# Compactness of the space of measure-preserving maps

Let $$X, Y$$ be Polish spaces and $$\mu, \nu$$ be Borel probablity measures on $$X, Y$$, respectively. Assume that $$\mu$$ is non-atomic. Denote by $$\mathcal{M}(\mu,\nu)$$ a set of Borel measurable mappings $$T$$ from $$X$$ to $$Y$$ which preserve measures (i.e. for any Borel set $$B$$ of $$Y$$, we have $$\mu(T^{-1}(B))=\nu(B)$$). Is the set $$\mathcal{M}(\mu,\nu)$$ compact with respect to the convergence of $$\mu$$-measure?

• What do mean with 'with respect to the convergence of $\mu$-measure'? – p4sch Dec 7 '18 at 12:29
• I made some mistakes. Convergence ''in'' $\mu$-measure, not convergence ''of'' $\mu$-measure. We need to fix a distance $d$ on $Y$. A sequence ${T_n}$ of Borel measurable maps from $X$ to $Y$ converges in $\mu$-measure to a Borel measurable map $T\colon X\to Y$ if for any $\epsilon>0$, $\mu(\{x\in X \mid d(T_n(x),T(x))>\varepsilon\})$ tends to $0$ as $n$ tends to infinity. – Nick Dec 7 '18 at 13:04