I am not an expert in the field and would appreciate any help with this problem. Suppose $(\mathcal{X},\mu)$ is a probability space, and that the measure $\mu$ is continuous with respect to the Lebesgue measure and admits a density $f(x\in\mathbb{R}^n)$ which is strictly positive $f(x)>0$ and finite.
Let $T:\mathcal{X}\to\mathcal{X}$ be a measure preserving map in a sense that $\int_{\mathcal{H}}d\mu(T(x))=\int_{\mathcal{H}}d\mu(x)$ for all $\mathcal{H}\subseteq\mathcal{X}$. Can we say that $T$ is invertible? I was thinking along the line that if $J$ is the Jacobian of the transformation from $x$ to $T(x)$, one can argue $f(T(x))|\textrm{det}(J)|=f(x)$ and since $f$ is non-zero, the Jacobian $J$ should be invertible, and thus $T$. Does that make sense? Or is there a better way to to argue about invertibility of $T$?
(Would be more appropriate as a comment, cannot do that yet)
Measure preservingness is generally defined as having $$T^{-1}\mathcal{H}~=~ \mathcal{H},\quad\forall\mathcal{H}\subset\mathcal{A}\,\text{measurable}.$$ (Check whether that coincides with what you wrote.) This is because measurability of $T$ only insures that measurability of $\mathcal{H}$ implies measurability of $T^{-1}\mathcal{H}$, but doesn't say anything about measurability of $T\mathcal{H}$. Knowing $T$ is measure preserving (in the sense of the definition that I provided), doesn't tell you anything about $T$ being invertible.