$|ยท|$ is outer measure.
I only need to prove $|A\cup B|\geq|A|+|B|$.
I've proved that if $B$ is open, then $|A\cup B|=|A|+|B|$.
Since $B$ is Lebesuge measurable, there exists an open set $O\supset B$ such that $|O-B|<\epsilon$.
My initial idea was to approximate $B$ with $O$,then
$$|A\cup B|\geq |A\cup O|-\epsilon=|A|+|O|-\epsilon$$
But I've found that $O$ doesn't have to be disjoint with $A$. So it doesn't feel like it's gonna work. In summary, if I want to find an open set to cover $B$, there is no guarantee that this open set will not intersect $A$. I'm running out of other ideas, can anyone help me? Thanks in advance!
Best Answer
Proof 1: Suppose $A$ and $B$ are disjoint subsets of $\mathbb{R}$ and $B$ is Lebesgue measurable. Since $B$ is Lebesgue measurable, we have, for any subset $E$ of $\mathbb{R}$, $$ |E| = |E \cap B| + |E \cap B^c| $$ Take $E = A \cup B$. Since $A$ and $B$ are disjoint, we have $E \cap B =B$ and $E \cap B^c = A$. So we have $$ |A \cup B| = |B| + |A| $$ that is $|A\cup B|=|A|+|B|$.
$\square$
Since the OP would like a proof using GTM282 by Sheldon Axler (pages 50,51,52). Here it is
Proof 2: We know that
Now, suppose $A$ and $B$ are disjoint subsets of $\mathbb{R}$ and $B$ is Lebesgue measurable. Then, by the definition of Lebesgue measurable set, there is a Borel set $C \subseteq B$ such that $|B \setminus C| = 0$. Using the fact the $B \setminus C$ and $C$ are disjoint and $C$ is a Borel set, we can apply 2.66 and we have $$ |B| = |(B \setminus C) \cup C| = |B \setminus C| +|C| =|C| $$
Now, since $C \subseteq B$, we have that $A$ and $C$ are also disjoint. Using the monotonicity of outer measure, the result 2.66 again and the fact that $|B| =|C|$, we have $$ |A \cup B| \geqslant |A \cup C| = |A| +|C| = |A|+|B| $$ So $|A \cup B| \geqslant |A|+|B|$. It follows immediately that $$ |A \cup B| = |A|+|B| $$
$\square$