This is embarrassing, but I am unable to prove that $P(A^c \cap B) = P(B) - P(A \cap B)$. Any pointers?
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What you want to show is equivalent to:
$$P(A^c \cap B) + P(A \cap B) = P(B)$$
Note by definition $(A^c \cap A)=\emptyset \Rightarrow (A^c \cap B) \cap (A \cap B) = \emptyset$
Besides you know that $(A^c \cap B) \cup (A \cap B)=B$
Therefore the statement follows by using the $\sigma$-additivity of $P$, namely
$$(X \cap Y)=\emptyset \Rightarrow P(X \cup Y)=P(X)+P(Y)$$
I let you write out the last step.