Calculating the probability that the following random subsets of $\mathbb{R}^2$ are open. For each $t \in \mathbb{R}$, select an open interval $U_t \subseteq \mathbb{R}$ containing 0. What is the probability that the set $U = \{ (t,x) | x \in U_t\}$ will be an open subset of $\mathbb{R}^2$? 
I was just thinking about this off hand, and I don't see an obvious way to attack it. On the one hand, if $U_t = (-1,1)$ then $U$ must be open. On the other hand, we could do something like $(0,1)$ for $t<0$ and $(-1,1)$ for $t \geq 0$, and this obviously can't be open. 
 A: This question doesn't really make sense without specifying a probability measure on the space of open intervals of $\Bbb R$ containing zero. The measure can't be uniform because there is no uniform probability measure on $\Bbb R$.
We could fix this by specifying that our open intervals $U_t$ are of the form $(a, b)$ with $a \in [-N, 0)$ and $b \in (0, N]$ for some fixed $N > 0$, with $a$ and $b$ uniformly distributed in their sample spaces. Then consider a point $(t, x) \in U$, $x\neq 0$. For $U$ to be open, there must exist an open rectangle $$[t - \epsilon, t+\epsilon] \times [x-\epsilon, x+\epsilon] \subset U$$ for some $\epsilon > 0$. In particular, there must exist an $\epsilon > 0$ for which $x \in U_s$ for every $s \in [t-\epsilon, t+\epsilon]$. 
So what is the probability that this happens? Assuming the $U_t$ are independently chosen, the probability is obviously zero. These are uncountably many sets chosen independently, and the probability that any one of them in particular doesn't contain $x$ is $|x|/N$, a fixed positive number.
