I'm trying to prove the following statement and was wondering if what I have done is legit. Statement: If $A$ and $B$ are subsets of $\mathbb R$ with $|B|=0$, then $|A\cup B|=|A|$. My attempt for a proof:
Define intervals $I_k$ such that $A\subset\bigcup_{k=1}^\infty I_k$, and intervals $I^*_k$ such that $B\subset \bigcup_{k=1}^\infty I^*_k$, with $l(I^*_k)=\frac{\varepsilon}{2^k}$, $\varepsilon>0$, with the latter definition justified by the fact that $|B|=0$. Now $$\sum_{k=1}^\infty l(I_k\cup I^*_k)\leq \sum_{k=1}^\infty l(I_k)+\sum_{k=1}^\infty l(I^*_k)=\sum_{k=1}^\infty l(I_k)+\varepsilon,$$ and when taking the infimums of the sums, by the definition of the outer measure we get
$$|A\cup B|\leq \inf\left(\sum_{k=1}^\infty l(I_k)+\varepsilon\right)=\inf\left(\sum_{k=1}^\infty l(I_k)\right)+0=|A|.$$
On the other hand, because outer measure preserves order,
$$|A|\leq |A\cup B|\ \text{for}\ A\subset A\cup B,$$
and thus $|A\cup B|=|A|$.
What I'm myself not completely sure about is that it's ok to write $|A\cup B|$ as $\inf\left(\sum_{k=1}^\infty l(I_k\cup I^*_k)\right)$, but that of course may not be the only problem with the argument.
Best Answer
If $A$ is measurable as you state in the question then by the completeness of Lebesgue measure then $B$ is also measurable and then $A \cup B$ is measurable and the result is pretty obvious. Without the assumption that $A$ is measurable your proof is almost correct. You don't have to assume that $|A\cup B|$ is the infimum of $\sum_{k=1}^{\infty} l(I_k \cup l^*_k)$ though. If you consider the sets $$ A_1 = \left\lbrace \sum_{k=1}^{\infty}l(J_k): A \cup B \subseteq \bigcup J_k \right\rbrace$$ and $$ A_2 = \left\lbrace \sum_{k=1}^{\infty}l(I_k \cup I^*_k): A \subseteq \bigcup I_k, \;B \subseteq \bigcup I^*_k \right\rbrace$$ then it easy to notice than $A_2 \subseteq A_1$ and by the properties of infimum $$|A \cup B| = \inf A_1 \leq \inf A_2.$$ Then your proof works just fine.