# If $\gamma$ is a coupling of $\delta_x$ and $\delta_y$, can we show that $\int f\:{\rm d}\gamma=f(x,y)$?

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 $$\pi_2(\gamma)=\delta_y$$. • I guess that the closest you can come is the disintegration theorem, which relates the "product measure" with what the marginal does on the fibers. Jun 3 '20 at 10:50 • @Iwassuspendedfortalking Thank you for your comment. I think I was a bit hasty. I thought the claim would be almost trivial for$f=1_{A_1\times A_2}$,$(A_1,A_2)\in\mathcal E^2$, and hence follow in the usual way for arbitrary$f. However, it's actually not that clear to me that it holds in that case. Any idea how we can show it? Jun 3 '20 at 14:08 ## 1 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 $$\pi=\delta_x\otimes \delta_y$$. • Thank you for your answer. Everything you wrote is correct, but let me note that you've mistakenly wroteX$instead of$E$in the line where you consider$x\in A$and$y\in B$. Moreover, you are talking about "Borel sets" which doesn't make sense, since$E$is not a topological space and hence$\mathcal E$is not a Borel$\sigma$-algebra. But you didn't use that; the important thing is that$\mathcal E_1\otimes\mathcal E_2=\sigma(\mathcal E_1\times\mathcal E_2)$. Jun 3 '20 at 16:09 • And it's worth to mention that the result clearly extends to the case where$\delta_x$and$\delta_y\$ are Dirac measures on different measurable spaces. Jun 3 '20 at 16:09
• I've got a subsequent question: math.stackexchange.com/q/3735409/47771. Maybe you can take a look. Jun 26 '20 at 13:59