I wrote up the following proof for proving $L^\infty$ is complete. Please check if I made any mistakes, thank you!
There is a common/standard treatment for establishing the limit of a Cauchy sequence in $L^\infty$-norm using the approach suggested by Munroe, Aliprantis and Burkinshaw, Wheeden and Zygmund, Folland, and so on to take care the measurability based on the dependency and countability of measurable subsets that follow directly after applying the definition of $L^\infty$-norm and the convergence of Cauchy sequence in measure.
It appears implausible that all of these books are incorrect, which was what I was told by my graduate level analysis course professor.
But wait! I can't find a mistake in my thinking or the proofs provided by these books, so I'm hoping that someone will point out my error or that I can send some errata emails.
In a previous post, I mentioned an alternative approach for establishing the limit of the Cauchy sequence in $L^\infty$-norm which was also pointed out by the same professor that to start a proof with this is not logical and wasting one's time.
Here is for that specific method using subsequence of a sequence, since logically, if this proof is wrong that still does not imply the proof of Wheeden and Zygmund (which is also reproduced by several other authors) could be wrong/irrelevant.
The attempted proof starting with a diagonal argument:
Let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite measure space.
Step 1. Assume Cauchy in norm: $\left\Vert f_{n+p}-f_{n}\right\Vert_\infty\to 0$ when $n\to \infty$ and for all $p$ in $\mathbb{N}$ where $\{f_i \}$ is an arbitrarily chosen Cauchy sequence that converges in $L^\infty$-norm.
Since $X$ is $\sigma$-finite, $X=\bigcup\limits_{m=1}^\infty X_m$, where $\mu\left(X_m\right)<\infty$ and we can assume $X_i \cap X_j = \emptyset$ for all $i\neq j.$
The hypothesis $\left\Vert f_{n+p}-f_{n}\right\Vert_\infty\to 0$ implies that $$ \int_{X_m}\vert f_{n+p}(x)- f_n(x)\vert dx\to 0 $$ when $n\to \infty$ for all $p\in\mathbb{N}$.
Thus, according the completeness of $L^1(X_m)$ (please note that it is not according to the completeness of $L^\infty(X_m)$), there exists a vector $f^{(m)} \in L^1(X_m)$ such that $f_n\to f^{(m)}$ in $L^1(X_m)$ as $n\to \infty$, where the superscript "$(m)$" is for bookkeeping that it shows the dependency to the index $m$ of the sub set $X_m$ of $X$. In other words, $\exists f^{(m)}\in L^1(X_m)$ such that $$ \lim\limits_{n\to \infty}\left\Vert f_n -f^{(m)} \right\Vert_{1} = 0 $$ in $L^1(X_m)$-norm.
Hence, for all $m\in \mathbb{N}$, based on the assumption, there exists a subsequence of functions of the assumed Cauchy sequence $$ \{\ f_n^{(m)}\} \subseteq \{ f_n\} $$ such that
- $f_n^{(m)}(x)\to f^{(m)}(x)$ when $n\to \infty,$ a.e. $x\in X_m$, and
- for all $m>1$, $$\{\ f_n^{(m+1)}\} \subseteq \{ f_n^{(m)}\},$$ i.e. $\{\ f_n^{(m+1)}\}$ is a subsequence of functions the subsequence $\{ f_n^{(m)}\}$ which is also a subsequence of $\{ f_n\}$.
Why we are taking subsequence of subsequence? Why the diagonal argument can work? These questions can be answered by the following construction.
Let the following be the sequence or a subsequence of $\{f_n\}$: $$ \{f_n^{(1)}\}:=\{f_1^{(1)}, f_2^{(1)}, f_3^{(1)},...\}. $$
Take countable many terms in $\{f_n^{(1)}\}$ to be its subsequence: $$ \{f_n^{(2)}\}:=\{f_1^{(2)}, f_2^{(2)}, f_3^{(2)},...\}. $$
Again, take countable many terms in $\{f_n^{(2)}\}$ to be its subsequence: $$ \{f_n^{(3)}\}:=\{f_1^{(3)}, f_2^{(3)}, f_3^{(3)},...\}. $$
Thus, for the $m$-th iteration, we have $$ \{f_n^{(m)}\}:=\{f_1^{(m)}, f_2^{(m)}, f_3^{(m)},...\}, $$ and its subsequence to be the $(m+1)$-th iteration: $$ \{f_n^{(m+1)}\}:=\{f_1^{(m+1)}, f_2^{(m+1)}, f_3^{(m+1)},...\}. $$
Since they are all convergent sequence (converges for almost every $x\in X_m$), define the following limits for each subsequence:
| $f_1^{(1)}, f_2^{(1)},f_3^{(1)},...$, then its limit exists, and when an input is given the image of the limit function is defined to be $\lim_{n\to\infty}f_n^{(1)}(x):= f^{(1)}(x)$, for a.e. $x\in X_1$ |
|---|
| $f_1^{(2)}, f_2^{(2)},f_3^{(2)},...$, then its limit exists, and when an input is given the image of the limit function is defined to be $\lim_{n\to\infty}f_n^{(2)}(x):= f^{(2)}(x)$, for a.e. $x\in X_2$ |
| $f_1^{(3)}, f_2^{(3)},f_3^{(3)},...$, then its limit exists, and when an input is given the image of the limit function is defined to be $\lim_{n\to\infty}f_n^{(3)}(x):= f^{(3)}(x)$, for a.e. $x\in X_3$ |
| ...................................................................... |
| $f_1^{(m)}, f_2^{(m)},f_3^{(m)},...$, then its limit exists, and when an input is given the image of the limit function is defined to be $\lim_{n\to\infty}f_n^{(m)}(x):= f^{(m)}(x)$, for a.e. $x\in X_m$ |
| $f_1^{(m+1)}, f_2^{(m+1)},f_3^{(m+1)},...$, then its limit exists, and when an input is given the image of the limit function is defined to be $\lim_{n\to\infty}f_n^{(m+1)}(x):= f^{(m+1)}(x)$, for a.e. $x\in X_{m+1}$ |
The proof idea here is that since $\{f_n^{(2)}\}$ is a subsequence of $\{f_n^{(1)}\}$, we also have a limit function to take the input for a.e. $x\in X_1$, and the limit function is the limit of the sequence of functions $\{f_n^{(2)}\}$. That is, $\lim_{n\to \infty}f_n^{(2)}(x)$ converges to $g^{(1)}(x)$ for a.e. $x\in X_1$.
Likewise, since $\{f_n^{(m+1)}\}$ is a subsequence of $\{f_n^{(1)}\}$, $\{f_n^{(2)}\}$, $\{f_n^{(3)}\}$,...$\{f_n^{(m)}\}$, thus we also have corresponding limits for the sequence of functions $\{f_n^{(m+1)}\}$ for each different inputs, i.e.
- $\lim_{n\to \infty}f_n^{(m+1)}(x):= F^{(1)}(x)$ exists for a.e. $x\in X_1$,
- $\lim_{n\to \infty}f_n^{(m+1)}(x):= F^{(2)}(x)$ exists for a.e. $x\in X_2$,
- $\lim_{n\to \infty}f_n^{(m+1)}(x):= F^{(3)}(x)$ exists for a.e. $x\in X_3$,
- ........,
- $\lim_{n\to \infty}f_n^{(m+1)}(x):= F^{(m)}(x)$ exists for a.e. $x\in X_m$, and
- $\lim_{n\to \infty}f_n^{(m+1)}(x):= f^{(m+1)}(x)$ exists for a.e. $x\in X_{m+1}$.
That is, $\{f_n^{(m)}\}$ can also converge for almost every input $x$ in $X_i$ where $i< m$.
This is why the diagonal argument is applied, and there is one more step for this diagonal proof that is to take the diagonal terms (functions, not their images) to be a subsequence $\{f_n^{(n)}\}$ since this subsequence then can take care almost every input $x\in X = \cup_1^\infty X_m$. This is also why in several books, this convergent is called uniform convergence almost everywhere, because the $N=N_\epsilon$ does not depend on the input $x$.
To see this, denote all the images of limits of subsequences of functions to be $f(x):=f^{(m)}(x)$ for a.e. $x\in X_m$ for $m\in\{ 1,2,...\}$. Then the diagonal subsequence of images of functions converges to $f(x)$: $$ \lim\limits_{n\to\infty} f^{(n)}_{n}(x)= f(x) $$ for almost every $x\in X = \bigcup\limits_{m=1}^\infty X_m.$
Since $\lim\limits_{n\to\infty} f^{(n)}_{n}= f$, i.e. $f$ is the limit function of a convergent subsequence $\{f_n^{(n)}\}$ of $\{f_n\}$, then for almost every $x\in X = \bigcup\limits_{m=1}^\infty X_m$, we have $$ \lim\limits_{n\to\infty} f_{n}(x)= f(x) $$ for almost every $x\in X = \bigcup\limits_{m=1}^\infty X_m.$
Step 2. Next, prove: $f\in L^{\infty}(X)$ and $\left\Vert f_n-f\right\Vert_{\infty}\to 0$ as $n\to \infty$. According to the hypothesis, $\forall \epsilon>0$, $\exists N = N_\epsilon$ such that $\forall p\in\mathbb{N}$, we have $$ \left\Vert f_{n+p}-f_n \right\Vert_\infty < \epsilon. $$
Recall: An equivalent definition of $L^\infty$-norm of a function $v\in L^\infty(X,\mathcal{A},\mu)$: $$ \left\Vert v\right\Vert_\infty:=\inf\limits_{\mu\left(E_0\right)=0, E_0\subset X} \left( \sup\limits_{x\in X\setminus E_0} \left\{\vert v(x)\vert \right\}\right) $$
Thus, for all $p\in\mathbb{N}$, for all $n\geq N$, there exists $A_n^{(p)}\subset X$ such that $\mu\left( A_n^{(p)}\right) = 0,$ and $$ \sup\limits_{x\in X\setminus A_n^{(p)}}\left\vert f_{n+p}(x)- f_n(x)\right\vert< \epsilon. $$
Let $A_n := \bigcup\limits_{p=1}^\infty A_n^{(p)}$.
Then, $\mu\left( A_n \right)=0$, and for all $p\in\mathbb{N}$, and for all $n\geq N$, we have $$ \sup\limits_{x\in X\setminus A_n}\left\vert f_{n+p}(x)-f_n(x) \right\vert<\epsilon. $$
Take $p\to \infty$, then we can obtain $$ \sup\limits_{x\in X\setminus A_p}\left\vert f_n(x)-f(x)\right\vert\leq \epsilon. $$
It follows that $f\in L^\infty(X,\mu)$, and $\left\Vert f_n-f \right\Vert_\infty\leq \epsilon$ for all $n>N$.
The above proof was modified and based on a proof given in this book published in 1987.
So, am I missing anything here or have all of these folks gotten this detail incorrect (since I was taught that the step 1 is irrelevant/incorrect, and in step 2 the dependency of $p$ and $n$, i.e. the set $A_n$ and $A_n^{(p)}$, is irrelevant)?
Can someone kindly verify or correct anything I wrote in the above? Thank you.
I already wrote a related answer at here.
In brief, the thought process of the claim that to say step 1 is not logical/mathematical, or this kind of proofs are wasting time, or the result can follow without a need to further address why might due to the reasoning as follow:
Without a proof, from (ii) to (iii) is merely an analogy from topological/metric space to measure space by assuming results of Cauchy sequences in metric/topological spaces could also valid in a morally similar way, hence if without a proof, then (iii) becomes an assumption.
However, since (a) the condition "almost every $x\in\mathbb{R}$" exists in (ii), and (b) in (i) the $L^\infty$-norm is used, the measure $\mu$ of the measure space is invoked. In other words, if one says that what consequences Cauchy sequences have in metric/topological spaces will also be true in measure space, then one is using an analogy or wishful thinking which is not mathematics (at least yet). To be mathematical or mathematically logical, one should prove/disprove the analogy. That is why Munroe assigned this implication (from (ii) to (iii)) as an exercise, Aliprantis and Burkinshaw assigned it as an exercise, and Folland put this into a theorem.
Although the method is not uniques, we can still show the sufficiency and necessity of the proof (for the implication from (ii) to (iii)). For necessity (of the step 1), one can see from (ii) to (iii), one cannot bypass it without proving it by simply saying one can define Cauchy sequence in a different way, because this implication is not included in the Cauchy sequence in measure space or in $L^\infty$-norm which might be either (i) or (ii). The implication from (ii) to (iii) is not a part of a definition of Cauchy sequence - it could possibly be a result of the definition, but then this shows that a proof for this implication is necessary. To show the sufficiency (of the step 1), one just needs to verify to see if this proof is correct so that the result could be obtained.