Definition A continuous map $f\colon \mathbb{R}^d \to \mathbb{R}^d$ is volume-preserving if, for every Borel set $V\subset\mathbb{R}^d$, $\mathcal{L}^d(V) = \mathcal{L}^d(f^{-1}(V))$.
I am wondering if the following holds:
Suppose $f_n\colon \mathbb{R}^d \to \mathbb{R}^d$ is a volume-preserving homeomorphism for each $n\in\mathbb{N}$. If $f_n$ converges uniformly to $f$, then $f$ is a volume-preserving homeomorphism.
So far, we know that $f$ is volume-preserving for the following reason. Let $\phi \in C_c^\infty$. Because $f_n$ is volume-preserving, $\int \phi\circ f_n\,dx = \int \phi\,dx$. As $f_n \to f$ uniformly, one can show that $\int \phi\circ f_n\,dx \to \int \phi \circ f\,dx$. Now we know that $\int \phi \circ f\,dx = \int \phi\,dx$, and so $f$ is volume-preserving.