Let $\mu, \nu$ be two positive Borel measures on $\mathbb{R}^d$ with the same mass. A probability measure $\pi$ on $\mathbb{R}^d\times\mathbb{R}^d$ is called a transference plan from $\mu$ to $\nu$ if: $$|\mu|\int_{\mathbb{R}^d} \text{d} \pi(x,\cdot)= \text{d} \mu(x),\quad |\nu|\int_{\mathbb{R}^d} \text{d} \pi(\cdot,y)= \text{d} \nu(x). $$
Now, let $\mu'$ be another positive Borel measure and $\mu'\le \mu$, meaning that there exists a Radon-Nikodym derivative $f\in L^1(\mu)$ such that $0\le f\le 1$ and $\text{d} \mu'(x)=f \text{d}\mu(x) $.
My question is, what is $\nu'$, the image of $\mu'$ under $\pi$?
I encountered this concept in the proof of Proposition 1 of Generalized Wasserstein Distance and its Application to Transport Equations with Source. I tried the following construction: $$ \nu'(B)\triangleq |\mu|\int_{\mathbb{R}^d\times B} f(x)\text{d} \pi(x,y),\quad \forall B\in \mathscr{B}(\mathbb{R}^d),$$ which is indeed a measure less or equal to $\nu$ and with the same mass as $\mu'$. This construction could get me through several results in that paper until Proposition 3, whose proof seemingly requires certain symmetry of the construction. That is, if $\nu'$ is the image of $\mu'$ under $\pi$, then $\mu'$ is the image of $\nu'$ under $\pi$. A simple counterexample shows that my definition is not equipped with this symmetric property: if $\mu=\frac{1}{3}\delta_1+\frac{2}{3}\delta_2, \nu=\frac{3}{4}\delta_1+\frac{1}{4}\delta_2, \pi=\frac{1}{3}\delta_{(1,1)}+\frac{5}{12}\delta_{(2,1)}+\frac{1}{4}\delta_{(2,2)}$ ($\pi$ is optimal in the $2$-Wasserstein distance between $\mu$ and $\nu$) and $\mu'=\frac{1}{6}\delta_{1}$, then $\nu'=\frac{1}{6}\delta_{1}$ but the image of $\nu'$ is $\frac{2}{27}\delta_{1}+\frac{5}{54}\delta_{2}\ne \mu'$.
