Meaning of integral signs in classical physics When I began studying physics, by myself, on a universitary textbook,  F.J. Keller, W.E. Gettys , M.J. Skove, Physics, about one year ago, I believed that all the integrals that I was going to find in such elementary texts of classical physics were, as WolframMathworld says, Riemann integrals, although I had studied a very little bit of Lebesgue integration and functional analysis by following all the volume of A.N. Kolmogorov, S.V. Fomin, Элементы теории функций и функционального анализа (just a few contents more than Introductory real analysis).
Interpretating integrals as Riemann integrals seems to me to be consistent with the language of some physics texts when they talk about summing "infinitesimal" quantities, like "summing infinitesimal masses" $\rho dV$ to calculate the mass $\int_V\rho dV$ of a body, which I would read as a shorter way to say "calculating the limit $\lim_{\delta_P\to 0}\sum_i\bar{\rho}(\boldsymbol{\xi}_i)\Delta V_i$ where $\delta_P$ is the mesh of the partition and $\bar{\rho}(\boldsymbol{\xi}_i)\Delta V_i$, with $\boldsymbol{\xi}_i$ in the small parallelepiped whose volume is $\Delta V_i$, is an approximation of the mass contained the parallelepiped".
Nevertheless, I have recently found a use of the integral signs that confuses me. While studying elementary classical electrodynamics, in particular the Biot-Savart and Ampère's law, I find expressions such as the following one for the magnetic field $$\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\iiint_V \mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}\, d^3l$$which cannot be a proper Riemann if $\mathbf{r}\in \bar{V}$. From the way it is manipulated in derivations of Ampère's law such as that found in Wikipedia, I would be inclined to exclude that it is a Lebesgue integrals, either, because, provided that the same sign $\iiint_Vd^3l$ means the same thing in each step*, if $\iiint_V\mathbf{J}(\mathbf{l})\nabla_{r}(\|\mathbf{r}-\mathbf{l}\|^{-1})d^3l$ were a Lebesgue integral it would equates $\mathbf{0}$ and not $-4\pi\mathbf{J}(\mathbf{r})$ as desired, I would say. Since Dirac's $\delta$ appears in functional analysis, I would tend to suppose that $\iiint_Vd^3l$ is just a symbolic notation for a linear operator, but the only linear operator that I can imagine to be defined by $\iiint_V \frac{\mathbf{J}(\mathbf{l})\times(\mathbf{r}-\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|^3}\, d^3l$ is $$\mathbf{J}\mapsto\int_V \mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}\, d\mu_{\mathbf{l}}$$where the image of $\mathbf{J}$ is a Lebesgue integral, again, which brings us back to the same problem described above.
Is there any general principle valid to interpretate what integrals mean (i.e. whether they are Riemann, Lebesgue integrals or whatever else), if nothing is specified by the author, in texts of classical physics? I thank any answerer very much.

*If that were not true, that derivation would be even more complicated to understand, but I am not sure that we can rule the hypothesis that the sign $\iiint_Vd^3l$ is used for different things through the outline of proof out.
 A: I am not sure if I've understood your question, but I'm gonna try to show off some light on this. Read it slowly and carefully, it needs a bit concentration and abstract thinking.
Let's imagine that you integrate over a body, summing infinite infinitessimal partitions of this body over its volume. i.e. $\int_v dV$ as you well know. But, what that means? Let me show some examples to develop what you need to understand:


*

*Cube (or any 3D squared figure):


How do you manage to calculate de Volume of a Cube? It seems easy, assuming that the cube has $x$, $y$ and $z$ edges, just $V = xyz$. The definition of a perfect cube implies that $x=y=z$; so in this case $V = x^3$. All of this is assuming a discrete world, where the edge is always $x$.
Now we can think in a continous system, where we can find the volume of an infinitessimal cube with infinitessimal edges. This edges are definied as $dx$, $dy$ and - of course - $dz$. And the volume is $dV = dxdydz$. Now if we want to calculate the total volume of a volume a bit bigger than just infinitessimal, we must integrate it: $V = \int_v dV$. We are summing here infinitessimal little volumes to get a "total" volume. The interesting thing is that we said that $dV$ is defined with three independent and infinitessimal coordinates. Therefore, each one derserves its own integral. How? Well, is easy to figure that $x = \int dx$, $y = \int dy$ and $z = \int dz$. Then it's easy to see that $V = \int_v dV = \int dx\int dy \int dz$. And, if these coordinates are independent of each other, we can say: $V = \int \int \int dxdydz$.


*

*Sphere


Once you have understood the previous calculus, we can go forward to calcule the volume of a sphere.
In spherical coordinates:
$x = r\cos(\theta)\sin(\phi)$
$y = r\sin(\theta)\sin(\phi)$
$z = r\cos(\phi)$
You can calculate that $|\frac{d(x,y,z)}{d(r,\theta,\phi)}| = r^2\sin(\phi)$
Hence, $dxdydz = r^2\sin(\phi) dr d\phi d\theta$.
Now, as in the Cube case, calculate
$V = \int_0^r dr \int_0^{2\pi} d\theta \int_0^{\pi}\sin(\phi) d\phi$. You will get $V = \frac{4}{3} \pi r^3$.


*

*General


Generalizing all this, you can tell $d\vec{l} = (x,y,z)$ and integrate along a 1-dimensional structure, tell $d^2 l$ as the surface of any infinitessimal 2-dimensional structure, ..., tell $d^n l$ as the hypervolume of any infinitessimal n-dimensional structure.
Back to your case, what $d^3 l$ means? As you could infer, it represents three infinitessimal and independent variables. 
$dxdydz$ if you are in cartesian coordinates,
$r^2\sin(\phi)dr d\phi d\theta$ in spherical coordinates,
and so on.

I hope it could help you, I was not accurate to explain it as easy as possible. I you are interested, I would like to recommend you these books: Apostol - Calculus. Volume 1 & 2. It will help you for sure
