Density definition implies total mass integral Reading back through an old text on E&M today, I realized that while it makes intuitive sense to me that upon defining density as
$$\rho(x,y,z) = \frac{dm}{dV}$$
we get that
$$M = \int_V \rho dV$$
I don't know how to prove it.  (Note: in E&M it would be $dQ/dV$ but it doesn't really matter which type of density we're talking about so I just went with mass density here.)
Firstly, I'm not sure how the definition of $\rho$ really works as a derivative.  What function is it the derivative of?  It seems like what we really mean to define is
$$\rho(P) = \lim_{V\to \{P\}} \frac{m(V)}{|V|}$$
where $V$ is some subset of $\Bbb R^3$, $m$ is a function of type $m:\mathcal P(\Bbb R^3) \to \Bbb R$ that takes regions of $\Bbb R^3$ to numbers, $|V|$ is the notation I decided on for the volume of the region $V$, and $P$ is some point within the volume $V$.
This doesn't really fit the definition of the derivative I'm familiar with (from Calc I-III back in college).  But OK, let's go with that.  Then we should be able to show that
$$M := m(V) = \int_V\lim_{V'\to \{P\}} \frac{m(V')}{|V'|} dV$$
However, at this point, I'm at a loss for how to show this.  It certainly seems like it must be the case physically, but I'm not quite sure how to handle the integral of a limit.
How can I prove this formula (or what am I doing incorrectly)?
 A: Your question can be worded as follows: 

Lemma: Suppose we have a map $m(V) := \int_V \rho\, d \lambda$, where $\lambda$ is the appropriate Lebesgue Measure. Then: $$\rho(x) =
 \lim_{V \to x} \frac{m(V)}{|V|}$$ for (almost) every point $x$.

Note that once this is proved, we have
$$\int_V\lim_{V'\to \{P\}} \frac{m(V')}{|V'|} d\lambda = \int_V\rho \, d\lambda = m(V)$$
proving your claim. 
Now, the lemma above is highly non-trivial. It is known as the Lebesgue Differentiation Theorem, and a standard proof using the Hardy-Littlewood Maximal Function can be found on the linked Wikipedia page. 
Technically the result holds only almost-everywhere (it could fail on a set of measure zero), but in physical applications this might be meaningless. Additionally, the family of sets $\{V\}$ must be of bounded eccentricity for the limit to converge, and the limit is taken such that the diameters of the sets decrease
A: You are right, this definition is a little bit strange. I'd say that
$$\varrho = \frac{\mathrm{d}Q}{\mathrm{d}V}$$
means that (i.e. it's the definition) for all volume $\mathscr{V}$, we get that the charge inside $\mathscr{V}$ is:
$$Q(\mathscr{V}) = \int_\mathscr{V} \varrho \mathrm{d}V$$
From a mathematical point of view, $\varrho$ is usually not even a function (or at least in the textbook-exercises). In the case of a point-charge at $r_0$, with charge $q$, the  density is
$$\varrho(r)=q\delta(r-r_0)$$
where $\delta$ is the Dirac-delta. It makes sense because you will get that $Q(\mathscr{V})=q$ if $r_0 \in \mathscr{V}$, and zero otherwise.
