measure zero implies complement has full measure I'm doing a problem, where I've reduced the problem to seeing if this claim holds: suppose $R$ has lebesgue outer measure zero. Then $m(D\setminus R)=m(D)$.
For one direction, I just used monoticity of the integral, since $D\setminus R\subset D\implies m(D\setminus R)\leq m(D)$, but am having trouble proving the reverse inequality.
I feel like it's really obvious and I'm just not seeing it. -_-
 A: The intersection $D \cap R$ is contained in $R$, hence has (Lebesgue) measure zero.
Now, $D \setminus R = D \setminus (D \cap R)$, so 
$$
m(D \setminus R) = m(D \setminus (D \cap R) ) = m(D) - m(D \cap R) = m(D).
$$
P.S.  There is plenty of bad literature on measure theory, and the one and only book I would recommend is Vulikh's Brief Course in the Theory of Functions of a Real Variable.
A: Your question mentions Lebesgue outer measure, and you didn't state explicitly whether $m$ is assumed to be Lebesgue measure, or Lebesgue outer measure.
The answer given by avs assumes that $m$ is additive on disjoint sets, which is only guaranteed to be true if $m$ is Lebesgue measure (and therefore $D$ is implicitly assumed to be Lebesgue measurable).
But the result is in fact true if $m$ is Lebesgue outer measure and $D$ is an arbitrary subset of $\mathbb R$ (or $\mathbb R^d$).
Proof:
As you noted, the inequality
$$m(D \setminus R) \leq M(D)$$
follows immediately from monotonicity since $D \setminus R \subset D$.
For the opposite inequality, first note that $D = (D \setminus R)\cup(D \cap R)$. Then by subadditivity,
$$m(D) \leq m(D \setminus R) + m(D \cap R).$$
Now observe that $D \cap R \subset R$, so by monotonicity,
$$m(D \cap R) \leq m(R) = 0.$$
Putting these inequalities together gives us
$$m(D) \leq m(D \setminus R) + m(D \cap R) \leq m(D \setminus R) + m(R) \leq m(D \setminus R)$$
as desired.
