Show: $M\subset\mathbb{R}^n$ Jordan-measurable, iff $vol^*(\partial A)=0$ 

Show that a bounded subset $A\subset\mathbb{R}^n$ is Jordan-measurable iff and only if $\partial A$ is a Jordan null set, i.e. $vol^*(\partial A)=0$.


Here Show some properties of the (inner/ outer) Jordan-measure I tried to prove that
$$
vol^*(A)=vol^*(\overline{A}),~~~~~vol_*(A)=vol_*(A^°).~~~~~~~(*)
$$
I think I need this here:
Here is my try of a proof:
"$\Rightarrow$": Assume that $A$ is bounded and Jordan-measurable, i.e. $vol^*(A)=vol_*(A)$. Because $A$ is bounded, I think $A^°$ is bounded, too. Therefore I think it is $vol^*(A^°)<\infty$, so that I can calculate using (*)
$$
vol^*(\partial A)=vol^*(\overline{A}\setminus A^°)=vol^*(\overline{A})-vol^*(A^°)=vol^*(A)-vol_*(A^°)=vol^*(A)-vol_*(A)=0.
$$
Do not know exactly if this proof is okay. Especially the step
$$
vol^*(\overline{A}\setminus A^°)=vol^*(\overline{A})-vol^*(A^°)
$$
is not totally clear to me to be honest.
"$\Leftarrow$".
I think I have to assume that $A$ is bounded (is that right or do I have to prove that?).
It always is $vol^*(A)\geqslant vol_*(A)$.
Additionally it is using (*)
$$
vol^*(A)=vol^*(\overline{A})=vol^*(A^°\cup\partial A)=vol^*(A^°)+vol^*(\partial A)=vol^*(A^°)\geqslant vol_*(A^°)=vol_*(A).
$$
So together it is $vol^*(A)=vol_*(A)$.

Would be great to get a feedback!
Thank you!
math12
 A: 1. Regarding your first question:

Do not know exactly if this proof is okay. Especially the step
  $$vol^*(\overline{A}\setminus A^°)=vol^*(\overline{A})-vol^*(A^°)$$ is
  not totally clear to me to be honest.

The proof is almost correct.
In your other question,
you prove that $vol^*(A)=vol^*(\overline{A})$, and $vol_*(A)=vol_*(A^°)$.
Note that this implies in particular that if $A$ is Jordan measurable,
then so are $\overline{A}$ and $A^\circ$.
Indeed,
this is clear from the following inequalities:
$$vol_*(\overline A)\geq vol_*(A^\circ)=vol_*(A)=vol^*(A)=vol^*(\overline A);\text{ and }$$
$$vol^*(A^\circ)\leq vol^*(\overline A)= vol^*(A)=vol_*(A)=vol_*(A^{\circ}).$$
Now,
suppose that you have two bounded Jordan measurable sets $E\subset F$.
Then,
we see that $F=E\cup(F\setminus E)$,
and since Jordan measure is finitely additive on disjoint unions,
then it follows that
$$vol(F)=vol(E)+vol(F\setminus E).$$
Consequently,
the fact that $A$ is Jordan measurable implies that
$$vol^*(\overline A\setminus A^\circ)=vol(\overline A\setminus A^\circ)=vol(\overline A)-vol(A^\circ),$$
from which the result follows.

2. For the second question:
If I remember my courses on Jordan measure,
I believe that the Jordan outer/inner measures are not finitely additive,
that is,
if $A,B$ are disjoint,
it is not true in general that
$$vol_*(A\cup B)=vol_*(A)+vol_*(B)$$
and
$$vol^*(A\cup B)=vol^*(A)+vol^*(B).$$
Furthermore,
in your calculations,
it seems to me that you prove that $vol^*(A)\geq vol_*(A)$,
whereas I think you should be proving that $vol^*(A)\leq vol_*(A)$,
which,
together with $vol^*(A)\geq vol_*(A)$,
gives you the equality.
