On the conformal dimension of product measures

Given a compact set $E \subset \mathbb{R}^{d - 1}$, $d \geq 1$, write $K_{E} := [0,1] \times E \subset \mathbb{R}^{d}$. A theorem of C. Bishop and J. Tyson states that any set of the form $K_{E}$ is minimal for conformal dimension: if $(X,d)$ is a metric space and $f \colon K_{E} \to (X,d)$ is a quasisymmetric homeomorphism, then $$\dim_{\mathrm{H}} f(K_{E}) \geq \dim_{\mathrm{H}} K_{E}.$$ We prove that the measure-theoretic analogue of the result is not true. For any $d \geq 2$ and $0 \leq s < d - 1$, there exist compact sets $E \subset \mathbb{R}^{d - 1}$ with $0 < \mathcal{H}^{s}(E) < \infty$ such that the conformal dimension of $\nu$, the restriction of the $(1 + s)$-dimensional Hausdorff measure on $K_{E}$, is zero. More precisely, for any $\epsilon > 0$, there exists a quasisymmetric embedding $F \colon K_{E} \to \mathbb{R}^{d}$ such that $\dim_{\mathrm{H}} F_{\sharp}\nu < \epsilon$.