This posting is related to a recent question asked in [MSE](https://math.stackexchange.com/q/4640948/121671): Suppose $(X,\mathscr{B},\mu)$ is a $\sigma$-finite measure space. If $\nu$ is another measure on $\mathscr{B}$, $\nu(X)=\mu(X)$, and $\nu\ll\mu$, is there a measurable map $T:(X,\mathscr{B})\rightarrow(X,\mathscr{B})$ such that $\nu=\mu\circ T^{-1}$? (Here $\mu\circ T^{-1}$ is the push-forward of $\mu$ by $T$, that is $(\mu\circ T^{-1})(A):=\mu(T^{-1}(A))$).


The answer is yes for Borel spaces $X$ when $(X,\mathscr{B},\mu)$ is isomorphic to the standard Lebesgue space $((0,1),\mathscr{B}((0,1)),\lambda)$ ( $\lambda$ is Lebesgue measure restricted to the unit interval with the Borel $\sigma$-algebra) regardless of whether $\nu\ll\mu$ or not. Indeed, on the standard Lebesgue space, one can use the quantile function $Q\_\nu(q):=\inf\{x: \nu(0,x]\geq q\}$ to get $\nu=\lambda\circ Q^{-1}$.


My question is whether the statement holds another measure $\mu$
on $((0,1),\mathscr{B}(0,1))$ that is not equivalent to $\lambda$ (i.e. either $\mu\not\ll\lambda$ or $\lambda\not\ll\mu$)?



A type of coupling problem I
$\newcommand\de\delta$The answer to your first question is no.


E.g., let $\mu:=2\de\_{1/3}+2\de\_{2/3}$ and $\nu:=\de\_{1/3}+3\de\_{2/3}$, where $\de\_a$ is the Dirac measure with support $\{a\}$. Then all the conditions imposed on $\mu$ and $\nu$ hold. However, $\nu(\{1/3\})=1$ is an odd integer, whereas all the values of the measure $\mu$ are even integers. So, $\nu\ne\mu T^{-1}$ for any $T$.




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MO posts should not have multiple questions. So, I suggest you post the other two questions separately.


