13

If you randomly select a subset of $[0,1]$, what is the probability that it will be measurable?

Edit: This question may be unanswerable as asked. If additional assumptions could be made to make it answerable, I would be interested in learning about them.

Paul Orland
  • 6,888
  • 6
    Or is this even a valid question as currently phrased? – Paul Orland Dec 06 '12 at 17:52
  • 2
    That's like asking what's the measure of all non-measurable sets... – gnometorule Dec 06 '12 at 17:54
  • 2
    It is! Has that question been answered? – Paul Orland Dec 06 '12 at 17:55
  • How would you measure a non-measurable set? As first commenter pointed out, need to think about how to phrase/well-define this. – gnometorule Dec 06 '12 at 17:57
  • Well, I'm not asking how to measure the nonmeasurable subsets of $[0,1]$ themselves. I'm asking how to measure a specific subset of $2^{[0,1]}$ consisting of only these objects. – Paul Orland Dec 06 '12 at 18:39
  • 3
    What's your process for randomly selecting a subset of $[0,1]$? That is, what is your measure on $2^{[0,1]}$? – Jonathan Christensen Dec 06 '12 at 18:46
  • I guess that is the underlying question. Is there no canonical way to do this? – Paul Orland Dec 06 '12 at 19:53
  • There is a somewhat canonical way, using the Kolmogorov extension theorem, to get a ${0,1}$-valued stochastic process $X_t$ on $[0,1]$, such that its finite-dimensional distributions are equidistributed. These correspond to subsets of $U \subseteq [0,1]$, via $x\in U$ iff $X_t = 1$. I am not sure what the answer to the question for this construction is. – Lukas Geyer Dec 07 '12 at 01:57
  • TBH, I am fairly interested in this. I have been thinking about this: We know that the set $A\subset Pot([0,1])$ of all measurable sets is (truly) larger than $\mathbb{R}$, iow., there is no injection from A to $\mathbb{R}$. Using the continuum hypothesis, we know that there must exist a bijection between A and $Pot([0,1])\setminus A$, wich, in turn, proves there must exist a limit $\in(0,1)$ to this probability? Or am I mistaken? – CBenni Dec 17 '12 at 18:05
  • @CBenni you're mistaken on a few fronts; firstly, in invoking the notion of 'limit' when you haven't even defined what a limit would be with respect to, and secondly in assuming that any such would be in (0,1) simply because both sides are 'equinumerous'. Consider e.g. $\mathbb{N}$ and $\mathbb{N}\setminus{n^2:n\in\mathbb{N}}$. – Steven Stadnicki Dec 27 '12 at 05:28
  • 1
    Measurable in what sense? Borel? Lebesgue? Counting measure? Are you assuming the axiom of choice? – Asaf Karagila Dec 27 '12 at 06:19
  • 1
    @CBenni: What does CH have to do with subsets of $\mathcal P(\mathbb R)$? – Asaf Karagila Dec 27 '12 at 06:48
  • @StevenStadnicki true... I didnt think about that. I would still be interested in the answer to this question ;) I am not sure, but I think I read that Borel proved that almost all (all but for a null set) subsets of (0,1) are non-measurable. Therefore, we would have the probability 1. – CBenni Dec 27 '12 at 10:58
  • 3
    @orlandpm http://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable – CBenni Dec 27 '12 at 11:17
  • @CBenni: I think you could post this as an answer. – Nate Eldredge Dec 27 '12 at 18:47
  • @NateEldredge not so sure. But I will go for it ^.^ – CBenni Dec 27 '12 at 22:13

3 Answers3

14

The following topological smallness (rather than measure smallness) result may be of interest.

Let ${\mathbb P}[0,1]$ be the collection of all subsets of $[0,1]$ modulo the equivalence relation $\sim$ defined by $E \sim F \Leftrightarrow {\lambda^{*}(E \Delta F)} = 0,$ where $\lambda^{*}$ is Lebesgue outer measure and $\Delta$ is the symmetric difference operation on sets. The set ${\mathbb P}[0,1]$ can be made into a complete metric space by defining the distance function $d,$ where $d(E,F) = {\lambda^{*} (E \Delta F)}.$ In the paper cited below it is proved that the collection of measurable subsets of $[0,1]$ is a perfect nowhere dense set in ${\mathbb P}[0,1].$

Thus, in this setting, the collection of measurable subsets of $[0,1]$ makes up a very tiny part of the collection of all the subsets of $[0,1].$ Note that in the space ${\mathbb P}[0,1]$ the collection of measurable subsets is not just a first category subset of ${\mathbb P}[0,1]$ (this alone would make the collection a tiny subset of ${\mathbb P}[0,1]$), but in fact the collection of measurable subsets is actually a nowhere dense subset of ${\mathbb P}[0,1]$ (hence my saying the collection is a very tiny subset of ${\mathbb P}[0,1]$).

Nobuyuki Kato, Tadashi Kanzo, and Oharu Shinnosuke, A note on the measure problem, International Journal of Mathematical Education in Science and Technology 19 (1988), 315-318.

Dave L. Renfro
  • 36,843
  • Is it possible to give a free link to the paper that's cited in your answer. I found this link; however, I'm not involved with any university and I can't get it for free, plus my caretaker isn't willing to spend 56 dollars. Hopefully you have a copy that you could share in your answer. – Arbuja May 23 '23 at 17:56
  • 1
    @Davis Johnson: I don't have online access to this journal (in fact, none except MAA journals via my MAA membership), but I do have a photocopy of the paper (made many years ago in a university library) that I can .pdf-scan and email to you. Send me an email for me to reply. – Dave L. Renfro May 23 '23 at 18:25
  • @Davis Johnson: Yes, but delete your comment, as I don't want to wind up getting hundreds of spam emails if my email gets harvested somewhere. – Dave L. Renfro May 24 '23 at 14:36
  • Deleted the comment. – Arbuja May 24 '23 at 14:48
4

Almost the same Question was asked on mathoverflow: https://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable

The conclusion was that the set of all measurable sets was not measurable; Therefore, no Probability can be provided.

Community
  • 1
CBenni
  • 1,940
2

I am assuming that we have fixed some $\sigma$-field of subsets of the interval and "measureable" subset means a subset from that $\sigma$-field.

Your question is meaningless until we specify what does "randomly select a subset" mean. And to do that we have to choose a $\sigma$-field of subsets of the set $2^{[0,1]}$, say $\mathcal{A}$, and a probability measure $\mathbb{P}:\mathcal{A}\rightarrow [0,1]$. It is easy to see that now when we know what does "randomly select a subset" mean we can answer your question. And what is the answer? Depending on the choice of $\mathcal{A}$ and $\mathbb{P}$ it can be any number between $0$ and $1$, and none of those answers is better than the other. (From the point of view of probability theory)

Godot
  • 2,082