12
$\begingroup$

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.

  • 5
    Or is this even a valid question as currently phrased?2012-12-06
  • 1
    That's like asking what's the measure of all non-measurable sets...2012-12-06
  • 1
    It is! Has that question been answered?2012-12-06
  • 0
    How would you measure a non-measurable set? As first commenter pointed out, need to think about how to phrase/well-define this.2012-12-06
  • 0
    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.2012-12-06
  • 3
    What's your process for randomly selecting a subset of $[0,1]$? That is, what is your measure on $2^{[0,1]}$?2012-12-06
  • 0
    I guess that is the underlying question. Is there no canonical way to do this?2012-12-06
  • 0
    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.2012-12-07
  • 0
    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?2012-12-17
  • 0
    @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 $\mathbb{N}\setminus\{n^2:n\in\mathbb{N}\}$.2012-12-27
  • 1
    Measurable in what sense? Borel? Lebesgue? Counting measure? Are you assuming the axiom of choice?2012-12-27
  • 1
    @CBenni: What does CH have to do with subsets of $\mathcal P(\mathbb R)$?2012-12-27
  • 0
    @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 1.2012-12-27
  • 3
    @orlandpm http://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable2012-12-27
  • 0
    @CBenni: I think you could post this as an answer.2012-12-27
  • 0
    @NateEldredge not so sure. But I will go for it ^.^2012-12-27

3 Answers 3

9

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.

3

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.

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)