Mass conservation in a distribution sense.

Question

I have been reading T. Tao's physical derivation of Navier-Stokes equation (https://terrytao.wordpress.com/2018/09/03/254a-notes-0-physical-derivation-of-the-incompressible-euler-and-navier-stokes-equations/) and have some trouble with starting point (11) and its momentum counterpart. As I understand, due to mass conservation, you have that $ \int_{R^{d}}\psi(t,x)d\mu_{mass}(t)(x) = C$, ($\psi$ is a test function and C constant) so deriving and integrating you obtain (11). But I can not realize why this (equation I typed above) should be a consequence of mass conservation. Is there a physical interpretation of test function? Any help would be appreciated.