Derivative and divergence of measures, product of measures and vector field I am having trouble understanding the following notation: this comes from optimal transport theory and the continuity equation. $\mu: [0,1] \to P_2(\mathbb{R}^d)$ is a curve in space of probability measures having finite second moment in $\mathbb{R}^d$ and $v:[0,1] \times \mathbb{R}^d \to \mathbb{R}^d$ is a vector field. Then the continuity equation is
$$
\partial_t \mu_t + \nabla\cdot (\mu_t v_t) =0
$$
in the sense of distribution.
What does this mean? I do not understand taking derivative of a measure, multiplying a measure and a vector field, and taking the divergence of the result. Any help or reference?
Thank you!
 A: A distribution is a linear functional on $C_c^\infty(\mathbb{R}^d),$ the space of infinitely differentiable functions with compact support. The action of a distribution $u$ on the test function $\varphi \in C_c^\infty(\mathbb{R}^d)$ is often denoted by $\langle u, \varphi \rangle.$
A measure $\mu$ induces a distribution by
$\langle \mu, \varphi \rangle := \int \varphi \, d\mu.$
For $\mu : [0,1] \to P_2(\mathbb{R}^d)$ we define $\partial_t \mu_t$ by
$$
\langle \partial_t \mu_t, \varphi \rangle 
:= \frac{d}{dt} \langle \mu_t, \varphi \rangle
= \frac{d}{dt} \int \varphi \, d\mu_t,
$$
define $\mu_t v_t$ by
$$
\langle \mu_t v_t, \varphi \rangle
= \langle \mu_t, v_t \varphi \rangle
= \langle \mu_t, (v_t^1, \ldots, v_t^d) \varphi \rangle
= (\langle \mu_t, v_t^1 \varphi \rangle, \ldots, \langle \mu_t, v_t^d \varphi \rangle),
$$
and
$$
\langle \nabla\cdot(\mu_t v_t), \varphi \rangle
= - \langle \mu_t, \nabla\cdot(v_t \varphi) \rangle
= - \langle \mu_t, (\nabla\cdot v_t) \varphi + v_t \cdot \nabla\varphi \rangle
.
$$
Note: For the above to make sense with respect to the definition of distributions as being linear functionals on $C_c^\infty$ the vector field $v_t$ needs to be $C^\infty.$ But I guess that $\mu_t$ having finite second moment allows us to reduce the requirements on the test functions and on $v_t.$
