# Infinite dimensional Banach lattice $L^\infty(X)$ is not order continuous

Consider an arbitrary measure space $(X,\Sigma,\mu)$, with the only assumption being that $L^\infty(X)$ is infinite dimensional. Consider $L^\infty(X)$ as a Banach lattice with the usual ordering. As an exercise in Operator Theoretic Aspects of Ergodic Theory by Eisner et. al we are asked to show that $L^\infty(X)$ does not have an order continuous norm. By order continuous norm I mean that if $(f_n)\subset L^\infty(X)$ is a sequence such that $f_{n+1}\leq f_n$ and $\inf f_n=0$ implies $\|f_n\|\to 0$.

Now it is easy to construct a specific example that is not order continuous. Take for example $X=[0,1]$, $\Sigma$ the Borel $\sigma$-algebra and $\mu=\lambda$. If we consider $(f_n)=(\mathbf{1}_{[0,1/n]})$ we see that this space cannot be order continuous.

Now the way the question is asked leads me to believe that the authors are asking to find such a sequence contradicting order continuity for any arbitrary infinite dimensional $L^\infty(X)$. To do so I would like to just adjust the above example, but critically the above example relies on there being an infinite nested sequence of measurable sets such that the only set contained in all elements of the sequence has measure $0$, but all sets in the sequence have positive measure. Unfortunately I am unable to tell whether $L^\infty(X)$ being infinite dimensional places such a restriction on the measure space. So I'm basically looking for a proof of this fact, or any other consequences of the infinite dimensionality of $L^\infty(X)$that I could exploit.

• Do you also mean to assume that $f_{n+1}\le f_n$ in the definition? Sep 18, 2018 at 17:27
• @DavidC.Ullrich Yes absolutely. Thanks for pointing it out Sep 18, 2018 at 21:35
• By an "arbitrary infinite dimensional $L^\infty(X)$ space" you mean that $(X, \mu)$ can be a general measure space, right? Sep 19, 2018 at 9:42

Given a measurable set $A$, let $\sigma(A)$ be the algebra of measurable subsets of $A$ modulo null sets. I think it's clear what I mean by that. if not I can state it more formally.

If $A$ has positive measure, say $A$ can be partitioned if $A=B\cup C$ where each of $B$ and $C$ has positive measure and $B\cap C=\emptyset$.

Suppose that $A$ cannot be partitioned. Then for every $B\subset A$ either $\mu(B)=0$ or $\mu(A\setminus B)=0$, which says that $\sigma(A)$ has only two elements, $\emptyset$ and $A$. A fortiori,

If $\sigma(A)$ is infinite then $A$ can be partitioned.

Now suppose $L^\infty(X)$ is infinite-dimensional. This certainly implies that $\sigma(X)$ is infinite, so $X$ can be partitioned: $X=B\cup C$ such that etc.

Now at least one of $\sigma(B)$ and $\sigma(C)$ must be infinite; let $B_1=B$ or $B_1=C$ in such a way that $\sigma(B_1)$ is infinite.

So $B_1$ can be partitioned: Write $B_1=B\cup C$ and set $B_2=B$ or $B_2=C$ so that $\sigma(B_2)$ is infinite.

Etc. Now let $B=\bigcap B_n$ and $A_n=B_n\setminus B$. Then $A_{n+1}\subset A_n$, $\mu(A_n\setminus A_{n+1})>0$, and $\bigcap A_n=\emptyset$.

• And the problem with that is what. exactly?\ Sep 20, 2018 at 13:40

You need the following lemma.

Lemma If $(X,\mu)$ is a $\sigma$-finite measure space such that $L^\infty(X)$ is infinite dimensional then there is a infinite numerable sequence of measurable non-null disjoint sets $(A_n)_{n \geq 0}$ of $X$.

With that lemma the proof then is trivial since you can take $f_k = \chi_{\cup_{n \geq k} A_n}$.

The proof of the lemma is more or less equivalent to the proof of the fact that if $X$ is a measure space which is not equivalent to a finite measure space then it contains an infinite number of disjoint sets. I can complete the proof later if I have time.

• Thanks I am aware of this, but unfortunately there are no assumptions on the measure space, so we cannot use a lemma relying on $\sigma$-finiteness Sep 19, 2018 at 10:49