Wasserstein distance from a Dirac measure

http://en.wikipedia.org/wiki/Wasserstein_metric

I would like to prove that $$W^1(μ,δx_0)=∫d(x_0,y) μ(dy)$$ let $$γ∈Γ(μ,δx_0)$$ Can we say that it is the product of its marginal distributions $$γ=μ×δx_0 $$ and then apply Fubini's theorem? But being the product of its marginal distributions would mean $$γ({(x,y)/x∈A ,y∈B})=γ({(x,y)/x∈A})*γ({(x,y)/y∈B}) $$ which doesn't hold in general. So how can we use marginal distributions to compute $$∫d(x,y) γ(dx,dy)$$

Solution 1:

Indeed, in general, if $\mu_1$ and $\mu_2$ are two probability measures and $\nu$ has marginals $\mu_1$ and $\mu_2$ respectively, it is not true that $\nu=\mu_1\times\mu_2$. However, in the case $\mu_2=\delta_{x_0}$, this becomes true.

Solution 2:

$W(\mu, \delta_a) = \inf_{X,Y|\operatorname{law}(X)=\mu,\;\operatorname{law}(Y)=\delta_a}\mathbb E d(X,Y)=\inf_{X|\operatorname{law}(X)=\mu}\mathbb Ed(X,a)=\mathbb E_{X\sim \mu}d(X,a)$.