Let $(E,\mathcal E)$ be a measurable space, $\pi_i$ denote the projection of $E^2$ onto the $i$th coordinate, $\delta_x$ denote the Dirac measure on $(E,\mathcal E)$ at $x$ for $x\in E$ and $\gamma$ be a coupling$^1$ of $\delta_x$ and $\delta_y$ for some $x,y\in E$.
Let $f:E^2\to[0,\infty)$ be $\mathcal E^{\otimes2}$-measurable. How can we show that $$\int f\:{\rm d}\gamma=f(x,y)?\tag1$$
Can we even show that $\gamma$ is the Dirac measure $\delta_{(x,\:y)}$ on $(E^2,\mathcal E^{\otimes 2})$ at $(x,y)? In that case, $(1)$ would clearly follow.
$^1$ i.e. $\gamma$ is a probability measure on $(E^2,\mathcal E^{\otimes2})$ with $\pi_1(\gamma)=\delta_x$ and $\pi_2(\gamma)=\delta_y$.
Best Answer
Let us prove that the only coupling between $\delta_x$ and $\delta_y$ is the product measure $\delta_x\otimes \delta_y$. Consider such a coupling $\pi\in \Pi(\delta_x, \delta_y)$. Since the products of Borel sets form a pi-system, it suffices to check that $\forall (A,B)\in \mathcal B(E)\times \mathcal B(E), \pi(A\times B)=\delta_x\otimes \delta_y(A\times B)=\delta_x(A) \delta_y(B)$.
When $x\notin A$ or $y\notin B$, since $\pi(A\times B)\leq \pi(A\times E) = \delta_x(A)$ and $\pi(A\times B)\leq \pi(E\times B) = \delta_y(B)$, we have $\pi(A\times B) \leq \min(\delta_x(A),\delta_y(B))= 0$, hence $\pi(A\times B) = 0 = \delta_x(A) \delta_y(B)$.
When $x\in A$ and $y\in B$, we have $\begin{aligned}[t] 1-\pi(A\times B) &= \pi((A\times B)^c) = \pi((A^c\times E)\cup (X\times B^c))\\ &\leq \pi(A^c\times E) + \pi(X\times B^c)\\ &= \delta_x(A^c) + \delta_y(B^c) = 0 \end{aligned}$
Hence $\pi(A\times B)=1=\delta_x(A) \delta_y(B)$
This proves $\pi=\delta_x\otimes \delta_y$.