
> 
> On the other hand, if $\mu$ and $\nu$ are completely dependent then $\pi\_{x\_1}=\delta\_{f(x\_1)}$ for some function $f$. Then $W(\pi)=1$.
> 
> 
> 




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Note that
$$
\pi(dx, dy)=\pi\_x(dy)\mu(dx).
$$


If $\pi\_x(dy)=\delta\_{f(x)}(dy)$, then


$$
\pi(dx, dy)=\pi\_x(dy)\delta\_{f(x)}(dy).
$$



What does $\mu$ and $\nu$ "dependent" mean?
It makes no sense to say that two probability measures $\mu$ and $\nu$ are "completely dependent". Dependence (or lack thereof) is a property of random elements, not of probability measures.


Also, a disintegration $(\pi\_x)$ is an attribute, not of probability measures $\mu$ and $\nu$, but of a coupling $\pi$ of $\mu$ and $\nu$. On the other hand, $\mu$ and $(\pi\_x)$ determine $\pi$ and hence $\nu$.


E.g., if $\mu$ and $\nu$ are each the uniform distribution on $[0,1]$, then the formula $\pi\_x=\nu$ for $x\in[0,1]$ determines the "independent" coupling $\pi=\mu\otimes\nu$ of $\mu$ and $\nu$, whereas the formula $\pi\_x=\delta\_{f(x)}$ for $f(x):=x$ and $x\in[0,1]$ determines the "completely dependent" coupling $\pi$ (of $\mu$ and $\nu$) that is the uniform distribution on the diagonal $D:=\{(x,x)\colon x\in[0,1]\}$. Functions $f$ defined by formulas $f(x):=1-x$ and $f(x):=\lfloor2x\rfloor$ for $x\in[0,1]$ provide another two examples of "completely dependent" couplings $\pi$ of $\mu$ and $\nu$, with disintegration $\pi\_x=\delta\_{f(x)}$.


Note also that the author of the linked paper never defines "completely dependent" probability measures, and I have never seen such nonsense elsewhere. Apparently, what the author meant by



> 
> On the other hand, if $\mu$ and $\nu$ are completely dependent then $\pi\_{x\_1}=\delta\_{f(x\_1)}$ for some function $f$. Then $W(\pi)=1$.
> 
> 
> 


is just the following:



> 
> If the coupling $\pi$ of $\mu$ and $\nu$ is such that $\pi\_{x\_1}=\delta\_{f(x\_1)}$ for some Borel function $f$ and $\mu$-almost all $x\_1\in X$, then $W(\pi)=1$.
> 
> 
> 


