Proving $P(A) = P(A\cap B) + P(A\cap\overline{\text{B}})$ I am trying to prove that if A and B are events, $P(A) = P(A\cap B) + P(A\cap\overline{\text{B}})$.
To start, I'm using the identities $A = A\cap S$ and $S = B\cup \overline{\text{B}}$.
$$A = A\cap S\\
A = A\cap (B\cup \overline{\text{B}})\\
A = (A\cap B)\cup(A\cap\overline{\text{B}})$$
From here, I want to go to $P(A) = P(A\cap B) + P(A\cap\overline{\text{B}})$. In my book it says to refer to the following axioms:
$$1: P(A) \geq 0$$
$$2: P(S) = 1$$
$$3: P(A_1\cup A_2 ... \cup A_n) = \sum_{i=1}^\infty P(A_i)$$
(where S is a sample space associated with an experiment, and $A_1, A_2, ... A_n$ form a sequence of pairwise mutually exclusive events in S).
It seems simple, but I'm not sure how to move from sets to probability.
 A: More generally, given a partition $\mathcal{P} = \{P_{1},P_{2},\ldots,P_{n}\}$ of the sample space $\Omega$, one has that
\begin{align*}
\mathbb{P}(A) & = \mathbb{P}(A\cap \Omega)\\\\\
& = \mathbb{P}\left[A\cap\left(\bigcup_{k=1}^{n}P_{k}\right)\right]\\\\
& = \mathbb{P}\left[\bigcup_{k=1}^{n}(A\cap P_{k})\right]\\\\
& = \sum_{k=1}^{n}\mathbb{P}(A\cap P_{k})
\end{align*}
Such result is a particular application of the finite aditivity property of the probability measure.
At your case, $\mathcal{P} = \{B,B^{c}\}$.
Hopefully this helps!
A: With mfl's hint you can use 3. with mutually exclusive sets $A\cap B, A\cap \overline{B}$ and $\overline{A}$ and 2. that $P(S)=1$ if $S = A\cup \overline{A}$ is the whole space. That yields
$$
1 = P(\overline{A} \cup A) = P(\overline{A}\cup (A\cap B) \cup (A\cap \overline{B})) = P(\overline{A}) + P(A\cap B) + P(A\cap\overline{B})
$$
Putting the fact that $P(A) = 1 - P(\overline{A})$ (can be directly derived from property 2) into above formula gives the wanted result
$$P(A\cap B) + P(A\cap \overline{B}) = 1 - P(\overline{A}) = P(A)$$
