Any Lebesgue probability space is mod. 0 isomorphic to some Polish probability space (with $\sigma$-algebra the completion of Borel algebra, and some Borel probability). I would like to see an example of a Lebesgue probability space which is not isomorphic to such a Polish space.

In other word, I want a Lebesgue probability space $(X,\mathcal A,\mu)$ for which there is no topology $\mathcal T$ such that

  1. $\mathcal A$ is the completion of the $\sigma$-algebra generated by $T$ and
  2. $X$, endowed with $T$, is a Polish space.

(For me a Lebesgue probability space is a complete probability space $(X,\mathcal A,\mu)$ such that there is some Hausdorff, second countable topology $\mathcal T$ which turns $\mu$ into an inner regular Borel probability measure on the completion of $\sigma(\mathcal T)$).

Edit. I just noticed that I forgot the most important hypothesis that $\mu$ must be inner regular in the definition of Lebesgue probability space. Sorry about that. So in particular $\mu$ is a compact measure. Also this definition is in fact equivalent to Rohlin's original definition.


2 Answers 2


When requesting inner regularity, you hit the ZFC - undecidable Cantor's continuum problem.

First note that a Lebesgue space has cardinality at most continuum since a countable collection of subsets separates points. A uncountable polish space has exactly cardinality of the continuum since it has a perfect subset (Cantor - Bendixon).

If the continuum hypothesis is false, in the compact metric $[0,1]$ with Dirac's delta measure concentrated in 0 take a subspace $X$ of cardinality $\aleph_1$ containing 0. You have a Radon probability measure on the metric second countable $X$ which cannot support polish topologies since it is uncountable but not of continuum cardinality.

If the continuum hypothesis is true, on a Lebesgue space $X$ take a "concassage" (Bourbaki, chap. IX); as measure space, $X$ is direct sum of compact (and metrisable, being second countable) subsets of positive measure and a remainder of (essential) measure 0. The remainder is at most countable or of continuum cardinality, hence one can put on it a polish (even compact metric) topology; the sum is countable (the sum of measures converges to 1), so one obtains a polish (even locally compact second countable) space.

[You might want to ask the set theory experts about what happens in the ZF models of Soloway resp. Selah, where dependent choice holds (so one still has the basic theorems about measures, like countable additivity of Lebesgue measure on the real line) but every subset of a Euclidean space is Lebesgue measurable resp. has the Baire property, so a uncountable subset of the continuum has continuum cardinality but cannot be well ordered]


You seem to confuse categories in your question. Lebesgue spaces are defined (I mean here the standard definition due to Rokhlin) in the category of measure spaces - and there is no topology whatsoever in their definition. According to a theorem of Rokhlin all purely non-atomic Lebesgue spaces are isomorphic to the unit interval endowed with the Lebesgue measure.

On the other hand, in what concerns the topological category, there is indeed yet another theorem of Rokhlin: all Polish spaces endowed with a Borel measure are Lebesgue.

However, I don't see how one can talk about a topology on a Lebesgue space.

  • $\begingroup$ It looks like there have been some edits that clarify the situation. $\endgroup$ Mar 1, 2014 at 0:31
  • 1
    $\begingroup$ Take the usual (Vitali) non-measurable subset of [0,1] with the (Caratheodory measurable sets for the) external Lebesgue measure. The measure is not compact (Pfangzagl Pierlo, compact systems of sets) lacking conditional probabilities, but measures on Polish spaces are compact. $\endgroup$
    – user46855
    Mar 1, 2014 at 4:15
  • 1
    $\begingroup$ R W: we have a set $X$ so we can speak of a topology on $X$. This is a good starting point for the theory of Lebesgue spaces, see Thierry de la Rue, Espaces de Lebesgue (numdam.org/numdam-bin/feuilleter?id=SPS_1993__27_). $\endgroup$
    – timofei
    Mar 1, 2014 at 11:00
  • $\begingroup$ user46855 this is very interesting. I think it answers my question but I have to check some details. $\endgroup$
    – timofei
    Mar 1, 2014 at 11:13
  • $\begingroup$ @timofei: Aha - I see. De la Rue (whom you apparently follow) uses the term "Lebesgue space" for something which is quite different from original Rokhlin's Lebesgue spaces (which are nowadays a part of the standard setup in ergodic theory). The point (and the principal advantage of Rokhlin's approach), once again, is that Rokhlin's definition is given in the measure category - there is not a single word about any topologies in this definition. $\endgroup$
    – R W
    Mar 1, 2014 at 12:27

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