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Suppose that you are working with the axiom of choice. If you randomly choose a subset of the real line, what is the probability that it will be measurable?

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    $\begingroup$ What probability distribution are you using? $\endgroup$ – Chris Eagle May 9 '11 at 21:45
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    $\begingroup$ The cardinality of the collection of all measurable sets is $2^{\mathfrak{c}}$ (at least all subsets of the Cantor set); the cardinality of the collection of all nonmeasurable sets is also at least $2^{\mathfrak{c}}$ (you have at least that many distinct Vitali sets). Hence both are of cardinality $2^{\mathfrak{c}}$. So you can't even argue as one sometimes does that a "random real" is almost certain to be transcendental (based on cardinality). I don't think the question can be given a reasonable answer. $\endgroup$ – Arturo Magidin May 9 '11 at 21:58
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    $\begingroup$ The answer has to be zero. If not, I don't believe in choice. $\endgroup$ – Michael Lugo May 9 '11 at 22:47
  • $\begingroup$ @Micheal: Why? Usually when one encounters a counter example, they turn out to be numerous... $\endgroup$ – Thomas Rot May 9 '11 at 22:54
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    $\begingroup$ What about the probability distribution where $P(x \in A)=1/2$ for each $x\in\mathbb{R}$? $\endgroup$ – mjqxxxx May 10 '11 at 14:21
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A reasonable way to select a random subset $A$ of the real line might be the following. For each $t\in\mathbb{R}$, decide (independently for each $t$) to either put $t$ in $A$ or not, each with probability 1/2. This is what @mjqxxxx suggested.

To make this precise, for each $t\in\mathbb{R}$, we have the basic Bernoulli 1/2 probability space: $\Omega_t=\{0,1\}$, $\mathcal{F}_t=2^\Omega$, and $P_t$ is defined by $P_t(\{0\})=P_t(\{1\})=1/2$. The overall probability space is the product of these: $\mathbb{X}=\prod_{t\in\mathbb{R}}\Omega_t$, $\mathbb{F}=\bigotimes_{t\in\mathbb{R}}\mathcal{F}_t$, and $\mathbb{P}$ is the product measure. Note that $\mathbb{X}$ is the collection of all subsets of $\mathbb{R}$, i.e. the power set of $\mathbb{R}$.

Let $\mathcal{M}\subset\mathbb{X}$ denote the collection of Lebesgue measurable sets. (The same thing can be done for Borel measurable sets as well.) We would like to compute $\mathbb{P}(\mathcal{M})$.

Unfortunately, as will be shown below, $\mathcal{M}\notin\mathbb{F}$, and so $\mathbb{P}(\mathcal{M})$ is undefined. In other words, the set of measurable sets is not measurable, so we cannot speak of the probability that a random set is measurable. (At least not in this model.)

To prove this, define $\pi_t:\mathbb{X}\to\{0,1\}$ by $\pi_t(A)=1_A(t)$. Then $\pi_t$ are the projections, and, by definition, $\mathbb{F}=\sigma(\pi_t:t\in\mathbb{R})$. By Proposition 4.6 in "Probability and Stochastics" by Erhan Çinlar, a function $F:\mathbb{X}\to\mathbb{R}$ is measurable if and only if there exists a sequence $(t_n)$ in $\mathbb{R}$ and a measurable function $f:\{0,1\}^\infty\to\mathbb{R}$ such that $$ F(A) = f(\pi_{t_1}(A), \pi_{t_2}(A), \ldots), $$ for all $A\in\mathbb{X}$.

Now suppose $\mathcal{M}\in\mathbb{F}$ and define $F=1_{\mathcal{M}}$. Choose a sequence $(t_n)$ and a measurable function $f$ such that $$ 1_{\mathcal{M}}(A) = f(\pi_{t_1}(A), \pi_{t_2}(A), \ldots), $$ for all $A\in\mathbb{X}$. Let $B=\{t_1,t_2,\ldots\}$. Since $B$ is countable, it is Lebesgue measurable, and so $$ 1 = 1_{\mathcal{M}}(B) = f(\pi_{t_1}(B), \pi_{t_2}(B), \ldots) = f(1,1,\ldots). $$ Let $C$ be any non-measurable set. Since $t_n\in C\cup B$ for all $n$, we have $$ 1_{\mathcal{M}}(C\cup B) = f(\pi_{t_1}(C\cup B), \pi_{t_2}(C\cup B), \ldots) = f(1,1,\ldots) = 1, $$ and so $C\cup B$ is measurable. But $C^c\cap B$ is countable, and therefore measurable, from which we deduce that $$ C = (C\cup B) \cap (C^c \cap B)^c $$ is a measurable set, which is a contradiction.

Edit:

I wanted to add some additional information which was too much for a comment, and which incidentally also addresses the comment from @Hendrik.

One might wonder if we can find a measurable collection $\mathcal{A}$ such that $\mathcal{A}\subset\mathcal{M}$. If we can and if, for example, we have $\mathbb{P}(\mathcal{A})=1$, then we can say that the random set is almost surely measurable. But it should be straightforward to adapt the above proof to show that if $\mathcal{A}\in\mathbb{F}$ and $\mathcal{A}\subset\mathcal{M}$, then $\mathcal{A}=\emptyset$. Similarly, if $\mathcal{A}\in\mathbb{F}$ and $\mathcal{M}\subset\mathcal{A}$, then $\mathcal{A}=\mathbb{X}$.

It follows from this that $\mathcal{M}$ is not in the completion of this probability space. To see this, suppose that $\mathcal{M}=\mathcal{A}\cup\mathcal{N}$, where $\mathcal{A}\in\mathbb{F}$ and $\mathcal{N}\subset\mathcal{B}$ for some $\mathcal{B}\in\mathbb{F}$ satisfying $\mathbb{P}(\mathcal{B})=0$. Then $\mathcal{A}\subset\mathcal{A}\cup\mathcal{N}=\mathcal{M}$, which implies $\mathcal{A}=\emptyset$. Thus, $\mathcal{M}=\mathcal{N}\subset\mathcal{B}$, which implies $\mathcal{B}=\mathbb{X}$. But this contradicts $\mathbb{P}(\mathcal{B})=0$.

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    $\begingroup$ What happens if you take the completion of the measure space $(\mathbb{X},\mathbb{P})$? $\endgroup$ – Hendrik Vogt Jun 30 '11 at 8:14

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