Let $x$ be a continuous random variable with distribution $D$ and pdf $P(x)$ supported by $\mathbb{R}$. Let $T:\mathbb{R} \to \mathbb{R}$ be an increasing invertible transformation. We denote by $T \circ D$ the distribution of $T(x)$ for $x \sim D$. Let $c:\mathbb{R} \to \mathbb{R}$ be differentiable function. Since $P(x)dx = P(y)dy$ for $y = T(x)$, we have: \begin{equation} \mathbb{E}_{x\sim D}[c(T(x))] = \int^{\infty}_{-\infty} c(T(x)) P(x) dx = \int^{\infty}_{-\infty} c(y) P(y) dy = \mathbb{E}_{y \sim T \circ D}[c(y)] \end{equation}
My question: is the equation $\mathbb{E}_{x\sim D}[c(T(x))] = \mathbb{E}_{y \sim T \circ D}[c(y)]$ true for similar conditions in the $m > 1$ dimensional case.