Why was this set $N_{n,m}$ defined?

Question

I'm trying to understand the proof below.

Why was the set $N_{n,m}$ of measure zero defined in this proof?

Why is the sequence uniformly convergent?

Prove that the normed space $L^{\infty}$ equipped with $\lVert\cdot\rVert_{\infty}$ is complete.

Answer 1

How are the sets $N_{n,m}$ defined?

1. By definition of the $L^\infty$ norm, for any $f\in L^\infty(X,\mathcal{A},\mu)$, there exists a set $B$ with $\mu(B)=0$ such that for every $x\in B^c$, $$ |f(x)|\le\|f\|_\infty. $$

2. By what we have above, one has for any fixed positive integers $m$ and $n$, there exists $N_{n,m}$ with $\mu(N_{n,m})=0$, such that for every $x\in N_{n,m}^c$, $$ |f_n(x)-f_m(x)|\le \|f_{n}-f_m\|_{\infty}. $$

Why does the sequence $(g_n)$ with $g_n:=\tilde{f}_n$ converge uniformly on $N^c$?

Given $\epsilon>0$, there exists $M>0$ such that for all $x\in N^c$ and for all $n,m\geq M$, one has $$ |f_{n}(x)-f_m(x)|\le \|f_{n}-f_m\|_{\infty}<\epsilon,\tag{1} $$ which implies that for every (fixed) $x\in N^c$, $\{f_n(x)\}_{n=1}^\infty$ is a Cauchy sequence in $\mathbb{R}$ and hence the pointwise limit $\lim_nf_n(x)$ exists (in $N^c$). Thus, we can define: $$ f(x):=\lim_{n\to\infty}f_n(x),\quad x\in N^c. $$

On the other hand, (1) implies that given $\epsilon>0$, there exists $M>0$, such that for all $n>M$, $$ \sup_{x\in N^c}|f(x)-f_n(x)|<\epsilon, $$ which, by the definition of uniform convergence, implies that $g_n$ converges uniformly to $f$ in $N^c$.

Answer 2

Elements of $L^{\infty}$ are equivalence classes of functions. To work through the proof, they are taking specific representatives $f_n$ which have pointwise definitions. Of course, there may be a set of measure 0 where things are not nice, but it can only be of measure 0. The sets $N_{n,m}$ are the potentially "bad" sets, where we cannot guarantee convergence. In the end, the limit $f$ is defined as the limit of the functions on the "good" set $N^c$ and defined to be $0$ outside. This is a single representative of the equivalence class.

update

$L^\infty$ or almost any function space defined in terms of measures consists of equivalence classes of functions. While not always discussed in detail, proofs often use single representatives of the class, and the results are independent of the specific function chosen.

In this case of $L^{\infty}$, $f\sim g$ ($f$ relates to $g$) if the set of $x$ where $f-g$ is not zero is a set of measure 0. The norm is not the supremum, but rather the essential supremum, i.e. the supremum excluding arbitrary sets of measure 0. More precisely, given a representative $f$ for an equivalence class $F$

$$ \|F\|_{\infty} = \inf_{\text{$E$ having measure 0}} \sup_{x\in E^c} \{ |f(x)| \} $$

The sets $E_{n,m}$ are indexed by the possible pairs of the sequence $\{f_n\}$. Each pair $f_n,f_m$ gives rise to a difference $f_n-f_m$ where we need to focus on the essential behavior, on the complement of measure 0 sets. The union of the $E_{n,m}$ is still measure 0, being a countable union, so we can ignore everything in it.