If $A,B\subset \mathbb{R}$ and $B$ has outer measure 0, show that $|A\cup B|=|A|$.

Question

If $A,B\subset \mathbb{R}$ and $B$ has outer measure $0$, show that $|A\cup B|=|A|$.

My attempt:

Let $\left\{I_{k}\right\}_{k=1}^{\infty}$ be an open cover for $A$, and let $\left\{J_{k}\right\}_{k=1}^{\infty}$ be an open cover for $B$. It follows that $\left\{I_{k}\cup J_{k}\right\}_{k=1}^{\infty}$ is an open cover for $A\cup B$.

Then by definition, $|A\cup B|=\inf\left\{\ell(I_{k}\cup J_{k}: I_{k}\cup J_{k}\in \left\{I_{k}\cup J_{k}\right\}_{k=1}^{\infty} \right\})$, where $\ell$ is the length of the interval.

I want to show this equals to $|A|$, but I am not sure how to go from here. My guess is that since $|B|=0$, the length of $I_{k}\cup J_{k}$ will be the same as the length of $I_{k}$, since the lengths of $J_{k}$ cause no effect. Any help appreciated.

Answer 1

$A \subset A \cup B$ and so $|A| \leq |A \cup B| \leq |A| + |B| = |A|$

where the first inequality is by monotonicity and the second is by subadditivity

it's been a minute since measure theory but I think this works

Answer 2

Since $A \subset A \cup B$, and using that outer measure preserves order, we have that $|A| \leq |A \cup B|$. By the countable subadditivity of outer measure, $|A \cup B| \leq |A| + |B| = |A| + 0 = |A|$. Thus $|A| \leq |A \cup B|$ and $|A \cup B| \leq |A|$ together imply that $|A \cup B| = |A|$.