Any cover of the rationals would be a collection of open sets containing all rationals in [0,1]. A specific example would be: take any enumeration {$q_1,q_2,..,q_n,...$} of all rationals in $\mathbb Q \cap [0,1]$, and the use the open sets $O_n:=(q_n-\frac{1}{2^n},q_n+\frac {1}{2^n})$
To determine the outer measure, you may want to scale each interval by a fixed $\epsilon>0$ and then add the widths of all the intervals (see what happens when you let $\epsilon>0$ become small).
The following proof is intuitive enough to be summarized in a brief sketch:
- (i) A closed elementary set containing $E$ necessarily contains $\bar E$
- (ii) An open elementary set contained in $E$ is necessarily contained in $E^\circ$
- (iii) On the one hand, find elementary set $A,B$ such that $A\subset E\subset B$ and $B\backslash A$ serves as an elementary set containing $\partial E$; on the other hand, given an elementary set $C$ containing $\partial E$, find the corresponding elementary sets $A,B$.
${\bf Proof:}$
${\bf (i)}$ By definition of outer measure, we can find an elementary set $B$ containing $E$. $B$ is by definition, the disjoint union of $N$ boxes $B_i$'s. It is apparent that $|B_i|=|B_i^\circ|=|\bar B_i|$. Therefore, $\bar B=\cup_{i=1}^N\bar B_i$ has the same Jordan measure as $B$. Also $\bar E\subset\bar B$. Therefore:
$$
m^{\star,(J)}(\bar E)\leq m(\bar B)=m(B)
$$
Take the infimum over all $B\in\varepsilon(\mathbb R^d)$ and we have:
$$
m^{\star,(J)}(\bar E)\leq m^{\star,(J)}(E)
$$
On the other hand, since $E\subset\bar E$, we have:
$$
m^{\star,(J)}(E)\leq m^{\star,(J)}(\bar E)
$$
and therefore $m^{\star,(J)}(E)=m^{\star,(J)}(\bar E)$;
${\bf (ii)}$ Similar to ${\bf (i)}$;
${\bf (iii)}$ If we have $E$ is Jordan measurable, there exists elementary sets $A,B$ such that $A\subset E\subset B$ and $m(B\backslash A)\leq\varepsilon>0$ for arbitrary small positive number $\varepsilon>0$. Recall $A,B$ is the disjoint union of finitely many boxes:
$$
A=\cup_i B_{A,i},\qquad B=\cup_j B_{B,j}
$$
Obviously:
$$
\underbrace{\cup_i B_{A,i}^\circ}_{A^\circ}\subset\cup_i B_{A,i}\subset E\subset\cup_j B_{B,j}\subset\underbrace{\cup_j\bar B_{B,j}}_{\bar B}
$$
w.l.o.g, we denote the two new elementary sets by $A^\circ\subset A$ and $\bar B\supset B$ to indicate they are open and closed respectively. It is easy to see that:
$$
\bar E\subset\bar B
$$
and
$$
A^\circ\subset E\quad\Rightarrow\quad E^c\subset (A^\circ)^c\quad\Rightarrow\quad \overline{E^c}\subset (A^\circ)^c
$$
since $(A^\circ)^c$ is closed. Therefore,
$$
\partial E=\bar E\cap\overline{E^c}\subset \bar B\cap(A^\circ)^c =\bar B\backslash A^\circ
$$
$\partial E$ has outer measure zero since
$$
m(\bar B\backslash A^\circ)=m(B\backslash A)\leq\varepsilon
$$
can be arbitrarily small.
Next, assume we already have an elementary set $C$ with arbitrarily small measure and $\partial E\subset C$. Now by Boolean closure, $C^c$ is also elementary, which means $C^c$ is the disjoint union of finitely many boxes $B_k$'s.
$$
C^c=\cup_kB_k
$$
We claim that:
$$
\text{either }B_k\subset E\text{ or }B_k\subset E^c
$$
For otherwise, exists $x\in E, y\in E^c$ such that $x,y\in B_k$ for some $B_k$. Since $B_k$ is convex, the line segment $\overline{xy}$ is contained in $B_k$ too. Construct the function:
$$
\gamma:[0,1]\to B_k\subset\mathbb R^d, \gamma(0)=x, \gamma(1)=y,\gamma(t)=x+t(y-x)
$$
which is obviously continuous. Now look at the preimage of $\bar E$ and $\overline{E^c}$. We have:
$$
\gamma^{-1}(\bar E)\cup\gamma^{-1}(\overline{E^c})=\gamma^{-1}(\bar E\cup\overline{E^c})=\gamma^{-1}(\mathbb R^d)=[0,1]
$$
and
$$
\gamma^{-1}(\bar E)\cap\gamma^{-1}(\overline{E^c})=\gamma^{-1}(\partial E)=\varnothing
$$
since $B_k\subset C^c$ and $C^c\cap\partial E=\varnothing$. We have $[0,1]$ as the disjoint union of two non-empty (why?) closed sets $\gamma^{-1}(\bar E)$ and $\gamma^{-1}(\overline{E^c})$, which contradicts the fact that $[0,1]$ is connected.
Now we are allowed to pick out all $B_{k}$'s in $C^c$ such that $B_{k}\subset E$ (denoted $B_{i,s}$'s) and simply discard all $B_{k}$'s outside $E$ (denoted $B_{o,s}$'s). Take the union of such boxes and call it $A$. To find $B$, simply take the disjoint union of $A$ with $C$: $B=A\cup C$. It covers $E$ since all the $B_i$'s we discarded are completely contained in $E^c$. Obviously $C=B\backslash A$. And we are done.
${\bf Remark:}$ The proof of ${\bf (iii)}$ is apparently motivated by a pictorial understanding...I am an engineer and this is what I do...
Edit: A technical mistake in the proof. One should not take the complement of $C$ w.r.t. $\mathbb R^d$, but w.r.t. a closed and bounded box containing $C$, for otherwise, the jordan measure is not well defined. I could no longer change the gif (damn I lost my .AI)...
Here an animated gif I made today for your consideration.
Best Answer
The inner measure is the supremum of measures of simple subsets of $M_1$. As $M_1$ contains no rectangle at all, we conclude $m_*(M_1)=0$.
The outer measure is the infimum of measures of simple supersets of $M_1$. But any simple set, being a finite union of rectangles must cover $[0,1]^n$ almost completely (i.e., up to lower dimensional faces; this is because the complement of a simple set, relative to the cube, is also simple), and that means must have measure at least $1$. We conclude $m^*(M_1)=1$.
The essential difference to Lebesgue measure is: There, we are allowed to have simple sets made of infinitely many rectangles, which allows us to make their total volume arbitrarily small by picking one of volume $\epsilon\cdot \frac1{n^2}$, say, around the $n$th of our countably many points