Decomposition of Unmeasurable Sets into Measurable and Purely Unmeasurable Sets Below is a proof I wrote up that unmeasurable sets can be written as a union of a measurable set and a "purely unmeasurable" set. Is this result well-known (it is not very advanced, obviously), and does it hold more generally than given here?
Let $\mu^*$ be an outer measure on a set $X$.
If $\mathcal{M} \subset \mathcal{P}(X)$ is the set of $\mu^*$-measurable subsets of
$X,$ then $(X,\mathcal{M},\mu)$ is a measure space, where $\mu = \mu^*|_{\mathcal{M}}$.
We say $\mu^*$ is $\sigma$-finite if $(X,\mathcal{M}, \mu)$ is a $\sigma$-finite
measure space,
i.e. $X = \cup_{j=1}^\infty X_j,$ where $\forall j \in \mathbb{Z}_{>0}$, $X_j \in \mathcal{M}$
and $\mu^*(X_j) < \infty.$
Definition 1.  A subset $Y \subset X$ is called purely unmeasurable
if $Y$ is not measurable, and
for every $Z \subset Y,$
\begin{equation*}
Z\ \text{measurable} \ \implies \  \mu^*(Z) = 0.
\end{equation*}
Proposition 1. Let $\mu^*$ be a $\sigma$-finite outer measure on $X$.
If $Y \subset X$ is not measurable, then $Y$ can be written
$Y = A \cup B,$ where $A$ is purely unmeasurable, $B$ is measurable,
and $A \cap B = \emptyset.$ \
\textit{Proof.} First, suppose $\mu^*$ is actually finite. We give an algorithm for forming $A$ and $B.$
This algorithm can, in principle, be applied whether or not $Y$ is not measurable.
Initialize $A= Y,$ $B = \emptyset.$
For $\epsilon>0,$ define
$$\mathcal{A}_\epsilon = \mathcal{M} \cap \mathcal{P}(A) \cap (\mu^*)^{-1}([\epsilon, \infty]) = \mu^{-1}([\epsilon, \infty]) \cap \mathcal{P}(A).$$
Then the algorithm proceeds as follows.
For $n = 1, 2, 3, ...$,
until $\mathcal{A}_{1/n} = \emptyset,$
choose $E \in \mathcal{A}_{1/n}$ and re-define $A := A \setminus E,$
$B = B \cup E.$
The $n$th such ``until" loop lasts at-most $\lfloor{n\mu^*(X)}\rfloor$ steps, since $\mu^*(B)$ increases by
at least $1/n$
with each iteration; the algorithm therefore calls for at most countably-many iterations.
Any $B$ which results from this process is a countable union of measurable sets, and is hence measurable.
Since $A = Y \setminus B,$ and $B$ is measurable, if $Y$ is not measurable, then $A$ is not measurable.
Let $E$ be a measurable subset of $A$ (at the end of the algorithm). 
If $\mu^*(E)>0,$ then there is an $N \in \mathbb{Z}_{>0}$ so large that $\mu^*(E) > 1/N.$
But then, at the start of the $N$th iteration of the ``for" loop, we had $E \in \mathcal{A}_{1/N}.$
The algorithm could not proceed to the $(N+1)$th iteration, and thus could not have been completed,
without the removal of all or part of $E$ from $A.$ In view of this contradiction,
we see that there are two possibilities: if $Y$ was measurable to begin with, then $\mu^*(A) = 0.$
If $Y$ is unmeasurable, then $A$ is \emph{purely} unmeasurable.
Now assume that $\mu^*$ is $\sigma$-finite,
and write $X = \cup_{j=1}^\infty X_j,$ 
where $\mu^*(X_j) < \infty$, and for $i \neq j, X_i \cap X_j = \emptyset.$
For $n = 1,2,3,...$, let $X_n \cap Y = A_n \cup B_n$
be a decomposition given by the above algorithm. Note that,
even though $Y$ is not measurable, it may happen that $X_n \cap Y$ is measurable.
This is harmless: in the case that $X_n \cap Y$ is measurable,
$A_n$ is a measurable set with
$\mu^*(A_n) = 0.$
We define $B = \cup_{n=1}^\infty B_n$ and $A = Y \setminus B.$
Since $B$ is a countable union of measurable sets, it's measurable;
since $Y$ is not measurable and $A = Y \setminus B,$
$A$ is also not measurable.
Let $E$ be a measurable subset of $A.$ Since $E \cap X_n$ is a measurable subset of $A_n,$
$\mu^*(E \cap X_n) = 0.$ Thus by subadditivity, $\mu^*(E) \leq \sum_{n=1}^\infty \mu^*(E \cap X_n) = 0.$
Hence $A$ is purely unmeasurable.
Edit:
Does anyone know whether this result holds for outer measures which are not sigma-finite (in the sense that the induced measure is not sigma finite)?
 A: Your argument is slightly unclear to me, but I suspect is basically similar to the following:
For $Y\subseteq X$, let $m(Y)=\sup\{\mu(Z): Z\subseteq Y, Z\in\mathcal{M}\}$. The key point is that this supremum is really a maximum: if we pick some sequence $Z_i$ ($i\in\mathbb{N}$) such that $Z_i\subseteq Y$, $Z_i\in\mathcal{M}$, and $\sup\{\mu(Z_i): i\in\mathbb{N}\}=m(Y),$ then considering the set $W=\bigcup_{i\in\mathbb{N}}Z_i$ we have $W\in\mathcal{M}$, $W\subseteq Y$, and $\mu(W)=m(Y)$.
So let $W$ be a measurable subset of $Y$ with maximal measure as above. I now claim that $Y\setminus W$ does not have a measurable non-null subset. For if $U\subseteq Y\setminus W$ were measurable and non-null, then $W\cup U$ would be measurable and we would have $$\mu(W\cup U)=\mu(W)+\mu(U)>\mu(W)=m(Y),$$ contradicting the fact that $\mu(W)=m(Y)$.
I do not know a reference for this; I recall having it as an exercise in an analysis class, however.
A: Let $\mathcal B$ be a maximal collection of pairwise disjoint positive-measure measurable subsets of $Y$. It easily follows from $\sigma$-finiteness that $\mathcal B$ is countable. Hence $B=\bigcup\mathcal B$ is a measurable subset of $Y$. It is obvious from the maximality of $\mathcal B$ that the set $A=Y\setminus B$ is "purely unmeasurable".
P.S. In the case of Lebesgue measure, we can take the elements of $\mathcal B$ to be closed sets of positive measure, so that the measurable set $B$ is in fact an $F_\sigma$-set.
