If you construct a set in real life, then you have to have some way of judging whether the dart is “in” or “out”. I reckon that any method you can think of will in fact give a measurable set.
Alternatively, there are several ways of making all sets measurable. One is to reject the Axiom of Choice. The AoC is what’s used to construct immeasurable sets. It’s consistent in ZF without AoC that all sets are Lebesgue measurable.
If you like the Axiom of Choice, then another alternative is to only demand that your probability measure be finitely additive. Then you can give a “measure” (such finitely additive measures are actually called “charges”) such that all sets are measurable. What’s more you can make your probability charge agree with Lebesgue measure on the Lebesgue measurable sets. (I think you need AoC for this though.)
In L.J. Savage’s “The Foundations of Statistics” the axioms of probability are justified from decision theory. He only ever manages to prove that probability should be finitely additive; so maybe it doesn’t have to be countably additive. One bonus of finite additivity for Bayesians is that lots of improper priors become proper. For example, there’s a uniform probability charge on the naturals.
I’ve been thinking about this a bit more. My current thinking is basically what Coscott said:
We only care about probabilities if we can be forced to make a bet.
In order for it to be possible to decide who won the bet, we need that (almost always) a measurement to some finite accuracy will suffice to determine whether the dart is in or out of the set.
Thus the set has a boundary of measure zero.
Thus the set is measurable.
What we have shown is that in any bet we’re actually faced with, the sets involved will be measurable.
(The steps from 2 to 3 and 3 to 4 are left as exercises. (I think you need Lebesgue measurable sets rather than just Borel measurable ones))
Note that the converse fails: I believe you can’t make a bet on whether or not the dart fell on a rational number, even though the rationals are measurable.
Here’s a variant which is slightly different, and perhaps stronger since it also allows some operations with “infinite accuracy”.
In order to decide who won the bet, we need a referee. A natural choice is to say that the referee is a Blum-Shub-Smale machine, i.e. a program that gets a single real number x∈[0,1] as input, and whose operations are: loading real number constants; (exact) addition, substraction, multiplication and division; and branching on whether a≤b (exactly).
Say you win if the machine accepts x in a finite number of steps. Now, I think it’s always the case that the set of numbers which are accepted after n steps is a finite union of (closed or open) intervals. So then the set of numbers that get accepted after any finite number of steps is a countable union of finite unions of intervals, hence Borel.
If you construct a set in real life, then you have to have some way of judging whether the dart is “in” or “out”. I reckon that any method you can think of will in fact give a measurable set.
Alternatively, there are several ways of making all sets measurable. One is to reject the Axiom of Choice. The AoC is what’s used to construct immeasurable sets. It’s consistent in ZF without AoC that all sets are Lebesgue measurable.
If you like the Axiom of Choice, then another alternative is to only demand that your probability measure be finitely additive. Then you can give a “measure” (such finitely additive measures are actually called “charges”) such that all sets are measurable. What’s more you can make your probability charge agree with Lebesgue measure on the Lebesgue measurable sets. (I think you need AoC for this though.)
In L.J. Savage’s “The Foundations of Statistics” the axioms of probability are justified from decision theory. He only ever manages to prove that probability should be finitely additive; so maybe it doesn’t have to be countably additive. One bonus of finite additivity for Bayesians is that lots of improper priors become proper. For example, there’s a uniform probability charge on the naturals.
I’ve been thinking about this a bit more. My current thinking is basically what Coscott said:
We only care about probabilities if we can be forced to make a bet.
In order for it to be possible to decide who won the bet, we need that (almost always) a measurement to some finite accuracy will suffice to determine whether the dart is in or out of the set.
Thus the set has a boundary of measure zero.
Thus the set is measurable.
What we have shown is that in any bet we’re actually faced with, the sets involved will be measurable.
(The steps from 2 to 3 and 3 to 4 are left as exercises. (I think you need Lebesgue measurable sets rather than just Borel measurable ones))
Note that the converse fails: I believe you can’t make a bet on whether or not the dart fell on a rational number, even though the rationals are measurable.
Here’s a variant which is slightly different, and perhaps stronger since it also allows some operations with “infinite accuracy”.
In order to decide who won the bet, we need a referee. A natural choice is to say that the referee is a Blum-Shub-Smale machine, i.e. a program that gets a single real number x∈[0,1] as input, and whose operations are: loading real number constants; (exact) addition, substraction, multiplication and division; and branching on whether a≤b (exactly).
Say you win if the machine accepts x in a finite number of steps. Now, I think it’s always the case that the set of numbers which are accepted after n steps is a finite union of (closed or open) intervals. So then the set of numbers that get accepted after any finite number of steps is a countable union of finite unions of intervals, hence Borel.