The exterior measure of a closed cube is equal to his volume.

Question

The proof goes like that:

Let $Q\subset \mathbb R^d$ a closed cube of $\mathbb R^d$. Since $Q$ covers itself, we must have $m*(Q)\leq Q$. Therefore, it suffice to prove the reverse equality. Let $Q\subset \bigcup_{j=1}^\infty Q_j$ where $Q_j$ are cubes. For a fixed $\varepsilon>0$, we choose for each $j$ an open cube $S_j$ which contains $Q_j$ and such that $|S_j|\leq (1+\varepsilon)|Q_j|$.

Q1) Why can we take such $S_j$ ? which theorem give us such an $\varepsilon$ ?

Q2) When we talk about cube, is it consider open cube ?

Answer 1

Cubes may be open, closed or neither. They are of the form $\prod _{i}\left \langle a_{i},b_{i} \right \rangle$ where the symbol $\left \langle \right \rangle$ refers to $\textit {any}$ type of interval.

You can always enclose a cube $Q$ of any of these types in an open cube $Q'$ that is as "close" to $Q$ as you want.

In fact there is a more general statement that includes your case:

Fix $Q$ and define $S=\left \{ x\in R^{d}:d(x,Q)<\epsilon \right \}$ where $d(x,Q)=\inf_{y\in Q}\left \{ d(x,y) \right \}$. If we can show that $f(x)=d(x,Q)$ is continuous, this will prove that $S=d^{-1}((0,\epsilon))$ is open which is what we want.

So, pick a point $z\in Q$ and note that by the triangle inequality, we have

$d(x,z)\leq d(x,y)+d(y,z)$. This inequality is true for all $z\in Q$ so in fact

$$\tag1d(x,Q)\leq d(x,y)+d(y,Q)$$ Likewise, we have

$d(y,z)\leq d(x,y)+d(x,z)$ and therefore $$\tag2d(y,Q)\leq d(x,y)+d(x,Q)$$

Combining $(1)$ and $(2)$, we have now

$$\tag3\left | d(x,Q)-d(y,Q) \right |\leq d(x,y)$$

and we see now that $(3)$ proves continuity as soon as we set $\delta =\epsilon$.

Answer 2

It suffices to take an open cube with the same center and oriented the same way but slightly larger. If the closed cube's volume is $r^n$ the open cube has measure $(r(1+e))^n$ where the positive value e can be arbitrarily small. For example, in $R^2$ , if $C=[-1,1]^2$, take $(-1-e,1+e)^2$ . This is what Euler88 says in his comment.