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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.

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Daniel Fischer
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g.a.l.l.e.t.a
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  • One needs $N_{n,m}$ to define the set $N$, out of which the sequence $(\tilde{f}_n)$ of functions converges uniformly. –  Jul 02 '19 at 17:28
  • :) I know @Jack, I know. But why? I mean why a set with such special characteristics (measure $0,$ uniform convergence outside the set). I think it's because of the almost everywhere condition – g.a.l.l.e.t.a Jul 02 '19 at 23:01
  • Let me rephrase your question: are you asking why the sequence converges uniformly on $N^c$ (the complement of $N$)? –  Jul 02 '19 at 23:27
  • @Jack no. That's another question that I have, tho not the one that I meant in my question. – g.a.l.l.e.t.a Jul 02 '19 at 23:30
  • Or I guess you meant to ask that how $N_{n,m}$ is defined in a way that (1) it has measure $0$ (2) the inequality $|f_n(x)-f_m(x)|\leq|f_n-f_m|{\infty}$ holds outside $N{n,m}$? –  Jul 02 '19 at 23:32
  • @Jack Yes. And I think it's because of the a.e. property and I also think this was taking into account because of the definition of $L^\infty$ space – g.a.l.l.e.t.a Jul 02 '19 at 23:36
  • Let us continue this discussion in chat. –  Jul 03 '19 at 02:35

2 Answers2

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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.

Dunham
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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 $N^c$.

  • Jack: $(g_n)$ is Cauchy sequence, right? – g.a.l.l.e.t.a Jul 04 '19 at 21:45
  • Why Cauchy sequence in $\mathbb R$ ? Isn't $\mathbb C$ because by definition of the functions in $L^\infty?$ – g.a.l.l.e.t.a Jul 04 '19 at 22:05
  • $\mathbb{C}$ would be OK too. It would not make much difference whether one consider real-valued or complex-valued functions. This would also depends on context: writing $L^\infty$ one may consider either real-valued or complex valued functions –  Jul 04 '19 at 22:07
  • How do you call or where does it come from this: Given >0, there exists >0 such that for all $∈^$ and for all ,≥, one has $|()−()|≤‖_−_‖_∞<?$ since its still not Cauchy (this implies $f_n$ is Cauchy). – g.a.l.l.e.t.a Jul 05 '19 at 00:09
  • @g.a.l.l.e.t.a: the existence of $M$ comes from the assumption that ${f_n}$ is a Cauchy sequence in $L^\infty$. –  Jul 05 '19 at 00:20
  • I understand now, thank you Jack – g.a.l.l.e.t.a Jul 05 '19 at 00:33