Show that $(\Omega , A , P)$ is a probability space

Question

Let $\Omega$ = $\mathbb{R}$ (the set of real numbers).

Let $A$ = { $A$ $\subseteq$ $\mathbb{R}$ : $A$ or $A^C$ is countable}.

Let $P$ : $A$ $\to [0,1] $ be

$$P(A)= \begin{cases} 0, & \text{if $A$ is countable } \\ 1 & \text{if $A^C$ is countable } \end{cases}$$

Show that $(\Omega , A , P)$ is a probability space.

Hints : A set is countable if either it is finite or has the same cardinality of the set of the intergers $\mathbb{N}$ ={0 ,1 ,2 , ...} (ie., there is a bijection / one - to - one and onto, between it and $\mathbb{N}$).

A subset of a countable set is countable. Countable unions of countable sets are countable.

Answer 1

To show that $(\Omega,\mathcal{A},P)$ is a probability space you need to show that

1. $\mathcal A$ is a $\sigma$-field on $\Omega$,

2. and $P$ is a probability measure on $\mathcal A$.

To show 1. you need to show that $\mathcal{A}$ satisfies the three properties of being a $\sigma$-field, that is:

• $\mathcal{A}$ is a non-empty,
• if $A\in\mathcal{A}$, then $A^c\in\mathcal{A}$,
• if $(A_n)_{n\geq 1}\subseteq\mathcal{A}$ as a sequence of sets from $\mathcal{A}$, then $\bigcup_{n\geq 1}A_n\in\mathcal{A}$.

To show 2. you need to show that

• $P(\varnothing)=0$,
• $P(\bigcup_{n\geq 1} A_n)=\sum_{n\geq 1}P(A_n)$ for any disjoint sequence $(A_n)_{n\geq 1}$ of sets belonging to $\mathcal{A}$.

Many of these properties are straightforward to show. For the properties involving a sequence $(A_n)_{n\geq 1}$ you might want to divide it into the following two cases:

• All of the $A_n$'s are countable

• $A_n^c$ is countable for at least one $n$.