A question on a step in proving $L^\infty$ is complete.

Question

Let $(X,\mathcal{A},\mu)$ be $\sigma$-finite measure space.

Claim: $L^\infty$ is complete.

Ideas of the proof:

• (i) Assume Cauchy in norm - let $\{f_n\}$ be a Cauchy sequence which converges in $L^\infty$ norm. That is, let $\epsilon>0$, and assume $\left\Vert f_n-f_m \right\Vert_\infty<\epsilon.$ Prove: $ \lim\limits_{n\to\infty}f_n(x) \text{ exists } $ for all $x\notin A=\bigcup\limits_{m,n}A_{m,n}\subset X$, where $\mu\left( A_{m,n}\right)=0$ and on $A_{m,n}$ $f$ could take any finite value. I found that this diagonal argument was used in proving $L^\infty$ is complete in a book written by Zhang Gongqing, but I was taught that (to start with or even include) this step in proving $L^\infty$ is complete is not logical/mathematical.

• (ii) Define $f:=\lim\limits_{n\to\infty}f_n(x)$ for all $x\notin A$. Prove: $f \in L^{\infty}$.

• (iii) Prove: $\left\Vert f_n-f\right\Vert_\infty<\epsilon,$ i.e. $\{f_n\}$ is uniformly convergent on $X\setminus A$.

When I presented the above outline to the professor of my analysis course, I was taught that (i) is not logical. The structure of the set $A$ should not be mentioned, and it is incorrect to include a step to show $\lim\limits_{n\to\infty}f_n(x)= f(x)$ for all $x\notin A$. Instead, one should directly define this limit (with the definition argue that the value of the limit is real, hence exists), and by starting with another definition of Cauchy sequence to prove (iii) directly:

$(f_n)$ is a Cauchy sequence, if $\exists N\in \mathbb{N} s.t. \forall m, n \in \mathbb{N}$ we have $$ \vert f_m(x)-f_n(x)\vert <\epsilon. $$ for almost every $x$.

I was taught to remove anything about $A_{n,m}$ and the existence of $\lim\limits_{n\to\infty}f_n(x)= f(x)$ for all $x\in A$. Later on, I found a similar proof in Weeden and Zygmund My question is that since I was suggested to remove anything about the dependency on $m$ and $n$ of the measure zero subset, could Weeden and Zygmund's proof still work after removing the highlighted words? This $m,n$ dependency comes in naturally if one start to prove this theorem by recalling the definition of $L^\infty$ norm first. The reason to recall this definition is the a normed vector space is defined to be complete if any Cauchy sequences in the space converge in norm. The norm is defined as $$ \left\Vert g \right\Vert_\infty = \inf\{a\geq 0:\mu\left(x: \vert g(x)\vert>a\right)=0 \}. $$ Then by replacing $g$ with $f_n-f_m$ and then, there exists $N\in\mathbb{N}$ such that $$ \left\Vert f_n-f_m \right\Vert_\infty = \inf\{a\geq 0:\mu\left(\{x: \vert f_n(x)-f_m(x)\vert>a\}\right)=0 \}<\epsilon, $$ for all $m,n>N$. Since $\epsilon$ is arbitrary, $$ \left\Vert f_n-f_m \right\Vert_\infty = \inf\{a\geq 0:\mu\{\left(x: \vert f_n(x)-f_m(x)\vert>a\}\right)=0 \}\to 0 $$ as $m,n\to \infty$. Then, this implies that $\forall \epsilon>0$, $$ \mu\left(\{x: \vert f_n(x)-f_m(x)\vert \geq \epsilon \} \right)\to 0 $$ as $m,n\to \infty,$ i.e. $\{f_n\}$ is Cauchy in measure. Next, we can define $A_{n,m}:=\{x: \vert f_n(x)-f_m(x)\vert \geq \epsilon \}.$ Therefore, the dependency on $m$, $n$ (and $\epsilon$) appears naturally. If not, then I assume one must somehow use a different definition of completeness of $L^\infty$, or there is an equivalent definition of a sequence is Cauchy in measure. Next, by Theorem 2.30 in Folland (basically when the proof of this theorem is specified to $L^\infty$, then it is equivalent to (i) immediately), Cauchy in measure implies

• (a) there is a measurable function $f$ such that $f_n\to f$ in measure, i.e. for all $\epsilon>0$, $$ \mu\left(\{x:\vert f_n(x)-f(x)\vert\geq \epsilon\} \right)\to 0 $$ as $n\to \infty$. Then, by applying the $L^\infty$-norm again, we have $\forall \epsilon>0,$ $\exists N$ such that $\forall n>N$, $\left\Vert f_n-f\right\Vert<\epsilon.$

• (b) there is a subsequence $\{f_{n_j}\}$ that converges to $f$ a.e. (This is where the diagonal argument comes in. By relabeling the subsequence using $\{f_n\}$, then one can have $f_n\to f$ a.e.) By the construction of $f$, we know that since $\mu(A)=0$, and $f$ takes real value, $f$ is essentially bounded, thus $f\in L^\infty.$

My questions:

• (1) In summary, again, could Weeden and Zygmund's proof (or any similar proofs in several other books) still work after removing the highlighted words (i.e. anything related to the dependency on $m$ and $n$ of a measure zero set of a Cauchy sequence?

• (2) Is it due to a hidden assumption that why if one uses the alternative definition of Cauchy sequence then one does not need to mention the results such as Theorem 2.30 in Folland, exercise 19.7 in Aliprantis and Burkinshaw, exercise 42-d in Munroe 2nd ed., exercise 10.1 in Richard Bass, and so on? To make this question be more clear, the following is from exercise 42-d in Munroe 2nd ed. page 223: Hence, in this book, to have the limit $\lim_{n\to\infty} f_n=f$, one has to solve this exercise first, because if one wants to use it, it is better to know how to prove it. However, in the opposite, I was taught that we should directly define the limit (so instead of proving anything, it becomes a definition) and this definition has nothing to do with the hint given by Munroe (if I mention that then my proof will become not logical, i.e. false), meaning that this if and only if statement was turned into a definition directly. What distinguishes the two theoretical frameworks from one another? (For this reason, I suppose there's a hidden assumption or that the two distinct methods for determining this limit result from the two different starting points/assumptions/definitions.)

Can someone kindly verify or correct anything I wrote in the above? Thank you.

Answer 1

Now, it is clear (I think) that an analogy is not a proof. The thought process of that claim to say it's not logical/mathematical; it's wasting time; and the result can follow without a need to further address why might due to the reasoning as follow:

• (1) Given $\epsilon>0, \exists N\in\mathbb{N}$, then $\left\Vert f_n-f_m \right\Vert_\infty<\epsilon$, $\forall m,n>N$.
• (2) Then, $\vert f_n(x)-f_m(x)\vert <\epsilon$ for almost every $x\in\mathbb{R}$.
• (3) Then, one can define $\lim\limits_{n\to\infty} f_n (x) = f(x)$ for a.e. $x\in\mathbb{R}$, since by definition, the previous line shows $f_n$ is Cauchy (which is not clear, and need a proof to reach this result, e.g. either using Munroe's approach, or Wheeden and Zygmund's way to briefly explain why, or like the step 1 in this proof, or Folland's Theorem 2.30, and so on).

Without a proof, from (2) to (3) 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 (3) becomes an assumption.

One can see from (2) to (3), 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 itself which might be either (1) or (2) themselves, but the implication from (2) to (3) is not a part of a definition of Cauchy - it could possibly be a result of the definition, but then this shows that a proof is necessary.

However, since (a) the condition "a.e. $x\in\mathbb{R}$, or almost every $x\in\mathbb{R}$" exists in (2), and (b) in (1) 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 transition (from (2) to (3)) as an exercise, Aliprantis and Burkinshaw assigned it as an exercise, and Folland put this into a theorem.

Instead of using Munroe's approach to establish the limit, we can use problem 19.7 in Aliprantis and Burkinshaw 2nd edition, or Theorem 2.30 in Folland: Assume that $\{f_n\}$ is Cauchy in measure, if and only if there exists a measurable function $f$ such that $\lim_{n}f_n=f$ in measure, i.e. for every $\epsilon > 0$, $$ \mu\left( \{ x:\vert f_n(x)-f(x)\vert\}\geq \epsilon \right)\to 0 $$ as $n\to \infty.$ However, the proof of the above theorem is exactly using the argument that the union of measure zero set that has dependency on $m,n$ still has measure zero. In other words, the following are equivalent:

• $\lim\limits_{n\to\infty} f_n = f$ in $L^\infty$-measure,
• if and only if there exists a set $A$ with $\mu(A)=0$ and $\lim_n f_n = f$ uniformly on $X\setminus A$,
• if and only if there exists a measurable function $f\in L^\infty$ such that $f_n\to f$ in $L^\infty$-measure, and
• if and only if $\{ f_n\}$ is Cauchy in $L^\infty$-measure.

However, we have to prove the if and only if relation before we can add each one of the above into the TFAE list.

It is not trivial to add each of these results into the list, since it is not always true in $L^\infty$ to have both directions, e.g. see this example.

Answer 2

There is nothing wrong with the proof in Weeden & Zygmund.

$$ \|g\|_\infty=\inf\{a\ge 0\colon\mu\left(\{x\colon \vert g(x)\vert>a\}\right)=0\} $$

Note that if $b>a$ then $\{x\colon |g(x)|>b\}\subseteq\{x\colon |g(x)|>a\}$. Therefore if $\|g\|_\infty<+\infty$ then $\mu(\{x\colon |g(x)|>a\})=0$ for all $a>\|g\|_\infty$.

Assume $\|g\|_\infty<+\infty$. For $n>0$ (an integer) let $$ \begin{align} A_n&=\{x\colon|g(x)|>\|g\|_\infty+1/n\}\\ A&=\{x\colon|g(x)|>\|g\|_\infty\} \end{align} $$ By definition of $\|g\|_\infty$, $\mu(A_n)=0$ for all $n>0$. If $x\in A$, there exists some $m>0$ such that $|g(x)|>\|g\|_\infty+1/m$ (this is due to the Archimedean property of $\Bbb{R}$) and therefore $x\in A_m$. This proves that $$ A\subseteq \bigcup_{n\in\Bbb{N}^+} A_n $$ and since $\mu(A_n)=0$ for all $n\in\Bbb{N}^+$ it follows by $\sigma$-sub-additivity that $\mu(A)=0$. The complement of $A$ is $\{x\colon |g(x)|\le\|g\|_\infty\}$. Therefore $|g(x)|\le\|g\|_\infty$ almost everywhere.