An $m \times n$ matrix $A$ is called *badly approximable* if there exists $c > 0$ such that for all integer vectors $p \in \mathbb Z^m$ and $q \in \mathbb Z^n-\{0\}$ we have

$$ \|A q + p \| \ge c \| q \|^{-\frac{n}{m}}. $$

We denote the collection of badly approximable $m \times n$ matrices by $\textbf{BA}(m,n)$.

Let $d=m+n$ and $X_d:=\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ be the space of unimodular lattices in $\mathbb R^d$, equipped with an $\text{SL}(d,\mathbb R)$-invariant measure $\mu$.

The classical proof that $m(\textbf{BA}(m,n))=0$, where $m$ is the Lebesgue measure on $\mathbb R^{mn}$, goes as follows:

For each $t \in \mathbb R$ and for each matrix $A$, let $g_t:=\begin{bmatrix} e^{t/m}I_m & A \\ 0 & e^{-t/n}I_n \end{bmatrix}$ and $u_A:=\begin{bmatrix} I_m & A \\ 0 & I_n \end{bmatrix}$, where $I_k$ denotes the $k$-dimensional identity matrix.

By Dani's Correspondence (see the paper I linked below), $A$ is badly approximable if and only if the trajectory $(g_tu_A\mathbb Z^d)$ is bounded. Using this and the ergodicity of the action of $(g_t)$, one can conclude that

$$\mu (\{u_A \mathbb Z^d: A\in \textbf{BA}(m,n) \})=0. (*)$$

My question is why $\mu (\{u_A \mathbb Z^d: A\in \textbf{BA}(m,n) \})=0$ implies $m(\textbf{BA}(m,n))=0$.

This *last step seems to be omitted in any literature I have seen*, including Dani's original paper I believe. They just argue that $(*)$ holds (or with $P^-$ matrix multiple) and take $m(\textbf{BA}(m,n))=0$ for granted from that (See the top of the page 66 from the paper I linked).

In essence, this question is about the relation between Haar measure and Lebesgue measure. Please note in general, $\mu(\{u_A \mathbb Z^d: A\in S \subset \mathbb R^{mn}\})=0$ does not imply $m(S)=0$ (take $S=\mathbb R^{mn}$ for example). I really wish someone could explain the Haar measure passing to Lebesgue measure part well. I think I am okay with dynamical system part.