Probability Set Theory Question I'm working on the following problem but I'm having a hard time figuring out how to do it:
Q: Let A and B be two arbitrary events in a sample space S. Prove or provide a counterexample:
If $P(A^c) = P(B) - P(A \cap B)$ then $P(B) = 1$
Drawing Venn diagrams I can see how this is true, as $A \subset B$, but I'm not sure how to formally prove this. Any help would be great!
 A: Not true. If $A^c\subseteq B$, then $B-A\cap B=A^c$, so $P(A^c)=P(B)-P(A\cap B)$ for any $B$.  This holds since $A^c$ and $A\cap B$ are mutually exclusive and add up to $B$.
A: Let $P$ be uniform on $S=\{1,2,3\}$ and let $A=\{1,2\}, B=\{2,3\}$.
$P(A \cap B) = P \{2\} = {1 \over 3}$.
$P(B) = P \{2,3\} = {2 \over 3}$.
$P(A^c) = P \{3\} = {1 \over 3}$.
Hence the equation holds but $P(B) \neq 1$.
All that one can really conclude is that $P (A \cup B)^c = 0$.
A: 
Drawing Venn diagrams I can see how this is true, as $A \subset B$, but I'm not sure how to formally prove this. Any help would be great!

Nope.  (Unless you have a typo.)
$$\begin{align}\mathsf P(A^{\small\complement})&= \mathsf P(B)-\mathsf P(A\cap B) \\[1ex]&=\mathsf P(B)-\mathsf P(B)\,\mathsf P(A\mid B)\\[1ex]&=\mathsf P(B)\,\big(1-\mathsf P(A\mid B)\big)\\[1ex]&=\mathsf P(B)\,\mathsf P(A^{\small\complement}\mid B)\\[1ex]&=\mathsf P(A^{\small\complement}\cap B)\end{align}$$ Therefore the statement indicates that $A^{\small\complement}\cap B^{\small\complement}$ is a null set (which is not quite that $A^{\small\complement}\subseteq B$ ).
It would only entail that $\mathsf P(B)=1~~$ when $~~0<\mathsf P(A^{\small\complement})=\mathsf P(A^{\small\complement}\mid B)$, which requires … .
