When is $L^2(X)$ separable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:49:20Z http://mathoverflow.net/feeds/question/42310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42310/when-is-l2x-separable When is $L^2(X)$ separable? Bugs Bunny 2010-10-15T17:13:35Z 2013-05-21T03:13:07Z <p>I have never studied any measure theory, so apologise in advance, if my question is easy:</p> <p>Let $X$ be a measure space. How can I decide whether $L^2(X)$ is separable?</p> <p>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...</p> <p>I cannot even decide at the moment for which locally compact groups $G$ with Haar measure, $L^2(G)$ is separable...</p> http://mathoverflow.net/questions/42310/when-is-l2x-separable/42314#42314 Answer by Byron Schmuland for When is $L^2(X)$ separable? Byron Schmuland 2010-10-15T17:51:01Z 2010-10-15T18:14:56Z <p>In addition to the measure $\mu$ being $\sigma$-finite, I think you also need some conditions on the measurable space $(X,{\cal A})$.</p> <p>Proposition 3.4.5 of Cohn's book <em>Measure Theory</em> says that $L^p(X,{\cal A},\mu)$ ($1\leq p &lt; \infty$) 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. </p> <p>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. </p> <p>I haven't checked the details, so take my answer with a grain of salt!</p> http://mathoverflow.net/questions/42310/when-is-l2x-separable/42316#42316 Answer by R W for When is $L^2(X)$ separable? R W 2010-10-15T18:18:58Z 2010-10-16T01:58:25Z <p>$\sigma$-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.</p> <p>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. </p> <p>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.</p> <p>$L^2$ on any Lebesgue space (be it finite or $\sigma$-finite) is separable in view of the above isomorphisms.</p> <p>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.</p> <p>ADD</p> <p>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 <a href="http://en.wikipedia.org/wiki/Standard_probability_space" rel="nofollow">article</a> about it.</p> http://mathoverflow.net/questions/42310/when-is-l2x-separable/42383#42383 Answer by Dmitri Pavlov for When is $L^2(X)$ separable? Dmitri Pavlov 2010-10-16T12:45:17Z 2013-05-21T03:13:07Z <p>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.</p> <p>The space L^2(X) is independent of the choice of a faithful measure and depends only on the underlying <a href="http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820" rel="nofollow">measurable space</a> of X.</p> <p>There is a complete classification of measurable spaces up to isomorphism (for simplicity I only consider measurable spaces that satisfy a certain countability assumption, because only these spaces can have separable L^2-spaces): every measurable space canonically decomposes as a disjoint union of its atomic and diffuse parts. The atomic part is simply a disjoint union of points, whereas the diffuse part is a (noncanonical) disjoint union of real lines. Finite nonempty disjoint unions of real lines are isomorphic to the countable union of real lines, but otherwise the cardinality of the family determines the diffuse part uniquely.</p> <p>Thus isomorphism classes of measurable spaces are in bijection with pairs of cardinal numbers (m,n), where n is either zero or infinite. (Here m is the number of points in the atomic part and n is the number of real lines in the diffuse part.)</p> <p>L^p(X) (p≥1) is separable if and only if both m and n are at most countable. Thus there are two families of measurable spaces whose L^p-spaces are separable: (1) Finite or countable disjoint unions of points; (2) Disjoint unions of the real line and a space of the type (1).</p> <p>Equivalent reformulations of the above condition: (1) L^p(X) is separable if and only if X admits a faithful finite measure. (2) L^p(X) is separable if and only if X admits a faithful σ-finite measure. (3) L^p(X) is separable if and only if every measure on X is σ-finite. (Here I disallow nonsemifinite measures, i.e., measures that are equal to infinity on a set of nonzero measure. Note also that X is assumed to satisfy the countability property mentioned above.)</p> <p>The underlying measurable space of a locally compact group G satisfies the above conditions if and only if G is second countable.</p> <p>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.</p> <p>More information on this subject can be found in this answer: <a href="http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820" rel="nofollow">http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820</a></p> <p>Bruckner, Bruckner, and Thomson discuss separability of L^p-spaces in Section 13.4 of their textbook Real Analysis: <a href="http://classicalrealanalysis.info/documents/BBT-AlllChapters-Landscape.pdf" rel="nofollow">http://classicalrealanalysis.info/documents/BBT-AlllChapters-Landscape.pdf</a></p>