There exists no probability measure on such that for each .

The proof of this impossibility theorem requires the well-ordering principle (equivalent to the axiom of choice) and also the continuum hypothesis. Let be the set of sequences of positive integers. Then has the power of the continuum. (Let the th partial sum of a sequence in be the position of the th in the nonterminating dyadic representation of a point in ; this gives a one-to-one correspondence.) By the continuum hypothesis, the elements of can be put in a one-to-one correspondence with the set of ordinals preceding the first uncountable ordinal. Carrying the well-ordering of these ordinals over to by means of the correspondence gives a well-ordering relation with the property that each element has only countably many predecessors.

For , write if for all . Say that rejects if and ; this is a transitive relation. Let be the set of unrejected elements of . Let be the set of elements that reject , and assume it is nonempty. If is the first element (with respect to ) of , then (if rejects , then it also rejects , and since , there is a contradiction). Therefore, if is rejected at all, it is rejected by an element of .

Suppose is countable, and let be an enumeration of its elements. If , then is not rejected by any and hence lies in , which is impossible because it is distinct from each . Thus, is uncountable and must, by the continuum hypothesis, have the power of .

Let be a one-to-one map of onto ; write the image of as . Let be the image under of the set of in for which . Since must have some value , . Assume that is countably additive and choose in in such a way that for . If then . If is shown to be countable, this will contradict the hypothesis that each singleton has probability .

Now, there is some in such that (if , take ; otherwise, is rejected by some in ). If for , then , and hence (since otherwise rejects ). This means that is contained in the countable set , and is indeed countable.