When $L^\infty$ is 1-injective

Question

It is known that when $\mu$ is $\sigma$-finte measure, then $L^\infty(\mu)$ is $1$-injective. But I want to know whether it is right for any $L^\infty$ spaces.

Answer 1

A useful sufficient condition is that $(X,\Sigma,\mu)$ is a localizable measure space. See 363R of Fremlin, Volume 3, p.308.

Answer 2

For a compact Hausdorff space $K$ the algebra $C(K)$ is $1$-injective iff $K$ is extremally disconnected iff its Boolean algebra of idempotents is complete. So to construct a counterexample we need an $L^\infty$ algebra with an incomplete Boolean algebra of measurable sets modulo null sets. (quite obviously, this Boolean algebra is always $\sigma$-complete, and it is also well-known to be complete if the measure is $\sigma$-finite).

Now let $X$ be an uncountable set, equipped with the $\sigma$-algebra $\mathcal{F}$ that consists of sets that are either countable or have countable complement, and the counting measure $\mu$. Then obviously the only null set is the empty set. Now it's easy to see that the Boolean algebra of countable-or-cocountable sets is incomplete: if $Y \subset X$ is a subset that is uncountable and has uncountable complement, then the family of all countable subsets of $Y$ has no supremum within $\mathcal{F}$.