I do not understand a point in the proof of completness of $L^{\infty}$

Question

do not understand a point in the proof of completness of $L^{\infty}$. I have this proof.

We consider the sets $$A_{n,m}=\{x\in E:|f_{n}(x)-f_{m}(x)\|\leq\|f_{n}-f_{m}\|_\infty\}$$ for all $n,m\in\mathbb{N}$. I have $\{f_k\}$ Cauchy.

for which it holds that $\mu (E\backslash A_ {n, m}) = 0 $. Also, $ A = \bigcap_{n, m} A_ {n, m} $ Then, $ \mu (E \backslash A) = 0$ For each, $ n, m \in \mathbb {N} $ it holds that: $$ | f_ {n} (x) -f_ {m} (x) | \leq \| f_n-f_m \|_{\infty} <\epsilon $$ since the sequence $ \{f_{k}\} $ is uniformly Cauchy. It is also uniformly convergent. So there is a measurable function $ f: E \rightarrow \mathbb {R} $ such that $ f_n \rightarrow f $ uniformly in $ A $. Ie, $$ \| f_{n} -f \|_{\infty} = \| (f_n-f)_{\chi_{A}} \|_{\infty} \leq \ sup_{x\in A} | f_{n} -f (x) |\rightarrow 0 $$ So, $ f_n\rightarrow f \in L_{\infty} (E) $

$$$$ $$$$

So i dont understand why $$\| (f_n-f)_{\chi_{A}} \|_{\infty} \leq \ sup_{x\in A} | f_{n} -f (x) |$$ Can anybody help me? Thanks

Answer 1

The left hand side is the essential supremum of $|f_n-f|$ on $A,$ i.e., the lowest essential upper bound of $|f_n-f|$ on $A.$

The right hand side is the (ordinary) supremum of $|f_n-f|$ on $A,$ i.e., the lowest upper bound of $|f_n-f|$ on $A.$

The inequality holds because every upper bound is an essential upper bound: the set where it does not hold is the empty set, which has measure zero.