Existence of a mapping that induces a certain push-forward measure

Question

Let $X \subset \mathbb{R}^n$ be compact and consider the measurable space $(X, \mathcal{B}(X))$ where $\mathcal{B}(X)$ is the Borel $\sigma-$algebra over $X$. Denote by $\mu$ the Lebesgue measure on this space. Let $\nu$ be another measure on this space such that, $\nu \ll \mu$, i.e., it's absolutely continuous w.r.t $\mu$.

Can we always find a measurable mapping $T: X \to X$ such that $T_\# \mu = \nu$, i.e., such the push-forward measure induced by $T$ is equivalent to $\nu$? If not, what conditions one needs to impose on $\nu$ for this to hold?

I am not sure how to approach this, but my intuition is that $T$, if it exists, should be non-singular.

Answer 1

Coincidence of the total measures is an obstruction; morally this should be the only obstruction (in a variety of settings (often when there is more regularity to worry about) this is often named a "Moser Lemma" ($\dagger$)).

Indeed, first let's assume that $X\subseteq \mathbb{R}^n$ is bounded, so that $\text{leb}(X)<\infty$; wlog we may also assume $\text{leb}(X)>0$. Then for any measure $\nu$ on $X$ with $\nu(X)=\text{leb}(X)$, there is a bimeasurable ($\triangle$) $T:X\to X$ such that $T_\ast(\nu)=\text{leb}|_X$ as a corollary of the theorem mentioned in my answer at Is there always a mapping between probability measures on the $n$-sphere?. More generally to apply this theorem all that is required is that $\nu$ assigns zero measure to any point in $X$ ($\star$).

In the case of unbounded $X$ the situation is more subtle. For finite $\nu$ the above argument again applies by one-point compactifying $\mathbb{R}^n$.

More generally if $\nu$ is $\sigma$-finite and if there is a partition $\{X_m\}_m$ of $X$ into countably many pieces with $\nu(X_m)=\text{leb}(X_m)$, or even two partitions $\{X_m\}_m$, $\{X_m'\}_m$with $\nu(X_m')=\text{leb}(X_m)$.

(In an earlier version of this answer I had claimed that $X_m = \{x\in X| m\leq |x|<m+1\}$ would work, which is not correct in general.)

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$(\dagger)$ Moser had shown in his paper "On the Volume Elements on a Manifold" the (roughly) following: Let $\omega,\eta$ be two volume elements on a smooth manifold $M$, there is a diffeomorphism $f:M\to M$ with $f^\ast(\omega)=\eta$ iff $\int_M \omega=\int_M \eta$.

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Added:

($\triangle$) Thus $T:X\to X$ is a measurable bijection such that $T^{-1}:X\to X$ is also measurable.

($\star$) That is, the condition "for any $x\in X$, $\nu(\{x\})=0$" is sufficient for the argument.

Answer 2

The statment seems to hold for probability measures on Borel spaces under the assumption that $\mu$ is really the Lebesgue measure on $(X,\mathscr{B})$.

Definition: A measure space $(X,\mathscr{B})$ is Borel if there is a bijective map $\phi:(0,1)\rightarrow X$ such that $\phi$ is $\mathcal{B}((0,1))/\mathscr{B}$ -measurable and $\phi^{-1}$ is $\mathscr{B}/\mathcal{B}(0,1)$ -measurable. Such a map $\phi$ is called measurable isomorphism.

Uncountable Polish spaces for example are Borel spaces (Parthasarathy, K.L., Probability on Metric spaces, AMS Chelsea, 2005 (reprint of 1967 edition)).

Suppose $(X,\mathscr{B})$ is a Borel space and let $\phi$ be as in the definition above. Let $\lambda$ denote the Lebesgue measure restricted to $((0,1),\mathcal{B}(0,1))$. A probability measure $\mu$ on $(X,\mathscr{B})$ is said to be of Lebesgue type if $\mu=\lambda\circ \phi^{-1}$.

For such $\mu$, the problem in the OP then reduces to considering the space $((0,1),\mathcal{B}(0,1),\lambda)$. For any probability measure $\nu$ define $F_\nu(x)=\nu((0,x])$, $0<x<1$, and define the quantile function $Q_\nu:(0,1)\rightarrow\mathbb{R}$ as $$Q_\nu(q)=\inf\{x: F_\nu(x)\geq q\}$$ It can be proven that $$F_\nu(x)\geq q \quad\text{iff}\quad Q_\nu(q)\leq x$$ Hence $$\lambda\big(\{q:Q_\nu(q)\leq x\}\big)=\lambda\big(\{q: F_\nu(x)\geq q\}\big)=F_\nu(x)$$ This means that $\lambda\circ Q^{-1} =\nu$. Notice that in fact $Q(0,1)\subset(0,1)$; thus the push forward of $\lambda$ by $Q$ is $\nu$. Notice that whether $\nu\ll\lambda$ or not does not play a role here.

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Edit: If $\mu$ is not equivalent to the Lebesgue measure on $((0,1),\mathcal{B}(0,1))$, then the answer may be no. Consider the measure $\mu=\frac12\lambda+\frac12\delta_{1/2}$ and let $\mu=\lambda$. Clearly $\lambda\ll\mu$. Suppose $T:((0,1),\mathcal{B}(0,1))\rightarrow((0,1),\mathcal{B}(0,1))$ and let $p_0=T(1/2)$. Then $$0=\lambda(\{p_0\})<\frac12\leq \mu\big(T^{-1}(\{p_0\})\big)$$ which shows that no push forward of $\mu$ is $\nu$.

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More interestingly perhaps are the cases:

1. $X$ is a topological locally compact Hausdorff group equipped with the Borel $\sigma$-algebra $\mathscr{B}(X)$, and that admits a finite Haar measure, say $\mu(X)=1$. Is any Borel measure $\nu$ on $(X,\mathscr{B}(X))$ a push forward of $\mu$? The answer in this case escapes me, even for under the assumption $\nu\ll \mu$.
2. $M$ is a compact manifold of dimension $m\geq 1$ embedded in $\mathbb{R}^n$, for some $n\geq m$, one my also look at the Hausdorff measure $\mu=H^m$ restricted to $(M,\mathscr{B}(M))$. In this case, the answer to the same question above also escapes me.