# Can we put a probability measure on every $\sigma$-algebra?

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.

• 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? Jan 16, 2012 at 20:01
• @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. Jan 16, 2012 at 20:16
• slaps my head really hard Thank you. Jan 16, 2012 at 20:17
• But it seems one can achieve that with countably many point masses, assigned on a countable dense set. Jan 16, 2012 at 21:29
• Of course "open" and "dense" do not exist in the measurable context. Jan 16, 2012 at 21:32

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.)

• 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/… Jan 17, 2012 at 13:52
• Naive question: What would be an example of a $\sigma$-algebra that is Dedekind complete but not isomorphic to a product of the measure algebras on various $2^\kappa$, and hence does not have a probability measure on it, to answer the original question? Jun 2, 2022 at 17:19

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 $\mu(A)=1$.

• Nowadays, "inaccessible" usually means "strongly inaccessible". But I think that Sierpinski's theorem talks about "weakly inaccessibles", i.e., regular limit cardinals. Feb 21, 2012 at 13:20
• @Goldstern: You are right, Sierpinski defines inaccessible as a regular $\aleph_\alpha$ for which $\alpha$ is a limit ordinal. I edited accordingly. Feb 21, 2012 at 15:04
• 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? Jul 31, 2019 at 2:52
• @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 $\aleph_1$. Aug 2, 2019 at 16:40
• @RamirodelaVega That's interesting. I guess the implications between these statements were less obvious back in that era. Aug 3, 2019 at 5:54