Probability Triples — Alex Beaudin

Notes from Rosenthal Chapter II

A probability triple is just a measure space where $\mu(\mathcal{X}) = 1$ .

Looking at this, it’s not at all obvious whether or not a probability triple exists or how to construct one for a given $\Omega$ . However, we’ll unpack the definition and get some insights on constructing triples when possible or proving their existence. In particular, we can always define a triple for a finite or countable set.

An important characteristic of finite sets is the trivial existence of a $\sigma$ -algebra $\mathcal{F}$ . How do we construct valid $\sigma$ -algebras when we don’t have ‘discrete’ sets?

Ok, well a semialgebra might be quite easy to define. Can we generate a $\sigma$ -algebra from a semialgebra? Consider the first attempt

\htmlId{eq-1}{\mathcal{B}_0 = \{ \text{all finite unions of elements of } \mathcal{J}\}.} \tag{1}

Unfortunately, $\mathcal{B}_0$ is not a $\sigma$ -algebra.

There is a further problem that of closure under complements, which means that extending $\mathcal{B}_0$ to include countable unions would not make it a $\sigma$ -algebra. In particular, the complement of the set we described has to be in any $\sigma$ -algebra containing $\mathcal{J}$ , which is the set of irrationals in $[0, 1]$ . No countable union of intervals can produce this set.

Extension Theorems

Ok, so then how do we get a $\sigma$ -algebra? We could just take the closure of the semialgebra with respect to the operations necessary for a $\sigma$ -algebra. However, we’d often run into a problem with our measure not being sub-additive. In the example above, we’d reintroduce the sets which cause the measure not to be sub-additive!

theorem.The Extension Theorem

Let $\mathcal{J}$ be a semialgebra of subsets of $\Omega$ . Let $\mathbf{P}: \mathcal{J} \to [0, 1]$ with $\mathbf{P}(\emptyset) = 0$ and $\mathbf{P}(\Omega) = 1$ , satisfying the super-additive property

\htmlId{eq-3}{\mathbf{P}\left( \bigcup_{i=1}^k A_i \right) \geq \sum_{i=1}^k \mathbf{P}(A_i) } \tag{3}

whenever $A_1, \dotsc A_k \in \mathcal{J}$ , and the A_i are disjoint, and the countable monotonicity property

\htmlId{eq-4}{\mathbf{P}(A) \leq \sum_{n} A_n } \tag{4}

for $A_n \in \mathcal{J}$ and where $A \subseteq \bigcup_n A_n$ . Then there is a $\sigma$ -algebra $\mathcal{M} \supseteq \mathcal{J}$ , and a countably additive probability measure $\mathbf{P}^*$ on $\mathcal{M}$ such that

\htmlId{eq-5}{\mathbf{P}^*(A) = \mathbf{P}(A) \text{ for all } A \in \mathcal{J}.} \tag{5}

In other words, $(\Omega, \mathcal{M}, \mathbf{P}^*)$ is a valid probability triple.