The Lebesgue measure on the reals has the following property: it assigns zero probability to every singleton. This is intuitively what we would expect of a uniform measure of the reals by the following line of thought: If I throw a uniformly distributed dart at the interval [0,1], the probability that I will land on any singleton is infinitely smaller than one, i.e., zero. Here is a generalization of this line of thought: If A is a subset of [0,1] of lower cardinality than [0,1] (i.e., lower cardinality than c, the cardinality of the continuum), then the probability that the dart will land in A is infinitely smaller than one, i.e., zero, since A is intuitively infinitely smaller than [0,1] (the union of infinitely many disjoint copies of A will still be smaller than [0,1], assuming the Axiom of Choice).
One might at this point ask: Does Lebesgue measure respect this intuition? If the Continuum Hypothesis is true, then of course it does. For then all subsets of lower cardinality than [0,1] are countable, and all countable subsets have null measure, since the singletons have null measure. Without the Continuum Hypothesis this may or may not be true. See the references here. However, even without the Continuum Hypothesis, but given Choice, it can be easily shown (see the previous link) that any Lebesgue measurable subset of lower cardinality than the continuum has null measure. But ZFC is consistent with the claim that all subsets of the reals of lower cardinality than the continuum have null measure as well as with the claim that some are non-measurable and hence do not have null measure.
Nonetheless, given Choice, the Lebesgue measure on the reals extends in a very natural way to a translation-invariant measure m* on a σ-algebra F* that contains all subsets of the reals that have cardinality less than that of the continuum. We can call this the lower-cardinality-nulling extension of the Lebesgue measure.
Because of the intuitions in the first paragraph, it seems to me that the lower-cardinality-nulling extension of Lebesgue measure (or some extension thereof) is the measure we should work with for confirmation theoretic purposes where normally Lebesgue measure is used. Also, while right now I don't know if there could be a translation-invariant extension of Lebesgue measure that assigns non-zero measure to some subset of cardinality less than c, it is easy to see that if there is such an extension, then it assigns measure 1 to some subset of [0,1] of lower cardinality, and that is surely intuitively unacceptable for the uniform results of dart throws and the like. Hence, every acceptable translation-invariant extension of Lebesgue measure assigns zero to every set of cardinality less than c, and since there is a translation-invariant such extension, so we have a good intuitive argument in favor of working with such a measure.
If this is right then, then the Brown two-dart argument against the Continuum Hypothesis (see references and helpful critique here) is misguided. For we should take the measure governing dart throws to be a lower-cardinality-nulling extension of Lebesgue measure. And once we do that, then the Brown two-dart argument works just as well without assuming the Continuum Hypothesis. Hence whatever problem it identifies is not specific to the Continuum Hypothesis.
Appendix: Construction of the lower-cardinality-nulling extension of the Lebesgue measure. Let F* be the σ algebra consisting of all the subsets of R that differ from a member F by a set whose cardinality is less than c. Suppose A is a member of F*. Then A can be represented as (U∪N1)−N2 where U is in F* and N1 and N2 have cardinality less than c. Let m*(A)=m(U). To see that m* is well-defined, suppose that (U∪N1)−N2=(V∪M1)−M2, where V is in F and M1 and M2 have cardinality less than c. Then U and V differ by sets of cardinality less than c, and hence their symmetric difference has null Lebesgue measure, and so m(U)=m(V), and m*(A) is well-defined. Now a countable union of sets An of cardinality less than c has cardinality less than c. This follows from the fact that (using the Axiom of Choice) c has uncountable cofinality, so that there is a cardinality K such that |An|≤K<c for all n, and hence, again by Choice, the union of the An must have cardinality at most K, if K is infinite, and at most countable if K is finite. Since F* is clearly closed under complements, it's a σ-algebra. To check that m* is a measure, we need only check it's countably additive. This easily follows once again from the fact that a countable union of sets of cardinality less than c has cardinality less than c. And translation invariance is obvious.
The above argument only uses the claim that measurable sets of cardinality less than c are null. This is true for n-dimensional Lebesgue measure. (Proof: Suppose that A in Rn is a bounded measurable set of cardinality less than c. Let A' be a projection of A onto any one axis. Then A' is a measurable one-dimensional set of cardinality less than c, and hence null. For large enough L, A will be a subset of the cartesian product of A' and an (n−1)-dimensional ball of radius L, and so A will also be null. But if all bounded measurable sets of cardinality less than c are null, then so are the unbounded ones.) And so the cardinality-nulling extension works in n-dimensions as well.
3 comments:
The extension result is a direct consequence of a lemma of Szpilrajn together with the fact that all Lebesgue measurable sets of cardinality less than c have measure zero.
There is a nice proof that no translation invariant extension of Lebesgue measure assigns non-zero measure to a set of cardinality less than c.
"Brown's argument" is really Freiling's. And Freiling noted that an argument like his can be used as an argument against the Axiom of Choice (what I said in the post about the argument working without the Continuum Hypothesis should have been qualified by adding: assuming Choice).
Post a Comment