measure zero implies complement has full measure

Question

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. -_-

Answer 1

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.

Answer 2

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.