Push forward of the Lebesgue measure is the Haar measure of the Carnot group I have the following problem. I have a Carnot group $(\mathbb G,*)$ which is a connected and simply connected Lie group whose Lie algebra $\mathfrak g$ is stratified as $\mathfrak g= V_1\oplus\dots\oplus V_r$ and $V_1$ generates the whole $\mathfrak g$ by commutation, i.e. $[V_1,V_{i}]=V_{i+1}$ and $[V_1,V_r]=0$ (see also here). Suppose that $\mathfrak g$ is $N$-dimensional.
Now, for these groups the exponential map $exp\colon \mathfrak g \to \mathbb G$ is a diffeomorphism, so one can identify a Carnot group $(\mathbb G, *)$ with a Lie group $(\mathbb R^N,\cdot)$ using exponential coordinates, i.e. $$(x_1,\dots,x_N)\mapsto exp(\sum_{i=1}^N x_iX_i)=exp\circ \pi^{-1}(x_1,\dots,x_N),$$
where $\{X_i\}_1^N$ is a basis of $\mathfrak g$ adapted to the stratification and $\pi$ is the projection of a vector field $\sum_1^N a_i X_i$ over this basis. "Adapted to the stratification" means that $\{X_1,\dots,X_{m_1}\}$ is a basis for $V_1$, $\{X_{m_1+1},\dots,X_{m_1+m_2}\}$ is a basis for $V_2$,... and $\{X_{m_1+\dots+m_{r-1}+1},\dots,X_{m_1+\dots+m_r}\}$ is a basis for $V_r$ (clearly $m_1+m_2+\dots+m_r=N$).
The composition law $\cdot$ takes the form $$x\cdot y=H(\sum_1^N x_iX_i,\sum_1^N y_i X_i),\quad x,y\in \mathbb R^N,$$ where $H$ is the Hausdorff-Campbell operation on $\mathfrak g$.
One can endow the group $\mathbb G$ with the so called Haar measure $vol_\mathbb G$, which, if I have understood correctly, is a measure which is invariant under $*$-left translation. 
Now, I know that the Haar measure of the diffeomorphic group $(\mathbb R^N,\cdot)$ is the Lebesgue measure (this can be seen studying the form of the composition law $\cdot$ and studying the form that the vector fields $X_1,\dots,X_N$ take under exponential coordinates).
Suppose that $Z_i$ is the vector field on $\mathbb R^N$ that corresponds to $X_i$, for each $1\leq i\leq N.$
Suppose that we know $X_i\chi_H=0$ in distributional sense for a set $H\subset \mathbb G,$ i.e. $$\int_H X_i\phi dvol_\mathbb G=0$$ for each $\phi\in C^\infty_c(\mathbb G).$ I was wondering if it is true that also $$\int_\overline H Z_i\varphi dx=0\quad (*)$$ for each $\varphi \in C^\infty _c(\mathbb R^N)$, that is $Z_i\chi_\overline H=0$ ($\overline H$ is $H$ seen in exponential coordinates).
Now, I think that if $vol_\mathbb G$ were the push-forward of the Lebesgue measure under the map $\Phi:=exp\circ \pi^{-1}$, then $(*)$ would be true. Indeed $$0=\int_H X_i \phi dvol_\mathbb G=\int_\overline H (X_i\phi)\circ \Phi dx.$$ Since we have that $X_i=(d\Phi) Z_i$, it follows that $[(X_i\phi)]\Phi(x)=[d\Phi_x(Z_i)](\phi)=Z(\phi\circ \Phi)(x)$, thus $$0=\int_\overline H (X_i\phi)\circ \Phi dx=\int_\overline H Z(\phi\circ \Phi) dx.$$
So it remains to prove that actually $vol_\mathbb G$ is the push forward measure of the Lebesgue measure $ \mathcal L^N$, i.e. $$\mathcal L^N(\Phi^{-1}(B))=vol_\mathbb G(B)$$ for each $B$ measurable set in $\mathbb G.$
And here I'm stuck.
 A: Yes, the Haar measure on any simply connected nilpotent real Lie group is just the push-forward by the exponential of the Lebesgue measure of the Lie algebra.
This amounts to showing that ($\sharp$) on a nilpotent Lie algebra, denoting by $\ast$ the BCH law, for each $y$, the map $R_y:x\mapsto x\ast y$ preserves the measure. (Indeed, this proves that the push-forward by the exponential of the Lebesgue measure is left-invariant, so by uniqueness of the Haar measure up to scalar multiplication, we are done).
To show ($\sharp$): Since $R_y$ is a diffeomorphism with inverse $R_{-y}$, it is enough to show that its differential has determinant 1 everywhere.
Writing BCH, we see that 
[R_y(x+t)=R_y(x)+t+J_y(x)(t)+o_{t\to 0}(\|t\|),]
where $J_y(x)$ belongs to the non-unital subalgebra generated by $\mathrm{ad}(x)$ and $\mathrm{ad}(y)$ in the algebra of linear endomorphisms of $\mathfrak{g}$. Since the adjoint representation can be put in strictly upper triangular form, we deduce that the endomorphism $t\mapsto t+J_y(x)(t)$ is unipotent. This is the differential of $R_y$ at $x$. So the latter has determinant 1 and ($\sharp$) is proved.
