$\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ has two interesting properties: on one hand it is non-compact, but on the other hand it admits a unique $\text{SL}(d,\mathbb R)$-invariant finite measure on $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$.

I wonder if there exists a bounded measurable subset $F$ of $\text{SL}(d,\mathbb R)$ (with subspace topology, of course) that is Borel isomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$?

By "Borel isomorphic" I mean: There exists a bijection $f:F\subset \text{SL}(d,\mathbb R) \to \text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ such that both $f$ and $f^{-1}$ are Borel measurable, with $F$ inheriting the subspace topology from $\text{SL}(d,\mathbb R)$. Of course, "up to a subset of measure zero" is always allowed.

If this is true, then it is good to know a constructive proof but the proof for the existence of such an $F$ will also greatly appreciated!

Here I am not assuming $F$ is a "fundamental domain" for $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$, whose definition itself has many interpretations and I asked $F$ to be bounded and thus may not be homeomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ (but as a possible approach one may want to construct such an $F$ whose closure is homeomorphic to the one-point compactification of $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ modulo a set of measure zero). I only want this $F$ to be measure theoretically isomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ but not topologically. Disproving this claim seems also very hard.