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I have never studied any measure theory, so apologise in advance, if my question is easy:

Let $X$ be a measure space. How can I decide whether $L^2(X)$ is separable?

In reality, I am interested in Borel sets on a locally compact space $X$. I can also assume that the support of the measure is $X$, if it helps...

I cannot even decide at the moment for which locally compact groups $G$ with Haar measure, $L^2(G)$ is separable...

Bugs Bunny
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    Dear BB: perhaps you want the Borel measure to be $\sigma$-regular, which loosely says that everything is determined by measures of compact subsets (which are required to be of finite measure); measures arising from functionals on $C_c(X)$ (e.g., Haar on $G$) are of this type. The precise def'n is on p. 256 of 3rd edition of Lang's real analysis book. A $\sigma$-regular Borel measure on a $\sigma$-finite $X$ is "regular" (Lang, p. 257), so the measure of any Borel set is sup of measures of compacts inside it. Thus, you win if loc. comp. $X$ has a countable base of opens with compact closure. – BCnrd Oct 15 '10 at 17:42
  • Given the flurry of answers, corrections and deletions to this question (some of them mine), this may be good fodder for http://mathoverflow.net/questions/23478/. – Nate Eldredge Oct 15 '10 at 18:35
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    Thanks, folks!! All voted up and I will chew on the answers for a day before accepting one... – Bugs Bunny Oct 17 '10 at 07:27

4 Answers4

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Without loss of generality we can assume that the support of the measure equals $X$ (i.e., the measure is faithful), because we can always pass to the subspace defined by the support of the measure.

The space $^2(X)$ is independent of the choice of a faithful measure and depends only on the underlying enhanced measurable space of $X$, i.e., measurable and negligible subsets of $X$.

There is a complete classification of measurable spaces up to isomorphism. Every measurable space canonically splits as a disjoint union of its ergodic subspaces, i.e., measurable spaces that do not admit measures invariant under all automorphisms.

Ergodic measurable spaces in their turn can be characterized using two cardinal invariants $(m,n)$, where either $m=0$ or both $m≥ℵ_0$ and $n≥ℵ_0$. The measurable space represented by $(m,n)$ is the disjoint union of $n$ copies of $2^m$, where $2=\{0,1\}$ is a measurable space consisting of two atoms and $2^m$ denotes the product of $m$ copies of 2. The case $m=0$ gives atomic measurable spaces (disjoint unions of points), whereas $m=ℵ_0$ gives disjoint unions of real lines (alias standard Borel spaces).

Thus isomorphism classes of measurable spaces are in bijection with functions M: Card'→Card, where Card denotes the class of cardinals and Card' denotes the subclass of Card consisting of infinite cardinals and 0. Additionally, if $m>0$, then $M(m)$ must belong to Card'.

The Banach space $^p(X)$ ($1≤p<∞$) is separable if and only if $M(0)$ and $M(ℵ_0)$ are at most countable and $M(m)=0$ for other $m$.

Thus there are two families of measurable spaces whose $^p$-spaces are separable:

  1. Finite or countable disjoint unions of points;
  2. The disjoint union of the above and the standard Borel space.

Equivalent reformulations of the above condition assuming $M(m)=0$ for $m>ℵ_0$:

  1. $^p(X)$ is separable if and only if $X$ admits a faithful finite measure.
  2. $^p(X)$ is separable if and only if $X$ admits a faithful $σ$-finite measure.
  3. $^p(X)$ is separable if and only if every (semifinite) measure on $X$ is $σ$-finite.

The underlying measurable space of a locally compact group $G$ satisfies the above conditions if and only if $G$ is second countable as a topological space.

The underlying measurable space of a paracompact Hausdorff smooth manifold $M$ satisfies the above conditions if and only if $M$ is second countable, i.e., the number of its connected components is finite or countable.

More information on this subject can be found in this answer: Is there an introduction to probability theory from a structuralist/categorical perspective?

Bruckner, Bruckner, and Thomson discuss separability of $^p$-spaces in Section 13.4 of their textbook Real Analysis: http://classicalrealanalysis.info/documents/BBT-AlllChapters-Landscape.pdf

Dmitri Pavlov
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    Thanks. This confused me at first because some authors use "measurable space" to refer to just a set $X$ equipped with a $\sigma$-algebra of measurable sets, whereas yours (as described in the link) also specifies the $\sigma$-ideal of measure zero sets. The former obviously would not suffice to determine the separability of $X$ (consider $[0,1]$ with its Borel $\sigma$-algebra and either Lebesgue or counting measure). – Nate Eldredge Oct 16 '10 at 14:28
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    Another question though: how does one define the support of a measure on an abstract measurable space (to ensure that it is "faithful")? The definition I know involves taking a closure, but here I suppose $X$ need not come with a topology. – Nate Eldredge Oct 16 '10 at 14:30
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    Dmitri and Nate: There is reason to be cautious here. Even a Borel probability measure on a compact Hausdorff space can fail to have "support". For instance, example 7.1.3 in Volume 2 of Bogachev's Measure Theory, where the underlying space is the set of all ordinals not exceeding the first uncountable ordinal.

    Here a measure "fails to have support" means that the union of all open null sets is not itself null.

    Bogachev goes on to show that all Radon measures have support, and likewise all Borel measures on a separable metric space.

     –  Oct 16 '10 at 14:56
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    Dmitri: It's a great answer, but I can't reconcile my compact counterexample with your statement that $L^2(G)$ will be separable if $G$ is $\sigma$-compact. Where did I go wrong? –  Oct 16 '10 at 15:04
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    @Nate: The support of a measure m on a measurable space is defined as follows. Consider the Boolean algebra of all measurable sets modulo null sets (i.e., two measurable sets are equivalent if their symmetric difference is a measure 0 set). This Boolean algebra is complete, as described in the link. Now take the supremum of all elements p of this algebra such that m vanishes on p. The complement of this supremum is the support of m. It is extremely important to factor out the null sets, otherwise the above procedure doesn't make sense. – Dmitri Pavlov Oct 16 '10 at 15:09
  • Thanks very much Dmitri! But I am still a little confused about the property of localizability, which is new to me. You say, for instance, that it is equivalent to the Radon-Nikodym theorem, but the statements and proofs of Radon-Nikodym I have seen don't seem to involve it (they start with just a set, a $\sigma$-algebra, and a $\sigma$-finite measure). What am I missing? And where is a good place to read more? – Nate Eldredge Oct 16 '10 at 18:00
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    @Nate: Every σ-finite space is localizable, but not every localizable space is σ-finite. A space is σ-finite if and only if it satisfies one of the equivalent statements above, which means that both m and n must be at most countable. The first six paragraphs of my answer in the link are devoted to an explanation of localizability. Most of the results mentioned there are proved in the following paper: Irving Segal, “Equivalences of measure spaces”. – Dmitri Pavlov Oct 18 '10 at 15:56
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    @Byron: For some reason I confused σ-compactness and second countability. – Dmitri Pavlov Dec 31 '11 at 20:32
  • I added a reference to a textbook by Bruckner, Bruckner, and Thomson, which discusses precisely the original question. – Dmitri Pavlov May 21 '13 at 03:14
  • @Dmitri: Is there any reference for the case of a locally compact group $G$? – Yulia Kuznetsova Oct 04 '14 at 20:53
  • @YuliaKuznetsova: Theorem 13.8 in the book by Bruckner-Bruckner-Thomson cited above can be applied here. It states that separability is equivalent to the separability of the metric space of equivalence classes of measurable sets modulo negligible sets, where the metric is the Haar measure of the symmetric difference. To construct a countable dense subset of the latter space, choose a countable basis for the topology (here second countability is used) and take finite unions of elements in the basis, which again form a countable set. – Dmitri Pavlov Oct 05 '14 at 11:16
  • @DmitriPavlov: Thank you, I see finally how it works. But it's a pity that's there is no ready and simple reference just for the groups, I would have used it. – Yulia Kuznetsova Oct 05 '14 at 21:10
  • I don't see how this can be. If $I$ is a large set and $G={1,-1}^I$ then $G$ is a compact abelian group. Haar measure is certainly faithful, isn't it? But $L^2(G)$ need not be separable. – David C. Ullrich Jan 05 '18 at 00:58
  • @DavidC.Ullrich: What specific claim are you questioning? For a large set I the topological space {1,-1}^I is not second countable and its underlying measurable space does not satisfy the countability assumption mentioned in the second paragraph. – Dmitri Pavlov Jan 05 '18 at 05:00
  • @DmitriPavlov The statement that $L^p(X)$ is separable if and only if $X$ admits a faithful measure. I didn't realize there were other assumptions there... – David C. Ullrich Jan 05 '18 at 13:41
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    @DavidC.Ullrich: The other assumption was also stated at the end of the list. I now moved it in front. – Dmitri Pavlov Jan 05 '18 at 19:35
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$\sigma$-finiteness of the measure has nothing to do. The only property which matters is the separability of the measure space itself, which (modulo some technicalities) means that there exists a countable family of measurable sets which separate points of $X$ (mod 0). Measure spaces with this property are called Lebesgue spaces (essentially, these are the only measure spaces one meets in the "real life"). Note that such a family of separating sets gives rise to an isomorphism of the original space with the countable product of 2-point sets.

Any Polish space (separable, metrizable, complete) endowed with a purely non-atomic Borel probability measure is isomorphic to the unit interval with the Lebesgue measure on it. In the same way, a Polish space endowed with a $\sigma$-finite purely non-atomic measure is isomorphic to the real line with the Lebesgue measure on it.

In the "Borel language" one talks about so-called standard Borel spaces. Any standard Borel space endowed with a $\sigma$-finite measure on the Borel $\sigma$-algebra is a Lebesgue space.

$L^2$ on any Lebesgue space (be it finite or $\sigma$-finite) is separable in view of the above isomorphisms.

On the other hand, if one takes a measure space which is not separable - like the uncountable product measure in the previous answer - then $L^2$ on this space is not separable either.

ADD

My answer was partially prompted by several comments which have since disappeared - otherwise I would have organized it in a somewhat different way. Unfortunately, the whole discussion illustrates the deplorable situation with teaching measure theory, as a result of which people, for instance, don't realize that in the measure category there is no difference between circles and intervals. A well-kept secret is the fact that there is (up to isomorphism) only one "reasonable" non-atomic probability space, and, consequently, only one reasonable non-atomic $\sigma$-finite space. There is a good Wikipedia article about it.

R W
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  • Since there are a lot of conflicting statements here, perhaps you could give a reference? – Nate Eldredge Oct 15 '10 at 18:33
  • I think you need to modify the answer so that the countable family doesn't just separate points mod 0 but also generates the $\sigma$-algebra. There are nonseparable extensions of Lebesgue measure and there is of course a countable separating family of intervals with rational endpoints. – Michael Greinecker Feb 27 '17 at 00:11
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I have just found an old article which contains the following result: for a locally compact group $G$, its topological weight $w(G)$ [the minimal cardinality of a topology base] is equal to the dimension of $L^2(G)$. Whence $L^2(G)$ is separable iff $G$ is second countable. This is Theorem 2 in: de Vries, J. The local weight of an effective locally compact transformation group and the dimension of $L^2(G)$. Colloq. Math. 39 (1978), no.2, 319-323.

Since I spent a considerable time searching for such a reference, I post it here. Worth noting also that the case of a compact $G$ is contained in Hewitt-Ross, Theorem 28.2.

Yulia Kuznetsova
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In addition to the measure $\mu$ being $\sigma$-finite, I think you also need some conditions on the measurable space $(X,{\cal A})$.

Proposition 3.4.5 of Cohn's book Measure Theory says that $L^p(X,{\cal A},\mu)$ ($1\leq p < \infty$) is separable if $\mu$ is $\sigma$-finite and $\cal A$ is countably generated. For example, it holds if $X$ is a complete separable metric space, and $\cal A$ is the Borel $\sigma$-algebra.

However, even for a compact group, you can make counterexamples like $[-1/2,1/2]^{[0,1]}$, an uncountable product of a circles. For the product measure, $\mu=\lambda^{[0,1]}$, the coordinate functions are orthogonal in $L^2$ but there are uncountably many.

I haven't checked the details, so take my answer with a grain of salt!

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    Your circles look a lot like intervals! :) – Mariano Suárez-Álvarez Oct 15 '10 at 18:01
  • Ha! I want the coordinate functions to be real-valued, but I want the topological circle. So I mentally put in or leave out details. No wonder my students find my lectures confusing! :) –  Oct 15 '10 at 18:04
  • Notice that separability is a condition you put on a topological space, while one would want conditions on the measured space for $L^2$ to be separable. – Mariano Suárez-Álvarez Oct 15 '10 at 18:10
  • Well, the measure condition is that $\cal A$ is countably generated. The topological condition of separability was an example only. I will edit my answer. –  Oct 15 '10 at 18:13