Is this partial converse of the excision property for outer measure true? From here on out let us assume $A \subsetneq B$ and that $B$ is measurable.
The excision property states that if $A$ is measurable and of finite measure, then $$m^*(B-A)=m^*(B)-m^*(A)\,\,\,\,(*)$$
I am interested in the converse (which I approach by contraposition). 
Question: If $A$ is not measurable, does $(*)$ fail for all measurable $B$ or just for some measurable $B$? 
The excision property relies on the fact that $m^*(B)=m^*(B\cap A) + m^*(B-A)$ for any set $B$ when $A$ is measurable. I know it must fail for some $B$ or else $A$ would be measurable, but I would like to conclude that $(*)$ fails for all $B$ when $A$ is a nonmeasurable subset of $B$.
What have I tried? Well I don't have any concrete examples of nonmeasurable sets in the sense that I don't know the outer measure of any nonmeasurable set. All I know is that they exists and how to make them, but since I rely on the axiom of choice to make such a set, I can't seem to do anything with it.
 A: In the following, I assume that your outer measure is induced by a premeasure given on a semiring $H$ as in this post Inequality with outer measure. With the usual definition of Lebesgue measure, this is the case (where $H$ usually is the family of all "boxes"). 
Also note that your identity trivially holds if $B$ is of infinite measure (why?). 
We will now see that your identity fails for all measurable $B$ of finite measure. Hence your claim is true in all interesting cases. 
For the proof note that the measurability of all elements of $H$ together with the definition of the associated outer measure implies that for each $C$, there is a "measurable envelope" $C^\ast$, i.e. a measurable set $C^\ast \supset C$ with $\mu^\ast (C^\ast)=\mu^\ast (C)$ (why exactly?). (I will write $\mu$ instead of $m$). 
Let $C = (B\setminus A)^\ast$ and note that we can assume $C\subset B$. By elementary considerations, we see
$$
C^c \subset (B\cap A^c)^c = B^c \cup A
$$
and hence
$$
B\setminus C \subset B \cap (B^c \cup A) = A \subset A^\ast. 
$$
But by your identity $(\ast)$ yields
$$
\mu^\ast (A^\ast)=\mu^\ast (A) =\mu^\ast (B)-\mu^\ast (B\setminus A) =\mu^\ast (B)-\mu^\ast (C) =\mu^\ast (B\setminus C)\leq \mu^\ast (A)=\mu^\ast (A^\ast). 
$$
This implies that $A^\ast \setminus (B\setminus C)$ is a null-set (hence measurable), so that also $A \setminus (B\setminus C)$ is a null-set (as a subset of the set mentioned before) and hence measurable. 
Finally,
$$
A = (B\setminus C)\cup [A\setminus (B \setminus C)]
$$
is measurable, a contradiction. (Recall from above that $B \setminus C \subset A$, so that the last equation is indeed true). 
