Conditional Probability Defines Probability Space

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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:

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

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

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:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle Q \left({\Omega}\right)$	$=$	$\displaystyle \Pr \left({\Omega \mid B}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {\Pr \left({\Omega \cap B}\right)}{\Pr \left({B}\right)}$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {\Pr \left({B}\right)}{\Pr \left({B}\right)}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Intersection with Universe
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $	as $\Pr \left({B}\right) > 0$

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

Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle Q \left({\bigcup_i A_i}\right)$	$=$	$\displaystyle \frac 1 {\Pr \left({B}\right)} \Pr \left({\left({\bigcup_i A_i}\right) \cap B}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac 1 {\Pr \left({B}\right)} \Pr \left({\bigcup_i A_i}\right) \left({A_i \cap B}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Intersection Distributes over Union
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {\Pr \left({B}\right)}{\sum \Pr \left({A_i \cap B}\right)}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	as $\Pr$ satisfies the Kolmogorov axioms
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \sum Q \left({A_i}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by definition.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle Q \left({\Omega}\right)\)	\(=\)	\(\displaystyle \Pr \left({\Omega \mid B}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\Pr \left({\Omega \cap B}\right)}{\Pr \left({B}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\Pr \left({B}\right)}{\Pr \left({B}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Intersection with Universe
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\Pr \left({B}\right) > 0$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle Q \left({\bigcup_i A_i}\right)\)	\(=\)	\(\displaystyle \frac 1 {\Pr \left({B}\right)} \Pr \left({\left({\bigcup_i A_i}\right) \cap B}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {\Pr \left({B}\right)} \Pr \left({\bigcup_i A_i}\right) \left({A_i \cap B}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Intersection Distributes over Union
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\Pr \left({B}\right)}{\sum \Pr \left({A_i \cap B}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\Pr$ satisfies the Kolmogorov axioms
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum Q \left({A_i}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition.