Find the probability distribution, knowing the conditional probability distribution. I have been working on some physics problem that I ‘translated’ into the following mathematical problem for which I need help to solve.

Suppose that $ \alpha $ is a random variable uniformly distributed on the interval $ [0,2 \pi] $. Then for each random (there are technicalities here) subset $ A $ of $ [0,2 \pi] $, we have
$$
\textbf{Pr}(\alpha \in A \mid \lambda(A) = x) = \frac{x}{2 \pi},
$$
where $ \lambda $ denotes the Lebesgue measure on $ [0,2 \pi] $.
Further suppose that we can randomly choose random subsets of $ [0,2 \pi] $ so that
$$
\{
A \mid
\text{$ A $ is a random subset of $ [0,2 \pi] $ and $ \lambda(A) \in [a,b] $}
\}
$$
is a random event for $ 0 \leq a \leq b \leq 2 \pi $ and that the random variable $ X $ defined by
$$
X \stackrel{\text{df}}{=}
\left\{ \begin{matrix}
\text{Random subsets of $ [0,2 \pi] $} & \to     & [0,2 \pi] \\
A                                      & \mapsto & \lambda(A)
\end{matrix} \right\}
$$
has a probability density function $ p $. Then I thought:
$$
  \textbf{Pr}(\alpha \in A)
= \int_{0}^{2 \pi} \frac{x}{2 \pi} \cdot p(x) ~ \mathrm{d}{x}.
$$

Now follow some questions that I have:


*

*Is the above formula actually correct?

*Is it possible to randomly ‘choose’ a random subset of $ [0,2 \pi] $ (or any other interval)?

 A: The second question is tantamount to asking if there exist a $ \sigma $-algebra $ \Sigma $ on the Borel $ \sigma $-algebra $ \mathscr{B}([0,2 \pi]) $ on $ [0,2 \pi] $ itself and a probability measure $ \mu: \Sigma \to [0,1] $ such that
$$
\forall x \in [0,2 \pi]: \quad
\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \} \in \Sigma,
$$
and the function $ p: [0,2 \pi] \to [0,1] $ defined by
$$
\forall x \in [0,2 \pi]: \quad
p(x) \stackrel{\text{df}}{=}
\mu(\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \})
$$
is a Lebesgue-integrable function with mass $ 1 $, i.e., $ \displaystyle \| p \|_{1} \stackrel{\text{df}}{=} \int_{[0,2 \pi]} p ~ \mathrm{d}{\lambda} = 1 $.
The answer is ‘no’. By way of contradiction, assume that $ \Sigma $ and $ \mu $ exist with the properties listed above. Observe that the sets
$$
\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \}, \quad x \in [0,2 \pi]
$$
are disjoint. As $ \| p \|_{1} = 1 $, we obviously have $ p(x) > 0 $ for uncountably many $ x \in [0,2 \pi] $. By the Infinite Pigeonhole Principle, there exist (i) an $ \epsilon > 0 $ and (ii) an uncountable subset $ S $ of $ [0,2 \pi] $ such that $ p(x) > \epsilon $ for every $ x \in S $. Let $ C \subsetneq S $ be countably infinite. Then by the $ \sigma $-additivity of $ \mu $, we have
\begin{align}
    \mu(\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) \in C \})
& = \mu \!
    \left(
    \bigcup_{x \in C} \{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \}
    \right) \\
& = \sum_{x \in C} \mu(\{ A \in \mathscr{B}([0,2 \pi]) \mid \lambda(A) = x \}) \\
& = \sum_{x \in C} p(x) \\
& > \sum_{x \in C} \epsilon \\
& = \infty.
\end{align}
This contradicts the assumption that $ \mu $ is a probability measure.
This automatically renders the first question meaningless.
