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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.

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Or is this even a valid question as currently phrased? –  orlandpm Dec 6 '12 at 17:52
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That's like asking what's the measure of all non-measurable sets... –  gnometorule Dec 6 '12 at 17:54
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It is! Has that question been answered? –  orlandpm Dec 6 '12 at 17:55
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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 6 '12 at 18:46
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@orlandpm mathoverflow.net/questions/102386/… –  CBenni Dec 27 '12 at 11:17

3 Answers 3

up vote 9 down vote accepted
+50

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.

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Almost the same Question was asked on mathoverflow: http://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.

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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)

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