I'm trying to show that $m^*(A-E)=m^*(A)$ when $m^*(E)=0$ where $m^*$ is the outer measure.
So my plan is to show that
$m^*(A-E) \geq m^*(A)$
and
$m^*(A-E) \leq m^*(A)$ (and this one is easy, because $A-E$ $\subset$ $A$)
I'm having a harder time showing $m^*(A-E) \geq m^*(A)$ . My strategy is to let $\epsilon>0$ and then to show that $m^*(A-\cup K)$ $> \epsilon + m^*(A)$ where $E \subset \cup K$. I might be on the wrong track here though. This measure theory stuff is hard, can any of you nerd genius's shed some light on this for me? Thanks!
As suggested by the comment, use the subadditivity of the outer measure to finish the question.
$$m^{\star}(A \setminus E) = m^{\star}(A \setminus E) +m^{\star}(E)\ge m^{\star}(A) \implies m^{\star}(A \setminus E) = m^{\star}(A)$$
To prove the subadditivity of the outer measure, let $\epsilon > 0$, and $A, B$ be two subsets of $\Bbb{R}$ having countable open interval covers $(I_i)_{i=1}^{\infty}$ and $(J_j)_{j=1}^{\infty}$ respectively so that $\sum_{i=1}^{\infty} \ell(I_i) < m^{\star}(A) + \frac\epsilon2$ and $\sum_{j=1}^{\infty} \ell(J_j) < m^{\star}(B) + \frac\epsilon2$. (Such covers exist since $m^\star$ is the infimum of the sum of length of covering open intervals.)
Clearly, $\{(I_i)_{i=1}^{\infty},(J_j)_{j=1}^{\infty}\}$ covers $A \cup B$, and the sum of length of covering open intervals
$$\sum_{i=1}^{\infty} \ell(I_i) + \sum_{j=1}^{\infty} \ell(J_j) < m^{\star}(A) + m^{\star}(B) + \epsilon$$
To finish the proof, on the LHS, take infimum over all countable open interval covers for $A \cup B$ so that $m^\star(A\cup B)$ appears on the LHS.
$$m^\star(A\cup B) < m^{\star}(A) + m^{\star}(B) + \epsilon$$
Since the choice of $\epsilon > 0$ is arbitrary, we conclude that
$$m^\star(A\cup B) \le m^{\star}(A) + m^{\star}(B).$$