Understanding Thm. 13.6 in Bruckner's Real Analysis (completeness of $L^{\infty}$)

Question

In the following is Theorem 13.6 from Bruckner's Real Analysis which I don't understand some claims on it :

Question in Blue: $\mu (|f_j(x)| > \|f_j\|_∞)=0$ and $\mu (|f_k(x)| > \|f_k\|_∞)=0$. But how that implies $\mu (|f_j(x) - f_k(x)| > \|f_k\|_∞)=0$?

Question in Green: It is clear that $f$ is bounded and is the pointwise convergence of $f_n$ to $f$. How that convergence is uniform?

Theorem 13.6 Let $(X, \mathcal{M}, \mu)$ be a measure space. Then the space $L_{\infty}(\mu)$ is a Banach space furnished with the norm $\|f\|_{\infty} .$

Proof. It is easy to see that a linear combination of essentially bounded functions remains essentially bounded, and so the space is linear. It is almost immediate that $\|f\|_{\infty}$ is a norm on this space. The triangle inequality, that $$ \|f+g\|_{\infty} \leq\|f\|_{\infty}+\|g\|_{\infty} $$ (which can also be considered as the extension of Minkowski's inequality to the case $p=\infty$ ), follows from the set inclusion $$ \begin{array}{c} \left\{x:|f(x)+g(x)|>\|f\|_{\infty}+\|g\|_{\infty}\right\} \\ \subset\left\{x:|f(x)|>\|f\|_{\infty}\right\} \cup\left\{x:|g(x)|>\|g\|_{\infty}\right\}. \end{array} $$ Exercise 13:3.2 shows that each of the sets on the right side of the inclusion has $\mu$ -measure zero and so, too, must the set on the left. This gives the triangle inequality.

$\quad$The completeness part of the proof is rather simpler than the completeness proof for the $L_{p}$ spaces with $1 \leq p<\infty$. Let $\left\{f_{n}\right\}$ be Cauchy in $L_{\infty}(\mu)$. Define $A_{i}$ to be the set of points $x$ in $X$ for which $\left|f_{i}(x)\right|>\left\|f_{i}\right\|_{\infty}$, and define $\color{blue}{B_{j, k}}$ to be the set of points $x$ in $X$ for which $\left|f_{j}(x)-f_{k}(x)\right|>\left\|f_{k}\right\|_{\infty}$. All these sets $\color{blue}{\text{have measure zero by definition}}$. Let $E$ be the totality of all these points, that is, the union of these sets taken over all integers $i, j, k$. Then $E$ has measure zero, and the sequence $\left\{f_{n}(x)\right\}$ converges for every $x \in X \backslash E$, and indeed $\color{green}{\text{it converges uniformly to some bounded function $f$}}$ defined on $X \backslash E .$ We can extend $f$ to all of $X$ in any arbitrary fashion [or simply set $f(x)=0$ for $x \in E]$, and it is easy to see that $f \in L_{\infty}(\mu)$ and that $\left\|f-f_{n}\right\|_{\infty} \rightarrow 0$ as $n \rightarrow \infty$.$\blacksquare$

(Original screenshot here)

Question 3 : Exercise 13:3.1 says that a sequence $f_n$ converges to a function $f$ in the space $L_∞(X, M,μ)$ if and only if there is a set $E ∈M$ with $μ(E)=0$ so that $f_n → f$ uniformly on $X \setminus E$. One direction is the "Green" question, for the other direction, isn't it a convention/definition rather than a theorem?

Answer 1

It seems that there is a typo in your textbook.

I am writing here a proof that is closed to your text's. Suppose $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence in $L_\infty(\mu)$. Then for any $\varepsilon>0$, there is $N$ such that $$\begin{align} \|f_n-f_m\|_\infty<\varepsilon,\qquad n,m\geq N\tag{0}\label{0}\end{align}$$ As it is mentioned in your book

1. $\mu(|f_n|>\|f_n\|_\infty)=0$
2. $\mu(|f_m-f_n|>\|f_n-f_m\|_\infty)=0$.

It follows that the set $A=\bigcup_n\{|f_n|>\|f_n\|_\infty)\cup\bigcup_{n,m}\{|f_n-f_m|>\|f_n-f_m\|_\infty\}$, being the union of countable sets of measure $0$, has measure $0$. Setting $B=X\setminus A$, we have from \eqref{0} that $$\begin{align} |f_n(x)-f_m(x)|\leq \|f_n-f_n\|<\varepsilon ,\qquad n, m\geq N,\,x\in B\tag{1}\label{1}\end{align}$$ This argument shows that on $B$, $f_n$ converges uniformly. Define $f(x)$ to be $0$ if $x\in A$ and $f(x)=\lim_{n\rightarrow\infty}f_n(X)$ for $x\in B$ (this is well defined since $\{f_n\}$ converges uniformly on $B$). Passing to the limit $m\rightarrow\infty$ in \eqref{1} gives you that $$|f_n(f)-f(x)|\leq\varepsilon,\qquad n\geq N,\, x\in B$$ Since $\mu(A)>0$, $\|f-f_n\|_\infty\leq\varepsilon$ for all $n\geq N$.

Since $\|f_n(x)\|\leq \varepsilon+\|f_N(x)\|_\infty:=C_\varepsilon$ for all $n\geq N$ and $x\in B$, $f$ is bounded by $C_\varepsilon$ in $B$. This means that $\|f\|_\infty\leq C_\varepsilon$.

Notes:

• If a function $f$ is bounded on a measurable set $E$ by say, a constant $C>0$, and $\mu(E^c)=0$, then $\mu(|f|>C)\leq\mu(E^c)=0$. Consequently, from the definition of $\|f\|_\infty$, it follows that

1. $f\in L_\infty(\mu)$,
2. $\|f\|_\infty\leq C$.

Proof: Recall that $\|f\|_\infty=\inf\{a>0:\mu(|f|>a)=0\}$. The assumption on $f$ implies that $C\in\mathcal{E}=\{a>0:\mu(|f|>a)=0\}$. Hence $\|f\|_\infty\leq C$.

• Another important property of $\|f\|_\infty$ that is used implicitly is the fact that $\mu(|f|>\|f\|_\infty)=0$.

Proof: Let $\mathcal{E}$ be as before. For any $a>0$, if $a>\|f\|_\infty$, then $a\in \mathcal{E}$. To see this, suppose $a>\|f\|_\infty$. Then, by definition of $\|f\|_\infty$, there is $b\in \mathcal{E}$ such that $\|f\|_\infty\leq b<a$. Then $\mu(|f|>a)\leq \mu(|f|>b)=0$; thus, $a\in \mathcal{E}$. Finally, let $a_n\in\mathcal{E}$ such that $a_n>\|f\|_\infty$ and $a_n\searrow\|f\|_\infty$. Then $\mu(|f|>\|f\|_\infty\}=\bigcup_n\{|f|>a_n\}$ whence we conclude that $$\mu(|f|>\|f\|_\infty)=\lim_n\mu(|f|>a_n)=0$$

Answer 2

$B_{j,k}$ needs not have measure zero. For, if $f_k$ is a constant function with $f_k =-1$, then $||f_k||_\infty = 1$. If $f_j$ is another constant function with $f_j=1$, then $ | f_j(x) - f_k(x) | = 2 > ||f_k||_\infty$ for all $x$.

There must be some typos (or even gaps) in the proof. You may need to fix them by yourself.

Remark: It is worth mentioning that elements in $L^\infty$ are not bounded measurable functions. They are equivalent classes of essentially bounded measurable functions. Let $([f_n])_n$ be a Cauchy sequence in $L^\infty$. When you pick a representative $f_n$ from the equivalent class $[f_n]$, you need to pay attention that your argument should be independent of choice of representative.

In order to have the duality $(L^1)^\ast = L^\infty$, if $\mu$ is not $\sigma$ finite, you need to modify the definition of "essential upper bound" by replacing $\mu$-null set with $\mu$-locally null set. See Definition of $L^\infty$.

Answer 3

An easier way to prove that the norm of a normed space is complete is to invoke the following result:

Let $(X,||\cdot||)$ be a norm space. Then the norm is complete if and only if for every sequence $(x_{n})$ in $X$, $\sum_{n=1}^{\infty}||x_{n}||<\infty\Rightarrow$ $\sum_{n=1}^{\infty}x_{n}$ converges.

Assume the above standard result in functional analysis. For your question: Let $(\alpha_{n})$ be a sequence in $L^{\infty}$ such that $\sum_{n=1}^{\infty}||\alpha_{n}||<\infty$. For each $n$, pick a representative $f_{n}\in\alpha_{n}$. Define $A_{n}=\{x\in X\mid|f_{n}(x)|>||\alpha_{n}||\}$, then $\mu(A_{n})=0$. (Warning! If $\mu$ is not $\sigma$-finite, $A_n$ is just locally null, in the sense that for any measurable set $C$ with $\mu(C)<\infty$, we have that $\mu(C\cap A_n)=0$. Moreover, in the following, $B^c$ is just a $\mu$-locally null set. In this proof, we assume that $\mu$ is $\sigma$-finite.)

Let $B=X\setminus\cup_{n}A_{n}$. For each $x\in B$, we have that $|f_{n}(x)|\leq||\alpha_{n}||$. Therefore, $\sum_{n=1}^{\infty}|f_{n}(x)|\leq\sum_{n=1}^{\infty}||\alpha_{n}||<\infty$. By completeness of $\mathbb{R}$, $\sum_{n=1}^{\infty}f_{n}(x)$ converges. Let $f=\sum_{n=1}^{\infty}f_{n}1_{B}$, which is measurable because it is pointwise limit of measurable functions. Moreover, for each $x\in X$, $|f(x)|\leq\sum_{n=1}^{\infty}||\alpha_{n}||<\infty$, so $f$ is bounded. Finally, go to show that $\sum_{n=1}^{N}\alpha_{n}\rightarrow[f]$ with respect to the norm $||\cdot||$.

Note that a representative of $\sum_{n=1}^{N}\alpha_{n}$ is $\sum_{n=1}^{N}f_{n}$. For each $x\in B$, we have that \begin{eqnarray*} & & \left|f(x)-\sum_{n=1}^{N}f_{n}(x)\right|\\ & = & \left|\sum_{n=1}^{\infty}f_{n}(x)-\sum_{n=1}^{N}f_{n}(x)\right|\\ & \leq & \sum_{n=N+1}^{\infty}|f_{n}(x)|\\ & \leq & \sum_{n=N+1}^{\infty}||\alpha_{n}||. \end{eqnarray*} Therefore, $\{x\in X\mid\left|f(x)-\sum_{n=1}^{N}f_{n}(x)\right|>\sum_{n=N+1}^{\infty}||\alpha_{n}||\}\subseteq B^{c}$ which has measure zero. That is, $\sum_{n=N+1}^{\infty}||\alpha_{n}||$ is an essential upper bound of $|f-\sum_{n=1}^{N}|$. Hence, $||[f]-\sum_{n=1}^{N}\alpha_{n}||\leq\sum_{n=N+1}^{\infty}||\alpha_{n}||\rightarrow0$ as $N\rightarrow\infty$. This shows that $\sum_{n=1}^{N}\alpha_{n}\rightarrow[f]$ with respect to $||\cdot||_{\infty}$-norm.