Total Probability Theorem/Conditional Probabilities

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.

Let $C \in \Sigma$ be an event independent to any of the $B_i$.

Then:
 * $\displaystyle \forall A \in \Sigma: \map \Pr {A \mid C} = \sum_i \map \Pr {A \mid C \cap B_i} \, \map \Pr {B_i}$

Proof
First define $Q_C := \map \Pr {\, \cdot \mid C}$.

Then, from Conditional Probability Defines Probability Space, $\struct {\Omega, \Sigma, Q_C}$ is a probability space.

Therefore the Total Probability Theorem also holds true.

Hence we have: