Sum of Measures of Two Subsets = Sum of Measures of Their Union & Intersection I have this solution to a problem in measure theory, I am posting it here to check my basic understanding:

Let $(X, \mathscr A, \mu)$ be measure space with $A, B \in \mathscr A$, show that
  $$\mu(A) + \mu(B) = \mu(A \cup B) + \mu ( A \cap B).$$

I would like to solve this problem by presenting three cases: 
(1) If $A$ and $B$ are disjoint, $A \cap B = \emptyset$:
$$\begin{align}
A + B &= (A \cup B) \tag{a}\\
\therefore \ \mu(A) + \mu(B) &= \mu (A \cup B) + 0 \tag{b}\\
&= \mu (A \cup B) + \mu (\emptyset) \tag{c}\\
&= \mu (A \cup B) + \mu(A \cap B) \tag{d}\\
\end{align}$$
(2) If $A$ is proper subset of $B$, $A \subset B$:
$$\begin{align}
B &= A \cup B \tag{a}\\
\therefore \mu(B) &= \mu (A \cup B)\tag{b}\\
A &= A \cap B \tag{c}\\
\therefore \mu(A) &= \mu (A \cap B)\tag{d}\\
\therefore \   \mu(A) + \mu(B)&= \mu (A \cup B) + \mu(A \cap B)\tag{e}\\
\end{align}$$
(3) If A intersects B, $A \cap B \neq \emptyset$,
$$\begin{align}
A \cup B &= (A + B) - (A \cap B)\tag{a}\\
\therefore \ \mu(A \cup B) &= \mu(A) + \mu(B) - \mu (A \cap B)\tag{b}\\
\mu(A \cup B) + \mu (A \cap B)&= \mu(A) + \mu(B) \tag{c}\\
\mu(A) + \mu(B) &= \mu(A \cup B) + \mu (A \cap B) \tag{d}\\
\end{align}$$
Here is my question: Am I making invalid conclusions by jumping from (1a) to (1b), from (2a) to (2b), from (2c) to (2d), etc.? Do let me know if you have a more elegant solution.
Thank you for your time and effort.
 A: (1) is flawed because $+$ is not defined on sets (at least in terms of this problem you're only using: union, intersection and complement (noting that set-difference can be written in terms of complements and intersections)). However (1b)- (1c) are true and follow from (1b) which is true by definition of measure and the fact that the sets are disjoint.
Jump (2a) to (2b): If $A$ is a proper subset of $B$, then $A = A \cup B$ is not true to begin with, because it's proper to begin with, which means there exists $b \in B$ such that $b \notin A$. If that were true however, the step from (2a) to (2b) would work, because the sets are equal, i.e., if $A = B$, then $\mu(A) = \mu(B)$ is true. Same reasoning holds from (2c) to (2d).
(3) is flawed for the same reason (1) is flawed. Once you have (3b), you can't necessarily go to (3c) if some of these measures are infinite, in particular $\mu(A \cap B)$. (3c) to (3d) is definitely true.

Here is an alternate approach:
Proof: Observe that $A = (A \setminus B) \cup (A \cap B)$ and thus $$\mu(A) = \mu\big(  (A \setminus B) \cup (A \cap B) \big) = \mu(A \setminus B) + \mu(A \cap B).$$ Observe that $B = (B \setminus A) \cup (A \cap B)$ and thus $$\mu(B) = \mu \big( (B \setminus A) \cup (A \cap B) \big) = \mu(B \setminus A) + \mu(A \cap B).$$ Now notice that
            \begin{align*}
    \mu(A) + \mu(B) &= \mu(A \setminus B) + \mu(A \cap B) + \mu(B \setminus A) + \mu(A \cap B) \\
    &= \mu(A \cap B) + \mu \big( (A \setminus B) \cup (B \setminus A) \cup (A \cap B) \big) \\
    &= \mu(A \cap B) + \mu \big( (A \cup B) \setminus (A \cap B) \cup (A \cap B) \big) \\
    &= \mu(A \cap B) + \mu(A \cup B).
   \end{align*}
A: A more elegant solution:
Define
$$
A_1 = A \cap B\\
A_2 = A \setminus A_1\\
A_3 = B \setminus A_1
$$
Note that by the definition of these set operations, the $A_i$ are disjoint.  By the general properties of a measure, the $A_i$ are measurable.
What you are attempting to show, then, is that
$$
\mu(A_1 \cup A_2) + \mu(A_1 \cup A_3) = \mu(A_1) + \mu(A_1 \cup A_2 \cup A_3)
$$
