# Existence of probability measure defined on all subsets

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

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.