Picking up a random element of $\mathcal P(\mathbb R)$: existence of probability measure?

Question

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 $X \subseteq \mathbb R$?

Answer 1

$\mathbb{P}$ 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.