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.
Best Answer
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 $$ \mu(A_1 \cup A_2) + \mu(A_1 \cup A_3) = \mu(A_1) + \mu(A_1 \cup A_2 \cup A_3) $$