Content $0$ implies boundary has content $0$ 
Problem: If a set $C$ has content $0$, then the boundary must also have content $0$.

The solution say

Solution: Suppose a finite set of rectangle $  U_i =  (a_{i1},b_{i1}) \times \dots   \times (a_{in}, b_{in})$ where $i \in \{1, \dots, m \}$ cover $C$ and have total volume less than $\epsilon/ 2$ where $\epsilon >0.$ Let $V_i = (a_{i1} - \delta ,b_{i1} + \delta)  \times \dots   \times (a_{in} - \delta , b_{in} + \delta)$ where 
$$\prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{ij} + 2\delta) - \prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{aj} ) < \epsilon/2m .$$

Or equivalently,

Then the union of $V_i$ cover the boundary of $C$ and have total volume less than $\epsilon$. Hence $\partial C$ is also of content $0$.

This is a really simple question, but does $V_i - U_i$ cover $\partial C$? $V_i - U_i$ means the part where we remove $U_i$ from $V_i$. 
Also is the last step because of $$ \sum_{i = 1}^m vol(\cup( V_i - U_i)) \leq \sum_{i = 1}^m\prod (b_{ij} - a_{aj} + 2\delta) - \prod (b_{ij} - a_{aj} ) < \sum_{i = 1}^m \epsilon/2m  = \epsilon /2.$$
or equivalently, 
$$ \sum_{i = 1}^m vol(\cup V_i ) \leq \sum_{i = 1}^m\prod (b_{ij} - a_{aj} + 2\delta) < \sum_{i = 1}^m \epsilon/2m  + \prod (b_{ij} - a_{aj} ) < \epsilon /2 + \epsilon/2 = \epsilon. $$
Am I also correct to understand the outline of this proof is 


*

*Use the finite  cover  that covers $C$, and take a $\delta$ cover over that finite cover just enough to cover the boundary of $C$.

*Using the finite cover of $C$, demand this cover so that it has volume less than $\epsilon/2$. Pick $\delta$ small enough so that we get
$$\prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{ij} + 2\delta) - \prod_{i\in \{ 1,\dots,n\}} (b_{ij} - a_{aj} ) < \epsilon/2m .$$


*

*Use finite subadditivitly to finish off the estimation. 

 A: $V_i-U_i$ will not cover $C$ in general. Note that $U_i$ is large enough to cover $C$, but it may be wasteful.
So the only trick is really to start with a sufficiently small finite cover for $C$, and blow it up slightly so that $V_i\supset \overline {U_i}$ and hence $\bigcup V_i\supset \bigcup \overline{U_i}=\overline{\bigcup U_i}\supset \overline C\supset \partial C$, and thus obtain an arbitrarily small cover of $\partial C$.
A: As this proof seems to raise a lot of questions, here is one that is in my opinion simpler.
If $C$ has content $0$, then for any $\epsilon > 0$ there is a finite collection $\{\bar{R_1},\ldots, \bar{R}_m\}$ of closed rectangles covering $C$ such that
$$\sum_{j=1}^m vol(\bar{R}_j) < \epsilon$$
This is frequently taken as the definition of zero content.  Proof of the equivalence of this definition with one using open rectangles is easy.
To finish, since $F = \bigcup_{j=1}^m \bar{R}_j$ is closed and $C \subset F$, then we must have $\partial C \subset F$. Otherwise, there is a point $x \in \partial C \cap F^c$, and since $F^c$ is open there is an open neighborhood $V_x$  with $x \in V_x \subset F^c$. As $x$ is a boundary point, the neighborhood $V_x$ must contain a point $y \in C\cap F^c$, a contradiction.
Thus, $\partial C \subset \bigcup_{j=1}^m \bar{R}_j$ where $\sum_{j=1}^m vol(\bar{R}_j) < \epsilon$ and $\partial C$ has content zero.
A: Am I missing something, or is the statement plain false? $\Bbb Q\cap[0,1]$ has Lebesgue Measure 0, but its boundary clearly does not. How does your concept of "content" differ from mine, such that the same isn't true for the content?
