I have the following statement from the first Chapter "Couplings and changes of variables" of the book Optimal Transport: Old and New by Cédric Villani
The increasing rearrangement on $\mathbb{R}$. Let $\mu, \nu$ be two probability measures on $\mathbb{R}$; define their cumulative distribution functions by $$ F(x)=\int_{-\infty}^x d \mu, \quad G(y)=\int_{-\infty}^y d \nu . $$ Further define their right-continuous inverses by $$ \begin{aligned} &F^{-1}(t)=\inf \{x \in \mathbb{R} ; \quad F(x)>t\} \\ &G^{-1}(t)=\inf \{y \in \mathbb{R} ; \quad G(y)>t\} \end{aligned} $$ and set $$ T=G^{-1} \circ F . $$ If $\mu$ does not have atoms, then $T_{\#} \mu=\nu$. This rearrangement is quite simple, explicit, as smooth as can be, and enjoys good geometric properties.
Why is that "If $\mu$ does not have atoms, then $T_{\#} \mu=\nu$"? Why do we need that $\mu$ does not have atoms?
Any help is appreciated.
I will use the left-continuous inverse $H^\leftarrow(p)=\inf\{x:H(x)\geq p\}$.
We know that if $Z$ has cdf $H$ (which is right-continuous and nondecreasing) and $U\sim \textrm{Uniform}(0,1)$ then:
You can find these facts in McNeil et al. (Proposition 5.2.(2) and Proposition A.3.(iv)).
If $X\sim \mu$ and it does not have atoms, then this implies it has a continuous cdf $F$. We can state: $$P(T(X)\leq y)=P(G^{\leftarrow}(F(X))\leq y)=P(F(X)\leq G(y))=G(y)$$ Now since $P(T(X)\leq y)=(T_\#\mu)((-\infty,y])$ and the measures $T_\#\mu,\nu$ agree on an $\cap$-stable generator of the Borel sets on the real line, we can state that $T_\#\mu=\nu$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$.