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.


3 Answers 3


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.

  • 1
    $\begingroup$ Can explain what do you mean by the term "infinite support" (or having "finite support") here? I know the definition of support when $X$ is topological space and support is defined as $\text{supp}(\mu):=\{x\in X|\ \mu(U)>0\text{ for all neighborhood of }x\}$. $\endgroup$
    – MathBS
    Nov 23, 2022 at 14:08
  • $\begingroup$ We do not need any topology. $X$ is a measure space. Two sets $A,B\subset X$ are treated as equal if $\mu(A\triangle B)=0.$ A set $A$ is treated as a point (an atom) if for every measurable subset $C\subset A$ such that $\mu(C)>0$ we have $\mu(A\setminus C)=0,$ i.e. $C$ is "equal" $A.$ The support is finite if $\mu$ is concentrated on finitely many points $A_1,A_2,\ldots A_n.$ Otherwise we say the support is infinite. The formal reasoning requires equivalence relation for measurable sets. $\endgroup$ Nov 23, 2022 at 16:38
  • $\begingroup$ Okay thanks for the definition. I understood the definition of finitely supported. But I am not able to understand: You have shown there does not exist pairwise disjoint sequence of sets $A_n$ such that each one of them has positive measure. How does that imply $\mu$ is concentrated on finitely many points/atoms $A_1,A_2,\ldots,A_n$? $\endgroup$
    – MathBS
    Nov 23, 2022 at 20:45
  • 1
    $\begingroup$ If there is a set $A$ of positive measure which does not contains an atom then it contains a set $A_1$ of positive measure such that $\mu(A\setminus A_1)>0.$ Since $A_1$ is no an atom then there exists $A_2\subset A_1$ such that $\mu(A_1\setminus A_2)>0 $ and $\mu (A_2)>0.$ We can continue thar way obtaining a sequence of pairwise disjoint sets $A_n\setminus A_{n+1}$ of positive measure as was claimed in my answer. The other case is that the measure is concentrated on atoms. If there are infintely many of them we get a family as claimed in my answer. Continued in the next comment. $\endgroup$ Nov 23, 2022 at 22:49
  • $\begingroup$ Otherwise the measure is concentrated on finitely many atoms and the space $L^\infty$ is finite dimensional. $\endgroup$ Nov 23, 2022 at 22:50

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 $[A\triangle B]\in\mathcal{P}$.


  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.

  • $\begingroup$ Thats great. But one question from the last part. Why $\mathcal{P}$ is uncountable when $\mathcal{P}$ is infinite? $\endgroup$
    – MathBS
    Nov 23, 2022 at 16:33
  • $\begingroup$ So I have to show if $I,J$ are two subsets of $\Bbb{N}$ with $I\ne J$ then we must have $\left[\bigcup\limits_{n\in I} A_n\right]\ne \left[\bigcup\limits_{n\in J} A_n\right]$. Is it right? $\endgroup$
    – MathBS
    Nov 23, 2022 at 17:02
  • $\begingroup$ And another query: I'm trying to show $\cup_m [A_m]=[\cup_m A_m]$ . Let $A\in$ LHS, so $A\in[A_n]$ for some $n$ i.e. $A\sim A_n$. Then I have to prove $A\sim\cup_m A_m$. But is it true always? Suppose all $A_m$ are disjoint. $A\sim A_n\implies \chi_A=\chi_{A_n}$. Now if $A\sim\cup_m A_m$, then $\chi_A=\sum_m \chi_{A_m}\implies \sum\limits_{m\ne n}\chi_{A_m}=0\implies \chi_{A_m}=0\forall m\ne n$, but we have choose all $A_m$ tohave positive measure. $\endgroup$
    – MathBS
    Nov 23, 2022 at 17:28
  • $\begingroup$ @OliverDíaz if $X = \mathbb R$ with Lebesgue measure, take $A_1 = [0, 1]$, $A_2 = [1, 2]$, $A_3 = [0, 2]$. Then none of them are equivalent, but $[A_i \cup A_2] = [A_3]$. $\endgroup$
    – mihaild
    Nov 23, 2022 at 18:33
  • 1
    $\begingroup$ @mihaild: Zorn's Lemma can be used to prove the existence of an infinite collection of pairwise disjoint sets of positive measure. From there, the non separability of $L_\infty$ is a couple of lines. $\endgroup$
    – Mittens
    Nov 24, 2022 at 6:07

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


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