Proof concerning outer measure Assume that $X:=[a,b]$ is a fixed interval in $\mathbb R$ and let $m^*$ be the outer measure on $X$. Suppose that $A \subset X$ is a null set, i.e. $m^*(A)=0.$ Show that for every $B\subset X$, 

$m^*(B \cup A)=m^*(B\setminus A)=m^*(B)$

I'm studying measure theory and looking for some direction on this proof. Thanks!
Take my hint from above if you would like just that. Below is a more complete solution.

$m^*(B\setminus A)\leq m^*(B)$ by monotonicity. $m^*(B)\leq m^*(B\cup A)$ by the same. $m^*(B\cup A)\leq m^*(B\setminus A)+m^*(A)$ by subadditivity.  Hence:
$$
m^*(B\setminus A)\leq m^*(B)\leq m^*(B\cup A)\leq m^*(B\setminus A)+m^*(A)=m^*(B\setminus A).
$$
So all the inequalities collapse into equalities.
$B \cup A = (B \setminus A) \cup A $, that is, it is the union of two disjoint sets so
$$m^*(B \cup A) = m^* (B \setminus A) + m^*(A)= m^*(B \setminus A). $$

Now since $A$ is a null set, $ B \setminus A= B$, hence $m^* (B \cup A)= m^*(B)$.