$(X,\mu)$ is a measure space. Show that, $L^\infty(X;\mu)$ is either finite dimensional or non-separable.

Question

Suppose $L^\infty(X;\mu)$ is not finite dimensional. We have to prove it is non-separable. Suppose not i.e. $L^\infty(X,\mu)$ is separable. Then $\mathcal{F}=\{f\in L^\infty(X;\mu):\ \text{Range}(f)\subseteq\{0,1\}\}=\{\chi_A:\ A\text{ is measurable}\}$ is separable in $\lVert\cdot\rVert_\infty$ norm.

I have proved that for $f,g\in\mathcal{F}$ with $f\ne g$, we have $\lVert f-g\rVert_\infty=1$. Hence, $\mathcal{F}$ is discrete space. As $\mathcal{F}$ is separable, $\mathcal{F}$ should be countable.

From here, I want to prove $L^\infty(X;\mu)$ is finite dimensional. Usually the cardinality of the set $\{f:X\to\Bbb{C}:\ \text{Range}(f)\subseteq\{0,1\}\}$ is $2^{|X|}$ but when we consider the functions as member of $L^\infty$, two distict maps $f,g$ (as set theoretic) may be equal more specifically, $\chi_A=\chi_B\iff \mu(A\triangle B)=0$.

Can anyone help me complete the proof? Thanks for your help in advance.

Answer 1

We say a measure $\mu$ has infinite support if there is a sequence of pairwise disjoint subsets $A_n$ such that $\mu(A_n)>0.$ For any subset $I\subset \mathbb{N}$ let $f_I$ denote the indicator function of $B_I=\displaystyle\bigcup_{i\in I}A_i.$ Then $\|f_I-f_J\|_\infty\ge 1$ for $I\neq J,$ as $ \mu (B_I\triangle B_J)>0.$ The cardinality of the family $I\subset \mathbb{N}$ is equal continuum. Therefore the space is not separable.

If the measure $\mu$ has finite support then $X$ can be decompsed into a finite family of disjoint subsets $A_1,A_2,\ldots,A_n$ of positive measure such that every set $A_j$ cannot be decomposed into two disjoint sets of positive measure. Then $L^\infty$ is $n$-dimensional.

Answer 2

Say family of measurable subsets to be good if all sets in it have positive measure and measure of intersection of any two different sets from the family is $0$. Also, say subset $A$ of $X$ is good if there are arbitrary large good families consisting of subsets of $A$.

There are formally three variants:

1. All good families have size of at most $k$. Then dimension of our space is also at most $k$ - take any good family of maximum size, and characteristic functions of sets from this family will form a basis.
2. There is an infinite good family. Then characteristic functions of all possible unions of sets from this family form uncountable discrete set.
3. There are arbitrary large good families, but no infinite good family.

Let us prove that variant 3 is impossible. We will need AC here: for any $A$ of positive measure let $f(A)$ be such that $f(A) \subset A$, $\mu(f(A)) > 0$, $\mu(A \setminus f(A)) > 0$ if such set exists. Note that if $A$ is good, then such set exists, and at least one of $f(A)$ and $A \setminus f(A)$ is also good.

Assuming there arbitrary large good families, we will build an infinite sequence of finite sequences of sets, $U_i^j$, $j = \overline{1, i}$, s.t. for any $i$, $\{U_i^j | 1 \leq j \leq i\}$ is a good family, and $U_i^j = U_{i + 1}^j$ for any $i$ and $j < i$. Then $\{U_{i + 1}^i | i \in \mathbb N\}$ is an infinite good family.

We will also maintain invariant that $U_i^i$ is good.

Now, the construction with all that definitions is very simple: $U_1^1 = X$, $U_{i + 1}^j = U_i^j$ for $j < i$. If $f(U_i^i)$ is good, then $U_{i + 1}^i = U_i^i \setminus f(U_i^i)$ and $U_{i + 1}^{i + 1} = f(U_i^i)$. Otherwise $U_i^i \setminus f(U_i^i)$ is good, and we can take $U_{i + 1}^{i} = f(U_i^i)$ and $U_{i + 1}^{i + 1} = U_i^i \setminus f(U_i^i)$.

Answer 3

Consider the collection $\mathcal{A}$ of equivalent classes ($A\sim B$ iff $\mu(A\triangle B)=\int|\mathbb{1}_A-\mathbb{1}_B|\,d\mu=0$) of measurable sets. For any given measurable set $A$, denote by $[A]$ its class of equivalence. Define $$\mathcal{P}=\{[A]:\mu(A)>0\}$$ Notice that

• for any class of equivalence $[A]$ ($A$ measurable), $\|\mathbb{1}_A\|_\infty\in\{0,1\}$, and $[A]\in\mathcal{P}$ iff $\|\mathbb{1}_A\|_\infty=1$.
• $\mathcal{P}$ is closed under countable unions, that is, if $\{[A_m]:m\in\mathbb{N}\}\subset\mathcal{P}$, then $[\bigcup_mA_m]\in\mathcal{P}$.
• If $[A], [B]\in\mathcal{P}$ and $[A]\neq[B]$, then $\mu(A\triangle B)=\int|\mathbb{1}_A-\mathbb{1}_B|\,d\mu>0$, that is $[A\triangle B]\in\mathcal{P}$.

Consequently,

1. If $\mathcal{P}$ is finite, then it is easy to see that $L_\infty(\mu)$ (in fact every $L_p(\mu),\, 0<p\leq \infty$) is finite dimensional.
2. If $\mathcal{P}$ is infinite, then $\mathcal{P}$ is uncountble: Let $\mathscr{C}$ be the collection of all $\mathcal{C}\subset 2^{\mathcal{P}}$ such that if $[A],[B]\in\mathcal{C}$ and $A\neq B$, then $A\cap B=\emptyset$, that is, $\mathscr{C}$ is the collection of all families of $\mu$-a.s pairwise disjoint sets in $\mathcal{P}$. Partially order $\mathscr{C}$ by inclusion. If $\mathscr{K}$ is a chain in $\mathscr{C}$, then $\bigcup\mathscr{K}$ is also in $\mathscr{C}$. By Zorn's Lemma, $\mathscr{C}$ has a maximal element $\mathcal{S}$. Since $\mathcal{P}$ is infinite, it follows that $\mathscr{S}$ is infinite. Thus, there is a sequence $\mathcal{Q}:=\{[A_n]:n\in\mathbb{N}\}\subset\mathcal{P}$ such that $A_n\cap A_m=\emptyset$ whenever $m\neq n$. There is a 1-1 correspondence between the collection of all countable unions of elements in $\mathcal{Q}$ and $2^{\mathbb{N}}$, namely for any $J\subset\mathbb{N}$, define $[A_J]=\big[\bigcup_{j\in J}A_j\big]$. This shows that $\mathcal{P}$ is uncountable. Notice that for any distinct $[A],[B]\in \mathcal{P}$, $\|\mathbb{1}_A-\mathbb{1}_B\|_\infty=1$; hence, the open balls $B(\mathbb{1}_A;1/2)$ and $B(\mathbb{1}_B;1/2)$ are disjoint, for if $\|f-\mathbb{1}_A\|_\infty<\frac12$, then $$\|f-\mathbb{1}_B\|_\infty=\|\mathbb{1}_A-\mathbb{1}_B-(\mathbb{1}_A-f)\|_\infty\geq\|\mathbb{1}_A-\mathbb{1}_B\|_\infty-\|\mathbb{1}_A-f\|_\infty>\frac12$$ Therefore, $L_\infty(\mu)$ is not separable.

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