Precise lengths

As usual, write [a,b] for the interval of the real line from a to b including both a and b, (a,b) for the interval of the real line from a to b excluding a and b, and [a, b) and (a, b] respectively for the intervals that include a and exclude b and vice versa.

Suppose that you want to measure the size m(I) of an interval I, but you have the conviction that single points matter, so [a,b] is bigger than (a,b), and you want to use infinitesimals to model that difference. Thus, m([a,b]) will be infinitesimally bigger than m((a,b)).

Thus at least some intervals will have lengths that aren’t real numbers: their length will be a real number plus or minus a (non-zero) infinitesimal.

At the same time, intuitively, some intervals from a to b should have length exactly b − a, which is a real number (assuming a and b are real). Which ones? The choices are [a,b], (a,b), [a, b) are (a, b].

Let α be the non-zero infinitesimal length of a single point. Then [a,a] is a single point. Its length thus will be α, and not a − a = 0. So [a,b] can’t always have real-number length b − a. But maybe at least it can in the case where a < b? No. For suppose that m([a,b]) = b − a whenever a < b. Then m((a,b]) = b − a − α whenever a < b, since (a, b] is missing exactly one point of [a,b]. But then let c = (a+b)/2 be the midpoint of [a,b]. Then:

1. m([a,b]) = m([a,c]) + m((c,b]) = (c−a) + (b−c−α) = b − a − α,

rather than m([a,b]) as was claimed.

What about (a,b)? Can that always have real number length b − a if a < b? No. For if we had that, then we would absurdly have:

2. m((a,b)) = m((a,c)) + α + m((c,b)) = c − a + α + b − c = b − a + α,

since (a,b) is equal to the disjoint union of (a,c), the point c and (c,b).

That leaves [a, b) and (a, b]. By symmetry if one has length b − a, surely so does the other. And in fact Milovich gave me a proof that there is no contradiction in supposing that m([a,b)) = m((b,a]) = b − a.

Magnifying sets of real numbers

If S is a set of real numbers and a is a real number, let aS={ax:x∈S} be the set you get by magnifying S by a factor a.

Here's a funny thing. Some sets get bigger when magnified and some get smaller. For instance, if we take the interval [0,1] and magnify it by a factor of two, we get the interval [0,2], which is intuitively "twice as big". But if we take the natural numbers N and magnify them by a factor of two, 2N will be the even natural numbers, and so 2N will intuitively be "twice as small" as N.

Next observe that if R−[0,1] is the real numbers outside the interval [0,1], then 2(R−[0,1])=R−[0,2] is smaller. Magnifying a set by a factor greater than 1 magnifies both the filled in parts of the set and the holes in the set. The effect of this on the intuitive "size" of the set will depend on the interaction between the holes and the filled in parts.

And if we take the Cantor set C, then magnifying it by a factor of three makes the set be intuitively twice as large. I.e., 3C=C∪(2+C). This makes it very intuitive that the dimension of the set is log 2 / log 3 (which is indeed its Hausdorff dimension). For intuitively if we have an n-dimensional set, and we magnify it by a factor of a, its size is aⁿ. So if n is the dimension of the Cantor set, then 2=3ⁿ, and so n is log 2 / log 3.

Absolutely negligible sets

In my last post, I was thinking about arguments for why sets of cardinality less than that of the continuum should be taken to have measure zero. Here's the sort of argument I was thinking about:

1. The members of a family F of subsets of Rⁿ should be counted as absolutely having measure zero when: (a) no invariant extension of Lebesgue measure assigns a non-zero measure to any of them and (b) every invariant extension of Lebesgue measure has a further invariant extension that assigns zero measure to every member of F.

The sets of cardinality less than that of the continuum satisfy (a) and (b). (The proof of (a) is here. The appendix in my post extends to prove (b) given (a).)

But (1) is actually false. Kharazishvili has introduced the notion of "absolutely negligible sets", which in the Lebesgue case are the sets whose singletons satisfy (a) and (b) (or a stronger condition involving quasi-invariant extensions), and has proved, inter alia, that Rⁿ is a countable union of negligible subsets (see this paper and references therein; the same point applies to the circle R/N and any other solvable uncountable group.) Consequently, we should not accept (1).

There is a philosophical lesson here: Even where there is an apparently very reasonable way to make a nonmeasurable measurable, we can't simultaneously make measurable all the sets that there is a reaonable way of making measurable. And all this stuff on extensions of Lebesgue measure leads to a question that I have not seen addressed in the philosophy literature: Which invariant extension of Borel measure should we take to model cases of things like dart throws and to use as the basis for our physics?

Nonetheless, I think (1) is evidence that the members of F are reasonably to be taken as having measure zero, so something of my previous argument survives.

A natural extension of Lebesgue measure, and Brown's argument against the Continuum Hypothesis

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 A_n 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 |A_n|≤K<c for all n, and hence, again by Choice, the union of the A_n 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.

Absolutely nonmeasurable sets

The ideal of a non-zero (point) probability assignment to all possibilities is incoherent for cardinality reasons. Moreover, as Alan Hajek has insisted, the existence of nonmeasurable sets provides further difficulties.

One might try to get around both issues by problem-specific Bayesianism, where one only insists on a probability assignment specific to a particular problem at hand. This gets around my no-go theorem, since that theorem shows that there is no single non-zero probability assignment to all the possibilities there are. But in any given probabilistic calculation, the collection of possibilities is restricted to some set, and then there could well be a generalized probability (e.g., satisfying the axioms here) for that problem.

One might even have some hope that problem-specific Bayesianism could handle the issue of nonmeasurable sets. For there are isometrically invariant extensions of Lebesgue measure (i.e., extensions invariant under translation, rotation and reflection) that make some Lebesgue nonmeasurable sets be measurable.

But no such luck. Start by noting that there are absolutely nonmeasurable sets. A bounded absolutely nonmeasurable set (I'm making up this technical term) is a subset A of n-dimensional Euclidean space Rⁿ such that there is no isometrically invariant probability measure that (a) makes A measurable, (b) assigns finite measure to every bounded measurable subset of Rⁿ, (c) assigns non-zero measure to some bounded subset of Rⁿ. The Hausdorff Paradox then shows that there is a bounded absolutely nonmeasurable set if n=3, assuming the Axiom of Choice.

In fact, from the Hausdorff Paradox we can prove that there is a bounded subset A of R³ such there is no isometrically invariant generalized finitely additive probability measure, e.g., in the sense of this post, on the cube [0,1]³ or on the three-dimensional ball of unit radius that makes A measurable.

So the problem-specific approach also runs into trouble, at least assuming the Axiom of Choice. And the Axiom of Choice (or, more weakly, the Boolean Prime Ideal Theorem--I don't know if this makes a difference, but in any case BPI has no intuitive support beyond the fact that AC implies it) is also assumed by hyperreal extensions of probability theory.

Of course, if one allows for interval-valued measures, that's different kettle of fish.

Non-measurable sets and interval-valued probabilities

I think there is nothing new here, but I want to collect together some facts that are interesting to me.

Suppose m is a countably additive measure on a set U. Then it's pretty easy to show that for any subset B of U, measurable or not, there exist measurable sets A and C such that:

• A is a subset of B and B is a subset of C
• A is maximal in measure among the measurable subsets of B: for every measurable subset A' of B, m(A')≤m(A)
• C is minimal in measure among the measurable supersets of B: for every measurable superset C' of B, m(C')≥m(C).

(Cf. Van Vleck.) We can now define the lower measure lm(B)=m(A) and the upper measure um(B)=m(C). Of course lm(B)≤um(B) for all B.

If m is a probability measure (i.e., m(U)=1), we can then extend the measure m on U to a complete measure (i.e., one such that any subset of a set of measure zero is also measurable) simply by taking as measurable all sets B such that lm(B)=um(B), and then setting m(B)=um(B).

From now on assume m is a complete probability measure. Then a set B is measurable if and only if lm(B)=um(B).

Suppose that X₁,X₂,... are independent random variables taking their values in U, with the probability that X_i is in B being equal to m(B). Let S_n(B) be the number of the variables X₁,...,X_n whose values are in B. If B is measurable, then the Strong Law of Large Numbers implies that almost surely (i.e., with probability 1), S_n(B)/n converges to m(B). It immediately follows that in general, whether or not B is measurable, almost surely

• lm(B)≤ liminf S_n(B)/n≤ limsup S_n(B)/n≤um(B).

In other words, almost surely, all the limit points of the asymptotic frequency S_n(B)/n fall between the lower and upper measures of B.

It would be interesting to see what else we can say about the limit points of the asymptotic frequency. One might speculate that lm(B) and um(B) are almost surely limit points of the asymptotic frequency, but I think that's not true in general. But could it be true in the special case where m is Lebesgue measure on an interval?

I've been thinking from time to time about this question: What do asymptotic frequencies of visits to a nonmeasurable set look like? Still no answer.

In any case, the above stuff suggests that dealing with nonmeasurable sets might be a good application for interval-valued probabilities, where we assign the interval [lm(B),um(B)] as the probability of B.

Oh, and finally, it's worth noting that Van Vleck has in effect shown that if m is Lebesgue measure on [0,1], then there is a subset of [0,1] whose lower measure is zero and whose upper measure is one.