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Let S be an uncountable set. Does there exist a probability measure which is defined on all subsets of S, with P({x}) = 0 for any element x of S ?

If I remove the condition P({x}) = 0, then I can trivially get a measure defined on all subsets as follows: Fix some a in S. For any subset U of S, define P(U ) = 1 if a is in U and 0 otherwise.

But what happens if I am not allowed to put nonzero probability on individual points ?

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I added the set-theory tag, since this topic is deeply connected with set-theoretic issues. –  Joel David Hamkins Mar 26 '10 at 13:01

5 Answers 5

up vote 14 down vote accepted

The existence of such a measure is equiconsistent to the existence of a measurable cardinal, one of the large cardinal notions, and if ZFC is consistent, cannot be proved in ZFC. (See the notion of real-valued measurable cardinal on the Wikipedia page.)

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Joel, could you please elaborate a little or give me a reference ? I am not very familiar with foundations of mathematics issues. Thanks a lot. –  Cosmonut Mar 26 '10 at 12:57
Ulam showed that there is no such measure on a set of power aleph_1 . When I am asked whether I believe the Continuum Hypothesis, I sometimes answer that, on the contrary, the cardinal of the continuum should be a real-valued measurable cardinal. –  Gerald Edgar Mar 26 '10 at 16:02

Here is some elaboration on Joel David Hamkins's answer.

A (two-valued) measurable cardinal is an uncountable cardinal $\kappa$ such that there is a continuous $({<\kappa})$-additive $\{0,1\}$-valued probability measure $\mu$ defined on all subsets of $\kappa$. To say that $\mu$ is $({<\kappa})$-additive means that if $(X_i)_{i \in I}$ is a family of pairwise disjoint subsets of $\kappa$ with $|I|<\kappa$, then $\mu\left(\bigcup_{i \in I} X_i\right) = \sum_{i\in I} \mu(X_i)$. Since $\mu$ can only take values $0$ and $1$, this is equivalent to saying that (a) at most one of the $X_i$ can have measure $1$, and (b) if they all have measure $0$ then so does $\bigcup_{i \in I} X_i$. (Some people say $\kappa$-additive instead of $({<\kappa})$-additive, but I prefer to use $\kappa$-additive to mean the above for families with index set of size equal to $\kappa$.)

The existence of a continuous countably additive $\{0,1\}$-valued probability measure $\mu$ defined on all subsets of a set $S$ implies the existence of a measurable cardinal. Indeed, I claim that if $\kappa \geq \aleph_1$ is the smallest cardinal such that $\mu$ is not $\kappa$-additive, then $\kappa$ is a measurable cardinal. To see this, let $(X_i)_{i<\kappa}$ be a family of pairwise disjoint measure $0$ sets such that $\bigcup_{i<\kappa} X_i$ has measure $1$ (i.e. the family contradicts $\kappa$-additivity in the only possible way). Defining $\bar{\mu}(I) = \mu\left(\bigcup_{i \in I} X_i\right)$ for every $I \subseteq \kappa$, we obtain a $({<\kappa})$-additive $\{0,1\}$-valued probability measure $\bar\mu$ defined on all subsets of $\kappa$.

So the existence of a continuous countably additive $\{0,1\}$-valued measure defined on all subsets of a set $S$ is exactly equivalent to the existence of a measurable cardinal. However, since a probability measure is also allowed to take values strictly between $0$ and $1$, this is not quite equivalent to the statement you asked about.

By analogy with the above, a real-valued measurable cardinal is an uncountable cardinal $\kappa$ such that there is a continuous $({<\kappa})$-additive probability measure defined on all subsets of $\kappa$. The existence of a real-valued measurable cardinal is equivalent to your statement by a variation of the trick used above.

In 1930, Ulam (Zur Masstheorie in der allgemeinen Mengenlehre, Fund. Math. 16) showed that if $\kappa$ is real-valued measurable then $\kappa \geq 2^{\aleph_0}$, and that if $\kappa > 2^{\aleph_0}$ is real-valued measurable then $\kappa$ is in fact measurable (with a possibly different measure). Ulam also showed that successor cardinals like $\aleph_1$ cannot be real-valued measurable.

In the 1960's, Solovay (MR290961) finally resolved the boundary case. He showed that if $\kappa = 2^{\aleph_0}$ is real-valued measurable then there is an inner model (namely $L[I]$ where $I$ is the ideal of null sets) wherein $\kappa$ is still real-valued measurable and GCH holds, therefore $\kappa$ is measurable in that inner model by Ulam's earlier results. While this doesn't mean that the existence of a real-valued measurable cardinal and the existence of a measurable cardinal are equivalent, it shows that the two statements are equiconsistent over ZFC.

Using forcing (and another result of Ulam), Solovay also showed that if there is a model with a measurable cardinal then there is a model in which the Lebesgue measure on $[0,1]$ can be extended to a probability measure defined on all subsets of $[0,1]$.

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Note that there is a lot of intermediate history between Ulam's 1930 paper and Solovay's 1971 paper that I completely skipped... –  François G. Dorais Mar 26 '10 at 18:24
Thanks, François, for fleshing out a fuller answer. –  Joel David Hamkins Mar 26 '10 at 23:15
No problem, I had time to spare and you seemed to be busy. –  François G. Dorais Mar 27 '10 at 0:59

Joel's answer is the correct one, but in some cases one only needs a finitely additive probability measure rather than a countably additive one, and in this case one can use an non-principal ultrafilter to create such a measure, which would give every set in the ultrafilter a measure of 1 and all the other sets a measure of zero. Indeed, one important way to think about ultrafilters is as a {0,1}-valued finitely additive probability measure on a set.

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This may be irrelevant but sounds interesting: There are finitely additive isometry invariant total extensions of Lebesgue measure on line and plane but not in higher dimensions (Banach Tarski paradox disallows it). –  Ashutosh Jun 5 '10 at 23:13

Let $S$ a noncountable set and $\mu$ a measure on $S$ such that $\mu(S)=1$ and $\mu$ definited on all subset of $S$.For $n\in \mathbb{N}$ let $E_n:=\{x\in S| \mu(\{x\})>1/n\}$ this is finite from $\mu(X)=1$, and from $\cup_nE_n=\{x\in S| \mu(\{x\})>0\}$ follow that $E:=\cup_nE_n$ is countable. Then either $\mu(E\backslash S)=0 $ ($\mu$ is essentially defined only in a countable subset) or $\mu(E\backslash S)\neq 0 $ we can restart from our inital data by the assumption: $\forall x\in S: \mu(\{x\})=0$.\

Now can be exist non trivial measure: Let $T \subset S $ a no countable subset, and define the measure $\mu$ as $\mu(A)=1$ if $T\backslash A$ is countable, $\mu(A)= 0$ if $T\backslash A$ isnt countable. This is a example of a $atomic\ measure$, for definitions a measure is atomic if exist a measurable $B$ such that $\mu(B) > 0 $ and for any measurable subset $A\subset B$ : $\mu(A)=0\ or\ \mu(B)=\mu(A)$, $B$ is said an $atom$ of $\mu$. From the "THERY OF CHARGES" Bashkara Rao AP 1983, Corollary 5.2.13 p. 149) follow an a disjoint union: $S= \bigcup_{N\geq n\geq 0} S_n $ (where can N can be infinite) such that: for $n>0$ any $S_n$ is an atom of $\mu$ and $\mu$ in non atomic on the field of subests of $S_0$ . The question now is: "there is a (null on sigletons) non atomic measure on the subesets of S?", this a old classic question studied by S. Ulam. From T.Jech "Set Theory" cap.10 if a such measure exit then exist a 0-1 valued measure by the some conditions above, and this is called a Ulam measure on the set S. And a cardinal is called "Measurable" if any set S with this cardinality has a ulam measure on all its subsets. Now a Ulam cardinal is also $strong\ inaccessible$ but from the Set theory you cannot prove the existence or (no existence) of $strong\ inaccesible$ cardinals (i.e there are model of Set theory by these cardinals, and others without these). Anyway A.Tarsky proved that the least noncountable stron-inaccessible cardinal is smaller than the least Ulam cardinal (if these exist). SSE: "From cardinals to chaos: reflections on the life and legacy of Stanislaw Ulam" (search on Google "set theory, existance of Ulam cardinals")

In "Rings of Continuous Functions" L. Gillman, M. Jerison, Cap.12, there is a nice study of these questions from a topological point of view (no Logical set theory foundation aspects), and show that there are a large collections of cardinal that cannot be Ulam cardinal.

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We can consider S as the unit cube in $R^3$ (by a bijection). THen by Borsuk-Ulam PAradox, we get an absurd.

Corrections: (1): Is the Banach-Tarsky paradox, (2): the (very strong) hypotesis of the invariance for euclidean traslations is excessive.

I'm shamed, excuse me.

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Your argument does not work. One of the reasons is that the common faces of cubes may have nonzero measure. Another one is that different cubes can have different measures, since we do not ask for translation invariance. –  Andrea Ferretti Jun 2 '10 at 17:37
And in any case I think you want to refer to the Banach-Tarski paradox. The only thing I know associated to Borsuk and Ulam is a theorem in algebraic topology. –  Andrea Ferretti Jun 2 '10 at 17:39
Yes, you're right. –  Buschi Sergio Jun 2 '10 at 20:38

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