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Given a set $X$, consider the sample space $Omega = mathcal P (X)$ and the $sigma$-algebra of events $mathcal F = mathcal P(Omega)$, where $mathcal P$ identifies the power set. For each $x in X$, define the event
begin{equation}
E(x) = {omega in Omega : x in omega} .
end{equation}
Then define $mathbb P: mathcal F to [0,1]$ the probability measure on $ mathcal F$ with the following properties:

1. for every $x in X$ the event $E(x)$ has probability $1/2$, therefore
   begin{equation}
   mathbb P(E(x)) = frac{1}{2} quad forall x in X ;
   end{equation}




2. all the events $E(x)$ are mutually stochastically independent, that is
   begin{equation}
   mathbb P (E(x_1) cap E(x_2)) = mathbb P (E(x_1)) mathbb P (E(x_2)) quad forall x_1, x_2 in X .
   end{equation}

If $X$ has a finite cardinality, it should be possible to prove that $mathbb P$ exists and is unique. However, what happens if $X$ has infinite cardinality? In particular:

• Does $mathbb P$ exist? If the answer is no, is it possible to prove it?


• If $mathbb P$ exists, is it also unique? If the answer is negative, can you find counter-examples?


• In case that existence and uniqueness are valid for any generic set $X$, let be $X = mathbb R$, or more generally a Lebesgue-measurable set of $mathbb R$. If $mathbb P$ exists and is unique also in this case, it is possible to calculate
  begin{equation}
  mathbb P ({omega in Omega : text {$omega$ is Lebesgue-measurable}}) quad ?
  end{equation}
  

Is it correct associating this measure to the (impossible) act of "picking up a random element of $mathcal P(X)$" also for $X subseteq mathbb R$?

$mathbb{P}$ does not exist when $mathcal{F} = mathcal{P}(Omega)$. I will sketch some of the details, but also see this MO question and the Fremlin & Talagrand paper linked there for more details.

It's useful to identify elements of $Omega$ with elements of ${0,1}^X$. Then, if $mathcal{F}$ is the product sigma-field on ${0,1}^X$, $mathbb{P}$, defined in your way on the $E(x)$, extends to a unique probability measure on $mathcal{F}$ in the usual way.

The reason that we cannot extend $mathbb{P}$ further to $mathcal{P}(Omega)$ is that the set $L={omega in Omega: omega text{is Lebesgue measurable}}$ is not measurable in the product sigma-field. (In fact, this set is maximally non-measurable in the sense that it has inner measure $0$ and outer measure $1$.) Intuitively, this is because sets in the product sigma-field have the property that they can be determined by sampling countably many points. But the Lebesgue-measurable sets do not have this property: you can add or remove countably many points from a Lebesgue-(non)measurable subset and preserve (non)measurability. (See Will Sawin's answer at the linked MO post for a more formal statement of these facts.)

If we allow $mathbb{P}$ to be a finitely additive probability measure, then $mathbb{P}$ extends to all of $mathcal{P}(Omega)$ quite easily (by the Hahn-Banach theorem, for instance). This is the situation treated in the Fremlin & Talagrand paper.