Suppose we have a Lie group on $\mathbb R^n$, something like $(\mathbb R^n, \cdot)$, where with $\cdot$ we denote the group law on $\mathbb R^n$. Call $\mathfrak g$ the Lie algebra of $(\mathbb R^n, \cdot)$, it contains all smooth $\cdot$-left invariant vector fields $X$ on $\mathbb R^n$.
Moreover, given a vector field $X\in \mathfrak g$, $exp(tX)$ denotes the point of $\mathbb R^n$ reached after the time $t$ by the flow line of $X$ starting at $0\in \mathbb R^n$. With $\Phi_X(p,t)$ we denote the flow of $X$, i.e. the point reached after the time $t$ by the flow line of $X$ starting from $p\in \mathbb R^n$.
For $X\in \mathfrak g$, we have that $\Phi_X(p,t)=p\cdot exp(tX)$.
Given $X\in \mathfrak g$, and $u\in L^1_{loc}(\mathbb R^n)$, we say that $Xu=f \in L^1_{loc}(\mathbb R^n)$ is the $X$-distributional derivative of $u$ if, by definition, $$\int_\mathbb {R^n} u~ Xv~ dx=-\int_{\mathbb R^n} v~ f~dx\quad \forall v\in C^\infty_c(\mathbb R^n).$$ We say that the Radon measure $\mu$ is the distributional derivative of $u$, and we keep writing $Xu=\mu$, if $$\int_\mathbb {R^n} u~ Xv~ dx=-\int_{\mathbb R^n} v~ d\mu\quad \forall v\in C^\infty_c(\mathbb R^n).$$
My question is: if $Xu=0$ in distributional sense, then is it true that $u$ is constant over the integral lines of $X$? That is, is $u(\bullet)=(u\circ \Phi_X)(\bullet,t)$ for each $t\in \mathbb R$? I was trying to read this article, where the result is proven in Theorem 2.12 over a general manifold $M$. But, because of this full generality, I haven't understood the proof.