Show some properties of the (inner/ outer) Jordan-measure 

Let $A\subset\mathbb{R}^N$ be a subset. Show that (1) $vol^*(A)=vol^*(\overline{A})$, (2) $vol_*(A)=vol_*(A^°)$.


Here $vol^*(A)$ of a bounded subset $M\subset\mathbb{R}^N$ is defined by
$$
vol^*(M):=\inf\left\{\sum_{k=0}^{m}vol(Q_k) | Q_1,\ldots,Q_m\subset\mathbb{R}^N\text{ compact N-dimensional intervalls with }M\subset Q_1\cup\ldots\cup Q_m\right\}
$$
and $vol_*(M)$ of any subset $M\subset\mathbb{R}^N$ is defined by
$$
vol_*(M):=\left\{\sum_{k=1}^{m}vol(Q_k) | Q_1,\ldots,Q_m\subset\mathbb{R}^N\text{ pairwise disjoint open N-dimensional intervalls with }Q_1\cup\ldots\cup Q_m\subset M\right\}
$$
and
$$
vol([a,b]):=\prod_{j=1}^{N}(b_j-a_j).
$$

Again Edited
(1)
I think I have to assume, that $A$ is bounded because otherwise the outer Jordan-measure is not defined.
It is $A\subset\overline{A}$, so any cover of $\overline{A}$ of the form $Q:=\bigcup_{i=1}^{m} Q_i$ with $Q_i, i=1,\ldots,m$ as in the definition of $vol^*$ is a cover of $A$, too. So it follows that
$$
vol^*(A)\leqslant vol^*(\overline{A}).
$$
Now let $\varepsilon > 0$ be arbitrary, then there is a covering $Q:=\bigcup_{i=1}^{m}Q_i$ $(Q_i$ as in the definition) with
$$
vol^*(A)\geqslant \sum_{i=1}^{m}vol(Q_i)-\varepsilon.
$$
Because $Q$ is a compact (and therefore especially closed) upperset of $A$, it has to be $\overline{A}\subset Q$, because $\overline{A}$ is the smallest closed upperset of $A$. From this it follows that $vol^*(\overline{A})\leqslant\sum_{i=1}^{m}vol(Q_i)$. So it is
$$
vol^*(A)\geqslant \sum_{i=1}^{m}vol(Q_i)-\varepsilon\geqslant vol^*(\overline{A})-\varepsilon.
$$
Because $\varepsilon$ was arbitrary, it follows that
$$
vol^*(A)\geqslant vol^*(\overline{A}).
$$
All together it is $vol^*(A)=vol^*(\overline{A})$ what was to be proved.
(2) (nearly the same)
It is $A^°\subset A$. So if $Q:=\bigcup_{i=1}^{m}$ with $Q_i, i=1,...,m$ as in the definition of the inner Jordan-measure, is any open subset of $A^°$, then it is a subset of $A$, too, so it follows that
$$
vol_*(A^°)\leqslant vol_*(A).
$$
Let $\varepsilon > 0$ be arbitrary. Then there is an open subset $Q:=\bigcup_{i=1}^{m}Q_i$ of the desired form of $A$ with
$$
vol_*(A)\leqslant\sum_{i=1}^{m}vol(Q_i)+\varepsilon
$$
and because $Q$ is an open subset of $A$ and $A^°$ is the biggest open subset of $A$, it is $Q\subset A^°$ what means that $vol_*(Q)\geqslant\sum_{i=1}^{m}vol(Q_i)$, i.e.
$$
vol_*(A)\leqslant\sum_{i=1}^{m}vol(Q_i)+\varepsilon\leqslant vol_*(A^°)+\varepsilon.
$$
With $\varepsilon\to 0$ all is proved.

Would be great to hear if my proof(s) are right!
With greetings and kind regards
math12
 A: Your first proof is correct, but could be shortened a little; as you noticed, a closed set $Q$ is a cover of $A$ iff it is a cover of $\overline{A}$. Thus, $vol_* (A) = vol_* (\overline{A})$, as you can approach both sets "from the outside" the same way. 
However, your idea doesn't transfer to the second statement the way you think. You're saying that you can choose $Q = \bigcup_i Q_i$ to be an open subset of $\mathbb{R}^n$, but $Q$ is the union of closed cuboids $Q_i$, as it is defined for the Jordan measure. So is it clear to you why can Q not be open?
Let's try an approach that's a little different, to account for the openness of $\mathring A$: Let $Q = \bigcup_i Q_i \subset A$ be a finite union of disjoint (and possibly empty) cuboids $Q_i$ with $vol_* (A) - vol_* (Q) = vol_*(A) - |Q| < \varepsilon/2$ for an arbitrary $\varepsilon > 0$ (it's not difficult to show that this Q exists if you remember that $vol_*$ is defined as an infimum). 
Now, shrink the size of every cuboid $Q_i$ in every dimension by a $\delta > 0$ so that you receive smaller cuboids $Q_{i,\delta}$ with $$Q_{i,\delta} \subset \mathring Q_i \implies \bigcup_i Q_{i,\delta} =: Q_\delta \subset \mathring Q \subset \mathring A$$ (that last relation is about $\mathring A$ being the largest open subset of $A$ which you should be familiar with!). $\delta$ obviously can be chosen small enough so that $|Q| - |Q_\delta| < \varepsilon /2$. Now, the following inequalities should be easy to understand:
$$
vol_*(A) \geq vol_*(\mathring A) \geq |Q_\delta| > |Q| - \varepsilon /2 > vol_*(A) - \varepsilon.
$$
Now, let $\varepsilon \to 0$ and all inequalities have to turn into equalities, and you're done.
