Immeasurable sets are not something in the real world that you can throw a dart at.
I can rephrase your problem to be:
“If I have an immeasurable set X in the unit interval, [0,1), and I generate a uniform random variable from that interval, what is the probability that that variable is in X?”
The problem is that a “uniform random variable” on a continuous interval is a more complicated concept than you think. Let me explain, by first giving an example where X is measurable, lets say X=[0,pi-3). We analyze random continuous variables by reducing to random discrete variables. We can think of a “uniform random variable” as a sequence of digits in a decimal expansion which are determined by rolling a 10 sided die. So for example, we can roll the die, and get 1,4,6,2,9,..., which would correspond to .14629..., which is not in the set X. Notice that while in principle we might have to roll the die arbitrarily many times, we actually only had to roll the die 3 times in this case, because once we got 1,4,6, we knew the number was too big to be in the set X. We can use this fact that we almost always have to roll the die only a finite number of times to get a definition of the “probability of being in X.” In this case, we know that the probability is between .141 and .142, by considering 3 die rolls, and if we consider more die rolls, we get more accuracy that converges to a single number, pi-3.
Now, let’s look at what goes wrong if X is not measurable. The problem here is that the set is so messy that even if we we know about the first finitely many digits of a random number, we wont be able to tell if the number is in X. This stops us from doing the procedure like above and defining what we mean.
EDIT: I retract the following. The problem with it is that Coscott is arguing that “something in the real world that you can throw a dart at” implies “measurable” and he does this by arguing that all sets which are “something in the real world that you can throw a dart at” have a certain property which implies measurability. My “counterexamples” are measurable sets which fail to have this property, but this is the opposite of what I would need to disprove him. I’d need to find a set with this property that isn’t measurable. In fact, I don’t think there is such a set; I think Coscott is right.
The sets with this property (that you can tell whether your number is in or out after only finitely many dice rolls) are the open sets, not the measurable sets. For example, the set [0,pi-3] is measurable but not open. If the die comes up (1,4,1,5,9,...) then you’ll never know if your number is in or out until you have all the digits. For an even worse example take the rational numbers: they’re measurable (measure zero) but any finite decimal expansion could be leading to a rational or an irrational.
The sets with this property (that you can tell whether your number is in or out after only finitely many dice rolls) are the open sets, not the measurable sets.
That doesn’t seem right to me. Take as my target the open set (0, pi-3). If I keep rolling zeros I’ll never be able to stop. (Edit: I know that the probability of rolling all zeros approaches 0 as the number of die rolls approaches infinity, but I assume that a demon can take over the die and start feeding me all zeros, or the digits of pi-3 or whatever. As I think about this more I’m thinking maybe what you said works if there is no demon. Edit 2: Or not. If there’s no demon and my first digit is 0 then I can stop, but that’s only because 0 is expressible as an integer divided by a power of ten. If there’s no demon and I roll the first few digits of pi-3, I know that I’ll eventually go over or under pi-3, but I don’t know which, and it doesn’t matter whether pi-3 itself is in my target set.)
Every die roll tells me that the random number I’m generating lies in the closed interval [x, x+1/10^n], where x is the decimal expansion I’ve generated so far and n is how many digits I’ve generated. If at some point I start rolling all 0s or all 9s I’ll be rolling forever if the number I’m generating is a limit point of the target set, even if it’s not in the target set.
I should have been more accurate and said “If the random number that you’ll eventually get does in fact lie in the set, then you’ll find out about this fact after a finite number of rolls.”
This really does define open sets, since for any point in an open set there’s an open ball of radius epsilon about it which is in the set, and then the interval [x, x+1/10^n] has to be in that ball once 1/10^n < epsilon/2.
EDIT: (and the converse also holds, I think, but it requires some painfully careful thinking because of the non-uniqueness of decimal expansions)
I think a more exact representation of what Coscott actually said is the following property: “We almost always only have to roll the die finitely many times to determine whether the point is in or out.”
This still doesn’t specify measurable sets (because of the counterexample given by the rationals). I think the type of set that this defines is “Sets with boundary of measure zero” where the boundary is the closure minus the interior. Note that the rationals in [0,1) have boundary everywhere (i.e. boundary of measure 1).
Ah, so if my target set is (0, pi-3) and the demon feeds me the digits of pi-3, I will be rolling forever, but if the demon feeds me the digits of pi-3-epsilon (or any other number in (0, pi-3)) I will be able to stop after a finite amount of rolls.
I think the type of set that this defines is “Sets with boundary of measure zero” where the boundary is the closure minus the interior.
That sounds right to me, although I don’t understand measure very well. I was informally thinking of this property as “continuousness”.
Yeah, but I can’t explain explain that without analysis not appropriate for a less wrong post. I remember that the probability class I took in undergrad dodged the measure theory questions by defining probabilities on open sets, which actually works for most reasonable questions. I think such a simplification is appropriate, but I should have had a disclaimer.
Immeasurable sets are not something in the real world that you can throw a dart at.
I can rephrase your problem to be: “If I have an immeasurable set X in the unit interval, [0,1), and I generate a uniform random variable from that interval, what is the probability that that variable is in X?”
The problem is that a “uniform random variable” on a continuous interval is a more complicated concept than you think. Let me explain, by first giving an example where X is measurable, lets say X=[0,pi-3). We analyze random continuous variables by reducing to random discrete variables. We can think of a “uniform random variable” as a sequence of digits in a decimal expansion which are determined by rolling a 10 sided die. So for example, we can roll the die, and get 1,4,6,2,9,..., which would correspond to .14629..., which is not in the set X. Notice that while in principle we might have to roll the die arbitrarily many times, we actually only had to roll the die 3 times in this case, because once we got 1,4,6, we knew the number was too big to be in the set X. We can use this fact that we almost always have to roll the die only a finite number of times to get a definition of the “probability of being in X.” In this case, we know that the probability is between .141 and .142, by considering 3 die rolls, and if we consider more die rolls, we get more accuracy that converges to a single number, pi-3.
Now, let’s look at what goes wrong if X is not measurable. The problem here is that the set is so messy that even if we we know about the first finitely many digits of a random number, we wont be able to tell if the number is in X. This stops us from doing the procedure like above and defining what we mean.
Is this clear?
EDIT: I retract the following. The problem with it is that Coscott is arguing that “something in the real world that you can throw a dart at” implies “measurable” and he does this by arguing that all sets which are “something in the real world that you can throw a dart at” have a certain property which implies measurability. My “counterexamples” are measurable sets which fail to have this property, but this is the opposite of what I would need to disprove him. I’d need to find a set with this property that isn’t measurable. In fact, I don’t think there is such a set; I think Coscott is right.
The sets with this property (that you can tell whether your number is in or out after only finitely many dice rolls) are the open sets, not the measurable sets. For example, the set [0,pi-3] is measurable but not open. If the die comes up (1,4,1,5,9,...) then you’ll never know if your number is in or out until you have all the digits. For an even worse example take the rational numbers: they’re measurable (measure zero) but any finite decimal expansion could be leading to a rational or an irrational.
That doesn’t seem right to me. Take as my target the open set (0, pi-3). If I keep rolling zeros I’ll never be able to stop. (Edit: I know that the probability of rolling all zeros approaches 0 as the number of die rolls approaches infinity, but I assume that a demon can take over the die and start feeding me all zeros, or the digits of pi-3 or whatever. As I think about this more I’m thinking maybe what you said works if there is no demon. Edit 2: Or not. If there’s no demon and my first digit is 0 then I can stop, but that’s only because 0 is expressible as an integer divided by a power of ten. If there’s no demon and I roll the first few digits of pi-3, I know that I’ll eventually go over or under pi-3, but I don’t know which, and it doesn’t matter whether pi-3 itself is in my target set.)
Every die roll tells me that the random number I’m generating lies in the closed interval [x, x+1/10^n], where x is the decimal expansion I’ve generated so far and n is how many digits I’ve generated. If at some point I start rolling all 0s or all 9s I’ll be rolling forever if the number I’m generating is a limit point of the target set, even if it’s not in the target set.
I should have been more accurate and said “If the random number that you’ll eventually get does in fact lie in the set, then you’ll find out about this fact after a finite number of rolls.”
This really does define open sets, since for any point in an open set there’s an open ball of radius epsilon about it which is in the set, and then the interval [x, x+1/10^n] has to be in that ball once 1/10^n < epsilon/2.
EDIT: (and the converse also holds, I think, but it requires some painfully careful thinking because of the non-uniqueness of decimal expansions)
I think a more exact representation of what Coscott actually said is the following property: “We almost always only have to roll the die finitely many times to determine whether the point is in or out.”
This still doesn’t specify measurable sets (because of the counterexample given by the rationals). I think the type of set that this defines is “Sets with boundary of measure zero” where the boundary is the closure minus the interior. Note that the rationals in [0,1) have boundary everywhere (i.e. boundary of measure 1).
Ah, so if my target set is (0, pi-3) and the demon feeds me the digits of pi-3, I will be rolling forever, but if the demon feeds me the digits of pi-3-epsilon (or any other number in (0, pi-3)) I will be able to stop after a finite amount of rolls.
That sounds right to me, although I don’t understand measure very well. I was informally thinking of this property as “continuousness”.
Yeah, but I can’t explain explain that without analysis not appropriate for a less wrong post. I remember that the probability class I took in undergrad dodged the measure theory questions by defining probabilities on open sets, which actually works for most reasonable questions. I think such a simplification is appropriate, but I should have had a disclaimer.