The outer measure is defined as
$|A|=\inf(\{\sum\limits_{i=1}^{\infty}l(I_i)\text{ with $I_1$,$I_2$,... s.t:} \hspace{0,2cm}A\subset\bigcup\limits_{i=1}^{\infty}I_i\}$
where $l$ is the length of an interval in the intuitive sense.
So I have to show that
"$A,B\subset\mathbb{R}$ and $|A|<\infty$, then $|B\setminus A|\geq|B|-|A|$".
Where $|\cdot|$ is the outer measure of a set. I'm struggling to find a way to manipulate the infimum of the open cover of $|B\setminus A|$ in order to show the inequality.
Let $(I_i)$ be a sequence of intervals covering $A$ and $(J_l)$ be a sequence of intervals covering $B \setminus A$. Then $(I_i)\cup (J_l)$ covers $B$. Hence $|B| \leq \sum_i l(I_i)+\sum_l l(J_l)$. Taking infimum over all covers $(I_i)$ and $(J_l)$ we get $|B| \leq |A|+|B \setminus A|$. Since $|A|< \infty$ we can subtract $|A|$ from both sides.