Every non-measurable $X \subseteq \mathbb{R}^n$ has non-measurable $Y \subseteq X$ such that $|Z|=0$ for every measurable $Z \subseteq Y$ I'm trying to argue that a Lebesgue non-measurable subset $X$ of $\mathbb{R}^n$ has a non-measurable subset $Y$ such that any measurable subset of $Y$ has measure zero.
Let $A_0$ be a measurable subset of $X$. If all such $A_0$ has measure zero, then let $Y = X$ and we are done. Otherwise, choose $A_0$ which has positive measure. By Vitali construction, there exists a non-measurable subset $A_1 \subset A_0$. Again, if $A_1$ doesn't have a measurable subset of positive measure, then set $Y = A_1$. Otherwise, we repeat what we did as above.
How does one rigorously conclude that the process will have to eventually terminate?
 A: The following turns my comments above into an answer (modulo exercises :P).
Throughout, I'll assume that $X$ is a non-measurable set which has a positive-measure measurable subset.
The process in the OP is too loosely-defined to be guaranteed to terminate - maybe we just happen to keep grabbing "inefficient" subsets of our starting set. Instead, it's better to try to carve out a "maximal(ish) measurable subset" (and think about what's left over).
Note my careful use of the suffix "ish" in the above. There is not going to be a literally maximal measurable subset of $X$: if $A\subseteq X$ is measurable and $a\in X\setminus A$, then $A\cup\{a\}$ is also a measurable subset of $X$. Similarly, we can't do something like taking the union of all the positive-measure subsets of $X$: that will just yield $X$ itself (again thinking about how singletons don't affect measurability), which isn't measurable. This second obstacle brings up a related issue, namely that measurability is fragile: if I take a union of a bunch of measurable sets, the result may not be measurable. Only countable unions are safe: the union of countably many measurable sets is always guaranteed to be measurable.
So we want to find a measurable subset of $X$ which in some sense is as big as possible, but $(i)$ that sense has to be somewhat nuanced and $(ii)$ we have to only take a small number (= countably many) of unions of measurable pieces. What can we do?
Well, besides literal maximality, we do have at hand a way of measuring when something is as big as possible: measure! Say that $A\subseteq X$ is a quasimaximal-measurable subset of $X$ (not an actual term) iff $A$ is measurable and, for every measurable $B$ with $A\subseteq B\subseteq X$, we have $m(A)=m(B)$ (or equivalently, if $B$ is any measurable subset of $X$ we have $m(A)\ge m(B)$). So maybe $A$ isn't literally as big as possible, but we can't do any better measure-wise.
Using basic facts about measurability, you should be able to prove the following statements:

*

*If $A$ is a quasimaximal-measurable subset of $X$, then $X\setminus A$ is non-null.


*If $A$ is a quasimaximal-measurable subset of $X$, then $X\setminus A$ does not have a positive-measure measurable subset.
At this point you're basically done ... except that you need to prove that quasimaximal-measurable subsets exist at all! HINT: consider the number $$\alpha=\sup\{m(A): A\subseteq X\mbox{ is measurable}\}$$ (for simplicity let's assume $\alpha<\infty$). Can you see how to build a measurable subset of $X$ with measure exactly $\alpha$ (think about countable unions!), and why such a subset must be quasimaximal-measurable?
