Is $L^\infty(\mu)$ a locally compact Hausdorff space?

Question

Here $\mu$ is a probability measure. Another similar question is: is it a $\sigma$-compact space? Thank you in advance!

Answer 1

This very rarely happens.

A real or complex topological vector space is locally compact if and only it is finite-dimensional. A Banach space is $\sigma$-compact if and only if it is finite-dimensional. For $L^\infty(\mu)$ of a probability measure this happens if and only if $\mu$ has finitely many atoms and no continuous part.

Answer 2

Let's assume $\mu$ is not a finite linear combination of point masses, in which case the answer is trivially yes.

Otherwise, the answer is no.

$L^\infty(\mu)$ is a Banach space, and we have that, in general for Banach spaces $B$,

$B$ locally compact $\iff$ $B$ is finite dimensional $\iff$ $B$ is sigma-compact.

There are a few ways to see this: the idea is to construct a sequence of unit vectors in the infinite dimensional space that can't have a convergent subsequence: If $B$ is locally compact then 0 has a precompact open neighbourhood U, this neighbourhood must contain an open ball around the origin whose closure is therefore compact.

Assuming there exists a countable disjoint collection $(A_n)$ of sets given positive weight by $\mu$, in $L^\infty(\mu)$ we can do this easily by setting considering the unit vectors $\mu(A_n)^{-1}1(A_n)$ which can easily be seen to not have any convergent subsequences.

The fact that infinite dimensional Banach spaces cannot be sigma compact follows from the fact that they cannot be locally compact - and hence compact sets in them must have empty interior - and then applying the Baire category theorem.

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Edit: More detail for the last part

A compact set in $B$ must have empty interior (i.e. it must not contain any open sets): if not, then there would be an open set in $B$ contained in a compact set, making it precompact, and then by translating this set we would have that every point in $B$ had a precompact open neighbourhood, and hence $B$ locally compact, a contradiction.

Now recall that the Baire category theorem implies that a complete metric space cannot be expressed as the countable union of closed sets with empty interior. Hence, it follows that an infinite dimensional Banach space cannot be written as a countable union of compact sets.