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

*

*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.

*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.


A: 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:

*

*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.

*There is an infinite good family. Then characteristic functions of all possible unions of sets from this family form uncountable discrete set.

*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 $U_{i + 1}^{i + 1} = U_i^i \setminus f(U_i^i)$.
