Non-zero total charge in a hollow conductor 
Given a closed, hollow conductor (as understood in electrostatics), denote by $V$ its cavity. Let
$$Q_V = \int_V \rho(x) \text{d}^3 x$$
be the total charge within the cavity. Is it always the case that $Q_V = 0$?

I am currently trying to understand some basic principles of electrostatics related to electrical fields, Gauss' law and the like (I am a math student, not a physicist). The above question and setting confuses me quite a lot. I started worrying about it because I came up with the following specific example (possibly wrong):
Assume that
$$A = \{x \in \mathbb{R}^3 \ : \ 1 \leq |x| \leq 2\}$$
is our "hollow" conductor. The "cavity" is then given by $V = \{ x \in \mathbb{R}^3 \ : \ |x| < 1\}$. As far as I understand, one usually assumes that for an electric field $E$, we have $E(x)=0$ for all points $x \in A$ since $A$ is a conductor. Now let $1<\alpha<2$ and let $S_{\alpha}$ be the sphere of radius $\alpha$ around the origin, i.e. $S_{\alpha} = \{ x \in \mathbb{R}^3 \ : \ |x|=\alpha \}$.
Because $S_{\alpha} \subseteq A$, we have $E=0$ on $S_{\alpha}$ and therefore, by Gauss' law, I would deduce
$$0=\int_{S_{\alpha}} (E \cdot n) \text{dS} = \frac{1}{\epsilon_0} \int_H \rho(x) \text{d}^3 x$$
where $H$ is the volume enclosed by $S_{\alpha}$. Since we can choose $\alpha$ as close to $1$ as we would like to, the final conclusion would be
$$Q_V = \int_V \rho(x) \text{d}^3 x =0.$$
That's the example which brought me to the question I asked above.
Here are more specific questions about the example:
My questions:

*

*Is the set $A$ from above a valid mathematical example of a hollow conductor as understood by physicists? I've never heard of such a thing before and I am therefore wondering what a mathematical model of such a thing might look like. If not, what is a mathematically precise example of a hollow conductor?


*Is the reasoning in my example correct? If so, does it generalize to any hollow conductor as I suggested in my question I asked at the beginning (at least if it is of finite size)?
 A: The underlying physical argument is that in a (resistive) conductor, charges will move according to any local electric field until they can no longer move (e.g. at a boundary).
Ask yourself: what happens when the conductor stops and the air starts?
Physically, the charges build up on the surface. In a real conductor, there will be a finite depth to this build up of charges (set by some weird surface physics, and possibly electronic repulsion), but in classical electromagnetism  we take the sheet of charge infinitesimally thin.
The more correct mathematical definition of a conductor would be
A region of space $V$ such that $\phi = constant$, where $\phi(x) = -\int_{x_0}^x E \cdot dl$, $x_0$ is an arbitrary point in space.
This means that if $\phi$ is differentiable you get $E=0$, but we do not necessarily have a smooth (or even defined) $E$ everywhere. Instead we have $E$ defined almost everywhere, i.e.e everywhere except on a set of measure zero.
This makes understanding what's going on a little nontrivial in spherical coordinates, so I'll simplify: Let half of the universe $H = \{(x,y,z) \in \mathbb{R}^3 | x>0\}$ be conductive, and the rest insulate. Apply an external electric field $E = E_0 \hat{x}$ in the insulating plane.
Then, choosing some volume $V=[-\epsilon,\epsilon]\times A$ with $A\subset \mathbb{R}^2$ closed (closedness/openness does not really matter since the difference is measure zero)
$$\int_{\partial V} \vec{E}\cdot d\vec{S} = -E_0 \iint_A dA = \frac{1}{ \epsilon_0}\iiint \rho(\vec{x}) dV  = \frac{1}{ \epsilon_0}\int dx \rho(x) \iint_AdA$$
$$ \Rightarrow \int \rho(x) dx = - \epsilon_0 E_0$$
Making $\epsilon$ arbitrary small, you see that there is always a residual charge. It follows that $\rho(x) = -\epsilon_0 E_0 \delta(x)$.
You're getting weird results (like $Q=0$, even with a charge) because you're treating everything as though it is a smooth, regular function. In general, the charge distribution $\rho$ lives in the space of distributions, so is not defined in any meaningful sense at any particular point. (see Bourbaki, Functional Analysis) It's more natural to think of it as an integration measure.
In loose terms, $\rho$ lives in the same space as Dirac deltas, in that they naturally "live under an integral sign", or should be thought of not as $\rho$ but as $\rho d^3x$ (in differential geometry language, a 3-form).
In your specific example, the sphere induces surface charges (of equal and opposite total charges, but different charge densities) on the inner and outer surface of the shell. Again, using the less confusing integral form of the Maxwell equations should tell you (write $B(R)$ for a solid ball of radius $R$)
$$ \int_{\partial B(R)} \vec{E} \cdot d\vec{S} = \iiint_{B(R)} \rho(\vec{x}) d^3 \vec{x}  = \frac{Q}{ \epsilon_0}\hspace{2em}\text{  if }R<1$$
$$ \Rightarrow \vec{E} = \frac{Q}{4\pi \epsilon_0 R} \hat{r}$$
Repeating for a thin shell (write $B(R)-B(r)$ for set removal)
$$ \iint_{\partial[B(R+\epsilon) - B(R-\epsilon)]} \vec{E} \cdot d\vec{S} = \iint_{B(R+\epsilon) - B(R-\epsilon)} \frac{\rho(\vec{x})}{\epsilon_0}$$
$$ \Rightarrow 0 - Q = \iiint_{B(R+\epsilon)} \rho(x) d^3x -\iiint_{B(R-\epsilon)} \rho(x) d^3x$$
$$ \Rightarrow  - Q = \iiint_{B(R+\epsilon)} \rho(x) d^3x - Q$$
$$ \Rightarrow \iiint_{B(R+\epsilon)}\rho(x) d^3x = 0$$
I.e. your answer from the start is completely valid - there is zero charge enclosed by the surface $\partial B(R+\epsilon)$, because there is a distributed surface charge of $-Q$ that has appeared on the interior shell.
Second Question
This is in no way specific to a sphere. Let $\partial V$ be an arbitrary smooth closed 2D surface, within which is air (the enclosed compact region $V$) and a single point charge $Q$. Then it's possible to find a chain of smooth surfaces such that $V_{in} \subset V \subset V_{out}$. (This follows from compactness.) Use cgs units with $\epsilon_0 = 1$ and assume $E = E(\vec{x}), \rho = \rho(\vec{x})$.
Then $\iint_{\partial V_{out}} E\cdot dS - \iint_{\partial V_{in}} E\cdot dS = \iiint_{V_{out}} \rho-\iiint_{V_{in}} \rho$. Since $\partial V_{out}$ is entirely within the conductor, you end up with
$$Q_in = \iint_{\partial V_{in}} E\cdot dS = -\iiint_{V_{out}-V_{in}} \rho dV$$
i.e. there is a charge of $-Q$ on the inner surface (in general it is not homogeneously distributed though).
