Consider a function $g(t,x): \mathbb{R} \times \mathbb{R} \mapsto \mathbb{R}$, and given the Lebesgue measure $\lambda$, consider the pushforward measure $\hat{\lambda}= \lambda \circ g(t)$, such that
$$ \int_I f(x) d \hat{\lambda} = \int_I f(g(t)) d \lambda \text{ (here $t$ is fixed}).$$
Now consider $\partial _t \hat{\lambda} \otimes dt$. It can be seen as a distribution over, say, $\mathcal{D} ( \mathbb{R}^+ \times \mathbb{R})= \mathcal{D}(Q)$ and so by taking a test function $\psi$:
$$ \langle \psi , \partial_t \hat{\lambda} \otimes dt \rangle = - \int_{Q} \partial_t \psi d \hat{\lambda} dt.$$
Now by definition of push forward this is equal to
$$- \int_Q \partial_t \psi (t, g(t)) d \lambda dt . $$
Consider the solution of $\dot{X}= b(X)$ with initial condition $X(0)=x$, and let $X(t,x)$ be the vector field associated to the equation. Assume $\hat{\lambda}$ to be the push forward of $\lambda$ with respect to $X(t)$. It is well known that this $\hat{\lambda}(t)$ satisfies
$$\partial_t \hat{\lambda} - div (b \hat{ \lambda})=0 \text{ in } (0, + \infty) \times \mathbb{R^n}$$
and $\hat{\lambda}(0)= \lambda$. Testing with $\psi \in C^\infty _c ((0, + \infty) \times \mathbb{R}^n)$ however I get (changing signs because of personal preference):
$$ \int (\partial_t \psi - b \cdot \nabla \psi) d \hat{\lambda} dt $$
and by definition of push forward this is equal to
$$ \int (\partial_t \psi (t, X(t)) - b(X(t)) \cdot \nabla \psi (t, X(t)) ) d \lambda dt$$
and I am stuck. If there were a plus sign, using that $b(X(t))= \dot{X} (t)$ I could rewrite the whole integrand as $\frac{d}{dt} \psi (t, X(t))$, exchange the order of integration and concludeby compactness of $\psi$. Where's my mistake?