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Yes your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x \in [0,1]$ under the uniform measure.

Let your random variable $M$ realizing $\mu$ be defined as follows. $M(x)\in \Omega$ is the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x.$$

Similarly for $\nu$ with $x+1/2$ mod 1: Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le (x+1/2$ mod 1).

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.

I suppose there could be some measurability concerns about $L$ but at least it works in the cases of discrete, or continuous real-valued, random variables.

Yes, your bound is optimal and here's how to attain it. 

Fix a linear ordering $L$ of $\Omega$. Pick a number $x \in [0,1]$ under the uniform measure and the value of our coupling shall be $(M(x),N(x))$.

Let $M(x)\in \Omega$ be the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x.$$

Let $N(x)\in \Omega$ be the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x+\frac12\mod 1.$$

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.

I suppose there could be some measurability concerns about $L$ but at least it works in the cases of discrete, or continuous real-valued, random variables.

Yes your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x$ in [0,1] under the uniform measure.

Let your random variable realizing $\mu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $a$ such that the $\mu$ probability of an outcome $L$-below $a$ is $\le x$.

Similarly for $\nu$ with $x+1/2$ mod 1:

Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le (x+1/2$ mod 1).

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.

I suppose there could be some measurability concerns about $L$ but at least it works in the cases of discrete, or continuous real-valued, random variables.

Yes your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x \in [0,1]$ under the uniform measure.

Let your random variable $M$ realizing $\mu$ be defined as follows. $M(x)\in \Omega$ is the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x.$$

Similarly for $\nu$ with $x+1/2$ mod 1: Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le (x+1/2$ mod 1).

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.

I suppose there could be some measurability concerns about $L$ but at least it works in the cases of discrete, or continuous real-valued, random variables.

Yes your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x$ in [0,1] under the uniform measure.

Let your random variable realizing $\mu$ be as follows. Your outcome in $\Omega$ is $a$ such that the $\mu$ probability of an outcome $L$-below $a$ is $x$.

Similarly for $\nu$ with $x+1/2$ mod 1:

Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le x+1/2$ mod 1.

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.

Yes your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x$ in [0,1] under the uniform measure.

Let your random variable realizing $\mu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $a$ such that the $\mu$ probability of an outcome $L$-below $a$ is $\le x$.

Similarly for $\nu$ with $x+1/2$ mod 1:

Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le (x+1/2$ mod 1).

This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.

I suppose there could be some measurability concerns about $L$ but at least it works in the cases of discrete, or continuous real-valued, random variables.