# Why does the integral of NDF over a solid angle equals the area where micronormals belong to that angle?

In Understanding the Masking-Shadowing Function in Microfacet-Based BRDFs, Section 2.2, Mr Eric Heitz defines the distribution of normals as:

And then, he goes on with the following assertion:

I fail to understand how can D have such a property. Is it an assumption based on common sense or on geometrical considerations? Or can this equality be proven by manipulating the formulation of the integral?

$D(\omega)$ 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!