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

where:
 * $$\Pr \left({A | B}\right) = \frac {\Pr \left({A \cap B}\right)}{\Pr \left({B}\right)}$$

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

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

Proof
To prove this assertion we need to show 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:

$$ $$ $$ $$