Let $G$ be the 3-dimensional Heisenberg group equipped with its Carnot-Caratheodory subriemannian metric $d_{G}$. Let $U$ be a domain in $G$ of the form $V \times I$, where $V$ is an open subset of $\mathbb{R}^{2}$ and $I$ is an open interval (so $I$ is the third coordinate in $G$ corresponding to the center of the group). I'm going to consider uniformly continuous maps $f: U \rightarrow V$. Let $H$ be the left-invariant contact distribution on $G$.
Let $d_{\mathbb{R}^{2}}$ denote Euclidean distance on $\mathbb{R}^{2}$. I want to consider a "quasiconformal" map $f: U \rightarrow \mathbb{R}^{2}$ in the following sense: there is a constant $C \geq 1$ independent of $r$ such that for each $x \in U$, $y \in H_{x} \cap U$, $$ \frac{\sup\{d_{\mathbb{R}^{2}}(f(x),f(y)): d_{G}(x,y) = r\}}{\inf\{d_{\mathbb{R}^{2}}(f(x),f(y)): d_{G}(x,y) = r\}} \leq C. $$ In other words, for each $x \in U$ $f$ maps concentric circles in $H_{x}$ to quasiconformally distorted circles around $f(x)$. I want to conclude that if $A$ is a measurable subset of $V$ above with $m_{G}(A \times I) = 0$ ($m_{G}$ the Haar measure on $G$) then $m_{\mathbb{R}^{2}}(f(A \times I)) = 0$. So a form of transversal absolute continuity. Is this known?
What makes me think this is true is that, if we replace $G$ by $\mathbb{R}^{3}$ with the Euclidean distance with $H_{x}$ now being parallel planes orthogonal to the third coordinate, the conclusion follows from standard quasiconformal mapping theory on the plane + Fubini's theorem. Unfortunately this tactic does not work with $G$ because there is no foliation by transverse parallel planes.
Note: Overhauled an earlier version of this question that was nonsense.
