Showing $m^*(A-E)=m^*(A)$ when $m^*(E)=0$ 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!
 A: 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
$$m^\star(A\cup B) \le m^{\star}(A) + m^{\star}(B).$$
