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The following question has puzzled me for some time:

Let $(\Omega,\Sigma)$ be a nonempty, measurable space. Does there necessarily exist a probability measure $\mu:\Sigma\to[0,1]$?

If there exists a nonempty measurable set $A$ such that no nonempty subset of $A$ is measurable (an atom), we can simply let $\mu(B)=1$ if $A\subseteq B$ and $\mu(B)=0$ otherwise. So the problem is only interesting if the $\sigma$-algebra has not atoms. This rules out every countably generated $\sigma$-algebra. An example of a $\sigma$-algebra that has no atoms but supports a probability measure is $\{0,1\}^\kappa$ for $\kappa$ uncountable, which we can endow with the coin-flipping probability measure.

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    $\begingroup$ Pick a point $x \in \Omega$ and define $\mu(A) = 1$ when $x \in A$ and $\mu(A) = 0$ when $x \notin A$. Maybe you're missing a nontriviality assumption? $\endgroup$ – François G. Dorais Jan 16 '12 at 20:01
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    $\begingroup$ @Francois, perhaps an appropriate nontriviality assumption might be $Supp(\mu)=\Omega$, in which case I think the answer is NO, given by letting $\Omega$ be uncountable with the discrete topology, and taking the Borel sigma algebra. $\endgroup$ – Otis Chodosh Jan 16 '12 at 20:16
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    $\begingroup$ slaps my head really hard Thank you. $\endgroup$ – Michael Greinecker Jan 16 '12 at 20:17
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    $\begingroup$ But it seems one can achieve that with countably many point masses, assigned on a countable dense set. $\endgroup$ – Joel David Hamkins Jan 16 '12 at 21:29
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    $\begingroup$ Of course "open" and "dense" do not exist in the measurable context. $\endgroup$ – Gerald Edgar Jan 16 '12 at 21:32
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You write "An example of a σ-algebra that has no atoms but supports a probability measure is $\{0,1\}^\kappa$ for $\kappa$ uncountable, which we can endow with the coin-flipping probability measure."

Maharam's theorem says that these are essentially the only ones. That is: Every Boolean algebra which is equipped with a probability measure (and is Dedekind complete, see below) is isomorphic to a product of the measure algebras on various $2^\kappa$ that you mentioned. (Including finite $\kappa$, to take care of measures with atoms.)

Dedekind complete means that every subset has a least upper bound. If you take a $\sigma$-algebra which carries a $\sigma$-additive probability measure, and divide by the ideal of null sets, then the resulting algebra is still a measure algebra and it will be Dedekind complete.

An exposition of Maharam's theorem can be found in Fremlin's book, volume 3. (The theorem I quoted can be generalized to algebras with a "semifinite" measure, which is more general than probability measure.)

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    $\begingroup$ I think one should make this more precise: Every atomless measure algebra is isomorphic to a countable convex combination of "coin flipping measures" with infinite "exponent". Dealing with atoms poses additional difficulties, you cannot construct a single unfair coin flip from fair coin flips. The original paper by Maharam can be found here: pnas.org/content/28/3/… $\endgroup$ – Michael Greinecker Jan 17 '12 at 13:52
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The following is a (corollary of a) theorem of Sierpinskii from 1933:

If $\mu:\mathcal{P}(\Omega) \to [0,1]$ is a probability measure and $|\Omega|$ is smaller than the first weakly-inaccessible cardinal, then there must be a countable $A \subseteq \Omega$ such that $\mu(A)=1$.

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    $\begingroup$ Nowadays, "inaccessible" usually means "strongly inaccessible". But I think that Sierpinski's theorem talks about "weakly inaccessibles", i.e., regular limit cardinals. $\endgroup$ – Goldstern Feb 21 '12 at 13:20
  • $\begingroup$ @Goldstern: You are right, Sierpinski defines inaccessible as a regular $\aleph_\alpha$ for which $\alpha$ is a limit ordinal. I edited accordingly. $\endgroup$ – Ramiro de la Vega Feb 21 '12 at 15:04
  • $\begingroup$ I thought Ulam proved this in 1930, in Satz (A) here: matwbn.icm.edu.pl/ksiazki/fm/fm16/fm16114.pdf Actually, I think what Ulam states is that you can't have all points have measure zero if $|\Omega|$ has no weakly inaccessible cardinal less than or equal to it. It takes a small argument with dependent choice to get from this to every measure having countable support. Is it the case that Sierpiński was the one who proved that part? $\endgroup$ – Robert Furber Jul 31 at 2:52
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    $\begingroup$ @RobertFurber, I can´t read german all that well but I think you are right about Satz(A) implying the proposition I wrote. I was using Sierpinskii´s Proposition U here: matwbn.icm.edu.pl/ksiazki/fm/fm20/fm20121.pdf which he attributes to Ulam for sets of size $\aleph_1$. $\endgroup$ – Ramiro de la Vega Aug 2 at 16:40
  • $\begingroup$ @RamirodelaVega That's interesting. I guess the implications between these statements were less obvious back in that era. $\endgroup$ – Robert Furber Aug 3 at 5:54

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