Let $\mu,\nu$ be probability measures on a metric space $(E,d)$ endowed with the Borel $\sigma$-algebra and $$\operatorname W_d(\mu,\nu):=\inf_{\gamma\in\mathcal C(\mu,\:\nu)}\int d\:{\rm d}\gamma,$$ where $\mathcal C(\mu,\nu)$ denotes the set of couplings of $\mu$ and $\nu$.
The Kantorovich-Rubinstein duality states that $$\operatorname W_d(\mu,\nu)=\sup_{\substack{f\::\:E\:\to\:\mathbb R\\|f|_{\operatorname{Lip}(d)}\:\le\:1}}\int f\:{\rm d}(\mu-\nu)\tag1,$$ where $$|f|_{\operatorname{Lip}(d)}:=\frac{|f(x)-f(y)|}{d(x,y)}\;\;\;\text{for }f:E\to\mathbb R.$$
The "$\ge$" part is almost trivial. Is there an elegant way to prove the other inequality?
In any case, I've got some questions on the statement.
- Does it matter whether we state it in the given form or if we replace the integral on the right-hand side by its absolute value?
- If $f:E\to\mathbb R$ satisfies $|f|_{\operatorname{Lip}(d)}<\infty$, then $f$ is clearly continuous and hence Borel measurable. But in order for the integral on the right-hand side of $(1)$ to be well-defined, it needs to be integrable (or at least quasi-integrable) as well. Is this the case? Clearly, by the reverse triangle inequality, $$\int|f|\:{\rm d}(\mu-\nu)\le|f|_{\operatorname{Lip}(d)}\int d\:{\rm d}\gamma\tag2,$$ where $\gamma$ is any coupling of $\mu$ and $\nu$. So, if we could show that there exists a coupling such that the integral on the right-hand side of $(2)$ is finite, we could conclude that $f$ is integrable. Is this the case?
- In any case, is $(1)$ equivalent to the same statement with requiring $f$ to be bounded as well?