0
$\begingroup$

Let $\Omega=\mathbb{R},$ \begin{equation}\mathcal{F} = \{A \subset \mathbb{R}: \text{either } A \text{ or } A^c \text{ is countable}\}\end{equation} where $P(A)=0$ if $A$ is countable and $P(A)=1$ if $A$ is uncountable.

Show that $P$ is a probability measure. [Hint: You need to show that if $\{A_n\}$ is any countable collection of pairwise disjoint sets in $\mathcal{F}$ then it can contain at most one uncountable set.]

Here's my attempt (I have already proved $\mathcal{F}$ is $\sigma-$algebra):

(i) Since $\Omega = \mathbb{R},$ we've got $P(\Omega)=1.$

(ii) Suppose $A_1,A_2,\ldots \in \mathcal{F}$ where $A_i \cap A_j = \varnothing \, \forall i \neq j.$

For (ii), I divided for 2 cases;

(a) If $A_1,A_2,\ldots$ are all countable, that is, $\cup_{n=1}^{\infty} A_n$ is countable. So, $P(\cup_{n=1}^{\infty} A_n) = 0 = \sum_{n=1}^{\infty} 0 = \sum_{n=1}^{\infty} P(A_n).$

(b) If there exists $m \in \mathbb{N}$ such that $A_m$ is uncountable. Since $A_m \in \mathcal{F}$, it means $A_m^c$ is countable. Because $A_j \cap A_m = \varnothing$, so $A_j \subset A_m^c$ where $j \neq m$. So, $A_j$ is countable $\forall j \neq m.$

Since $A_m$ is uncountable and $A_m \subseteq \cup_{n=1}^{\infty}A_n,$ that is, $\cup_{n=1}^{\infty}A_n$ is uncountable.

Thus, $P(\cup_{n=1}^{\infty} A_n) = 1 = P(A_m) + \sum_{j=1}^{\infty} P(A_j) = \sum_{j=1}^{\infty} P(A_j).$

But I have no clue about "contain at most one uncountable set". How could I show in that way. Thank you in advance.

$\endgroup$

1 Answer 1

0
$\begingroup$

This should be an application of De Morgan's laws. Towards a contradiction, suppose that there are $i,j \in \mathbb{N}$ with $i \neq j$ such that $A_i^c$ and $A_j^c$ are countable. As $A_i \cap A_j = \emptyset$, we have \begin{align*} \mathbb{R} = \emptyset^c = (A_i \cap A_j)^c =A_i^c \cup A_j^c, \end{align*} showing that $\mathbb{R}$ is countable. As this a contradiction, $A_m$ is the only set within $(A_i)_{i \in \mathbb{N}}$ having a countable complement. Hence, \begin{align*} \sum_{i = 1}^\infty P(A_i) = P(A_m) = 1. \end{align*}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .