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)?