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.
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 $m(A)=0$.