Boundary of surface Let $S$ be the region in $\mathbb{R}^2$ bounded by $x$-axis, $x=1$, and $y=x$. Define 
$$
f(x,y) = \begin{cases} 0 &\mbox{if } x = 0 \text{ or if $x$ or $y$ is irrational} \\ 
1/q & \mbox{if $x,y \in \mathbb{Q}$ abd $x=p/q$ in reduced form} \end{cases}
$$ on $S$.
[...]
Let $T=\left\{ (x,y,z) ; (x,y) \in S, f(x,y) \leq z \leq 1 \right\} \subset \mathbb{R}^3$. Let $\partial T$ be the boundary of $T$. Describe/sketch $\partial T$. Is $\partial T$ of measure / Jordan content zero? is $T$ a Jordan region?
My question for you: What exactly is $\partial T$ really here? 
My guess: well it seems to be $\{(x,y,1)\}$ plus an uncountable set of $\{(x,y,0)\}$, but I really don't know what to do with this yet. In a way one could argue that all of $T$ is its boundary, since $(x,y,1/2) \in T$ for appropriate $(x,y)$ does not have a neighbourhood in $T$ either. I am really not sure what the definition to work with is here.
Thanks
 A: I recall that a boundary $\partial{T} $of a set $T$ is defined as $\overline{T}\setminus\operatorname{int} T$.
Since the set $$\{(x,y):f(x,y)=0)\}=S\setminus {\Bbb Q}\times {\Bbb Q}$$ is dense in $S$, $$\overline{T}=S\times[0,1].$$
We have $$\operatorname{int} T\subset \operatorname{int}\overline{T}= S’\times (0,1),$$ where $S’$ is interior of $S$ in $\Bbb R^2$. From the other hand, $\operatorname{int} T\subset T$. Thus $$\operatorname{int} T\subset T\cap (S’\times (0,1)).$$ I claim that in fact these sets are equal. Indeed, let $$p=(x,y,z)\in T\cap (S’\times (0,1))$$ be an arbitrary point. Then $0<z<1$. Take a neighborhood $U$ in the plane $\Bbb R^2$ of the point $(x,y)$ so small that in contains no point $(p/q, p’/q’)\in\Bbb Q\times \Bbb Q $ (different from $(x,y)$ if both $x$ and $y$ are rational) such that $q<4/z$. Then $f(x’,y’)\le z/4$ for any point $(x’,y’)\in U$. Thus $$p\in U\times (z/2,1)\subset T.$$  
Therefore $$\partial{T}=S\times[0,1]\setminus (T\cap (S’\times (0,1)))=P\cup R,$$ where  
$$P=(S\times[0,1]\setminus (S’\times (0,1)))$$ is a surface of the prism $S$, and $$R=\bigcup\{{(x,y)}\times (0,f(x,y):(x,y)\in S’\cap {\Bbb Q}\times {\Bbb Q} \}$$ is a union of countable many vertical segments. The set $P$ has Jordan measure zero.  Since for each natural $n$ the set $$[0,1]\times [0,1]\times [0,1/n]\setminus R$$ is covered by a union $$\bigcup \{{p/q}\times [0,1] \times [0,1]: p,q\mbox{ are non-negative integers and } 0\le p\le q\le n \}$$ of finitely many parallelepipeds (of Jordan measure zero), it has Jordan measure zero. In particular, Lebesgue measure of both sets $P$ and $R$ is zero, so, according to my definition, $T$ is a Jordan region. 
