)|>\|f_j-f_k\|_\infty\quad\textrm{for all }x\in N_{j,k}.
$$
Suppose $\{f_k\}_{k=1}^\infty$ is Cauchy in $L^\infty$. Then for every $\color{blue}{\epsilon > 0}$, there exists $n(\epsilon)\in\mathbb{N}$ such that
$$
\|f_j-f_k\|_\infty\leq\epsilon\quad\textrm{for all }j,k\geq n(\epsilon).
$$
which implies that
\begin{align}  (*)  && && 
\left| f_j(x) - f_k(x) \right| &\leq \epsilon \quad \text{for all $j,k \geq n(\epsilon)$ and $x \in \color{blue}{N_{j,k}^c}$}
\end{align}
where $\color{blue}{N_{j,k}}$ is a null set.  Let 
$$
N = \color{blue}{\bigcup_{j,k \in \Bbb{N}} N_{j,k}}.
$$
Then $N$ is a null set. Note also that
$$
N^c = \color{blue}{\bigcap_{j,k \in \Bbb{N}} N^c_{j,k}}.
$$ 
For every $x \in N^c$ the sequence $\{f_k(x)\}_{k \in \Bbb{N}}$ is Cauchy in $\Bbb{R}$. 
Define a measurable function $f:X \rightarrow \Bbb{R}$ by 
$$
f(x) = \lim_{k \rightarrow \infty} f_k(x) \quad \text{for $x \in N^c$}
$$
and $f(x)=0$ for $x\in N$.
Taking $k \rightarrow \infty$ in $(*)$, for every $\color{blue}{\epsilon > 0}$, there exists an $n(\epsilon) \in \Bbb{N}$ such that $$  \left| f_j(x) - f(x) \right| \leq \epsilon \quad \text{for $j \geq n(\epsilon)$ and $x \in N^c$}  \text{.}  $$  
It follows that $f$ is essentially bounded and $f_j \rightarrow f$ in $L^\infty$ as $j \rightarrow \infty$.  
