If you randomly choose a subset of the real line, what is the probability that it will be measurable? 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?
 A: 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 $\mathbb{P}(\mathcal{B})=0$.
