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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.

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up vote 5 down vote accepted

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

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Thanks! A very useful reference. 363R gives a perfect result. – Lei Li Oct 10 '13 at 22:03

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 $\mathcal{F}$.

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Thanks Alexander – Lei Li Oct 10 '13 at 21:29

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