Conditional Probability Defines Probability Space

Theorem
Let $\left({\Omega, \Sigma, \Pr}\right)$ be a measure space.

Let $B \in \Sigma$ such that $\Pr \left({B}\right) > 0$.

Let $Q: \Sigma \to \R$ be the real-valued function defined as:


 * $Q \left({A}\right) = \Pr \left({A \mid B}\right)$

where:


 * $\Pr \left({A \mid B}\right) = \dfrac {\Pr \left({A \cap B}\right)}{\Pr \left({B}\right)}$

is the conditional probability of $A$ given $B$.

Then $\left({\Omega, \Sigma, Q}\right)$ is a probability space.

Proof
It is to be shown that $Q$ is a probability measure on $\left({\Omega, \Sigma}\right)$.

As $\Pr$ is a measure, we have that:


 * $\forall A \in \Omega: Q \left({A}\right) \ge 0$

Also, we have that:

Now, suppose that $A_1, A_2, \ldots$ are disjoint events in $\Sigma$.

Then: