Consider an infinitesimal mass element of mass $dm\left(\vec{r}\right)$ at position $\vec{r}$ and with velocity $\vec{v}$ . Its angular momentum is: $d\vec{L}=\vec{r}\times\vec{v}dm\left(\vec{r}\right)$
If there are multiple such masses, then we can sum over them to find the total angular momentum of the system. $$\vec{L}{total}=\sum{i=1}^{N}\vec{r}{i}\times\vec{v}{i}dm\left(\vec{r}_{i}\right)$$
Now, this summation can be written as an integral over the distribution of such masses. For a continuous mass distribution, the sum approaches the integral when $N\rightarrow\infty$ and the “mesh” of the mass distribution volume approaches zero. This statement is as rigorous as you want it to be and is the essence of the argument that is used to define Riemann integrals from sums of finite elements.
It is more intuitive to write $dm\left(\vec{r}\right)=m'\left(\vec{r}\right)dV=\rho\left(\vec{r}\right)dV$ and talk about Riemann integrals as opposed to their generalization, Riemann-Stieltjes. $\rho$ is identified as the volume density and in Cartesian coordinates $dV=dxdydz$ . 499
$$\vec{L}{total}=\sum{i=1}^{N}\vec{r}{i}\times\vec{v}{i}\rho\left(\vec{r}{i}\right)dV{i}$$
Now we can follow the arguments that lead from a discrete sum to a Riemann integral. In the limit that $N\rightarrow\infty$ we can write:$$\vec{L}{total}=\int{\Omega}\left(\vec{r}\times\vec{v}\right)\rho\left(\vec{r}\right)dV$$
Where $\Omega$ is a volume encompassing the mass distribution. This is the angular momentum of the system with no further constraints on the velocities.
If the object is a rigid body rotating around an axis, then all of the infinitesimal masses from which we started will have the same angular velocity $\vec{\omega}$ whose magnitude is given by:$\omega=\frac{v}{r}$ . Without loss of generality, take the axis of rotation to be the $z$-axis. In terms of polar coordinates: $\vec{v}=v\hat{\phi}$. $$ \vec{L}{total} = \int{\Omega}\left(r\hat{r}\times v\hat{\phi}\right)\rho\left(\vec{r}\right)dV = \int_{\Omega}\left(r\hat{r}\times r\omega\hat{\phi}\right)\rho\left(\vec{r}\right)dV = \left(\int_{\Omega}r^{2}\omega\rho\left(\vec{r}\right)dV\right)\hat{z} = \left(\omega\int_{\Omega}r^{2}\rho\left(\vec{r}\right)dV\right)\hat{z} = \omega I\hat{z}=I\vec{\omega} $$
Note: the sum over the masses can expressed as an integral over delta-functions: $$\vec{L}{total}=\int{\Omega}\sum_{i=1}^{N}\vec{r}\times\vec{v}\rho\left(\vec{r}\right)\delta^{3D}\left(\vec{r}-\vec{r}_{i}\right)dV$$
where $\delta^{3D}\left(\vec{r}-\vec{r}{i}\right)=\delta\left(x-x{i}\right)\delta\left(y-y_{i}\right)\delta\left(z-z_{i}\right)$ is the 3-dimentional delta function. 804