Why does the integral of NDF over a solid angle equals the area where micronormals belong to that angle?
$D(\omega)$ is defined as the area ($m^2$ unit in the numerator) of the microsurface with normals pointing in the direction $\omega$. $\mathcal{M}'$ is defined as the portion of the microsurface with normals point in the direction $\omega \in \Omega'$. So it's natural that the integral of $D(\omega)$ over $\Omega'$ gives the area of $\mathcal{M}'$.



Substituting the definition of $D$ into (4) gives:

$\int_{\Omega'} D(\omega_m) d\omega_m = \int_{\Omega'} \int_{\mathcal{M}} \delta_{\omega}(\omega_m(p_m)) d p_m d\omega_m$

A fundamental property of the Dirac delta function is that it integrates to 1 over its domain. Unfortunately, we are integrating over $\Omega'$, not $\Omega$. Here I will stop pretending that I am actually able to use formal mathematical terms, and I will just note that the integrand has non-zero values only where both $d p_m \in \mathcal{M'}$ and $d\omega_m \in \Omega'$, which is what (3) says. As long as at least one of our integration domains is restricted to the set we care about, we will get the same result. So I'll just move the prime around a tad:

$\int_{\Omega'} D(\omega_m) d\omega_m = \int_{\Omega} \int_{\mathcal{M}'} \delta_{\omega}(\omega_m(p_m)) d p_m d\omega_m$

Now we can apply the Dirac delta identity, since we're integrating over the whole domain:

$\int_{\Omega'} D(\omega_m) d\omega_m = \int_{\mathcal{M}'} d p_m$

Ta-da!