TITLE: Measure of non-measurable set is zero
QUESTION [6 upvotes]: I am working on a problem from Richard Bass' analysis, which to a slowpoke is just a jumbling of unrelated terms:

Suppose that set $N$ is non-measurable as defined in Section 4.4. Show that $m(A) = 0$ if $A \subset N$, where $A$ is Lebesgue measurable. 
  (Section 4.4 has this theorem: Let $m^*$ be defined by the usual outer measure definition, where $\mathcal C$ is the collection of intervals that are open on the left and closed on the right and $\mathscr l ((a, b]) = b - a$. Then $m^*$ is not a measure on the collection of all subsets of $\mathbb R$.)

Here are what I managed to salvage from the mess before my mind goes on strike: 
(1) Since $N$ is non-measurable per Section 4.4, which says that all subsets of $\mathbb R$ is non-measurable, then $N \subset \mathbb R.$ 
(2) Since $A \subset N$, isn't it $A \subset \mathbb R$ as well? 
(3) Since $A$ is Lebesgue measurable, then 
$$\forall B \subseteq \mathbb R, \ m(B) = m(B \cap A) + m(B \cap A^c).$$
Please let me know how and where I should go from here. Thank you for your time and effort.

REPLY [8 votes]: The statement is for a particular $N$ constructed in that book. 
The properties we need to use from $N$ is that its rational translations $N+r$ are disjoint and that $\bigcup_{r\in\mathbb{Q}\cap[0,1]}(N+r)\subset[-1,2]$.
Assume that $A\subset N$ is measurable. 
This implies that $A+r$ are disjoint for $r\in\mathbb{Q}\cap[0,1]$. So 
$$3\geq m([-1,2])\geq m(\bigcup_{r\in\mathbb{Q}\cap[0,1]}(A+r))=\sum_{r\in\mathbb{Q}\cap[0,1]}m(A+r)=\sum_{r\in\mathbb{Q}\cap[0,1]}m(A)$$
This can only happen if $m(A)=0$.

Notice this is the same as half of the proof of theorem 4.15. The only difference is that since $A$ is not all of $N$ we don't have the inclusion $\bigcup_{r\in\mathbb{Q}\cap[0,1]}(A+r)\supset[0,1]$ and therefore there is no inequality $\geq1$ from below preventing $m(A)=0$.