Let $\psi:U' \to \mathbb{R}^{N}$ be a smooth function, where $U' \supset \overline{U} \supset U \supset Q^{N-1}$ with $$Q^{N-1}=\{(x_1,\dots,x_{N-1}: x_1 \geq 0, \dots, x_{N-1}\geq 0 \hbox{ and } \sum_{i=1}^{N-1}x_i \leq 1\}.$$ Suppose that $U$ is bounded, $\psi$ is injective and $\psi'(x)$ has rank $N-1$ for all $x \in U'$ and define $S\doteq \psi(U).$
Consider the map $\vec{N}_\psi(t_0)=\frac{\partial \psi}{\partial t_1}(t_0) \times \cdots \times \frac{\partial \psi}{\partial t_{N-1}}(t_0)$ and define $$S \ni p \mapsto \vec{n}(p)=\frac{\vec{N}_\psi(t)}{|\vec{N}_\psi(t)|}, \quad \psi(t)=p.$$
Consider the $\sigma-$algebra of subsets of $S$ defined by $$\mathcal{M}_S=\{\psi(A):A \subset U, A \hbox{ is Lebesgue measurable}\}$$ and the measure $m_S:\mathcal{M}_S \to [0,\infty)$ given by $$m_S(\psi(A))\doteq \int_{A} |\vec{N}_\psi(t)|\, dm(t).$$
Let $L$ be the vector field given by $L=\sum_{j=1}^{N} a_j(x) \frac{\partial}{\partial x_j}$.
My question: How to prove that $$\int_{\psi(Q^{N-1})} L \cdot \vec{n} dm_S=\int_{Q^{N-1}} (\vec{a}(\psi(t)) \cdot \vec{N}_\psi(t))\, dt,$$ where $\vec{a}=(a_1,\dots,a_N).$
This seems to be related to the Radon-Nikodyn Theorem. I tried to develop the calculations but I didn't get it.
I think I managed to prove that $$\int_{S} \phi dm_S=\int_{U} \phi(\psi(t)) |\vec{N}_\psi(t)|dm(t)$$ when $\phi$ is a simple function. Indeed, if $\phi=\sum_{j=1}^{m}a_j \chi_{A_j}$, $S=\bigcup_{j=1}^{m}A_j$. Then, $U=\bigcup_{j=1}^{m}B_j$, where $B_j=\psi^{-1}(A_j)$. Thus,
\begin{eqnarray} \int_{S} \phi dm_S&=&\sum_{j=1}^{m} a_j m_S(A_j)\\ &=&\sum_{j=1}^{m} a_j m_S(\psi(B_j))\\ &=&\sum_{j=1}^{m} a_j \int_{B_j}|\vec{N}_\psi(t)|dm(t)\\ &=&\sum_{j=1}^{m} \int_{B_j}a_j|\vec{N}_\psi(t)|dm(t)\\ &=&\sum_{j=1}^{m} \int_{B_j}\phi(\psi(t))|\vec{N}_\psi(t)|dm(t)\\ &=&\int_{U}\phi(\psi(t))|\vec{N}_\psi(t)|dm(t), \end{eqnarray} since if $t \in B_j$ then $\psi(t) \in A_j$ which implies that $\phi(\psi(t))=a_j$. Now the idea is pass to the limit.