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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 $\pi_2(\gamma)=\delta_y$.

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  • $\begingroup$ 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. $\endgroup$ Jun 3 '20 at 10:50
  • $\begingroup$ @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? $\endgroup$
    – 0xbadf00d
    Jun 3 '20 at 14:08
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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 $\pi=\delta_x\otimes \delta_y$.

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  • $\begingroup$ Thank you for your answer. Everything you wrote is correct, but let me note that you've mistakenly wrote $X$ 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 $\mathcal E_1\otimes\mathcal E_2=\sigma(\mathcal E_1\times\mathcal E_2)$. $\endgroup$
    – 0xbadf00d
    Jun 3 '20 at 16:09
  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Jun 3 '20 at 16:09
  • $\begingroup$ I've got a subsequent question: math.stackexchange.com/q/3735409/47771. Maybe you can take a look. $\endgroup$
    – 0xbadf00d
    Jun 26 '20 at 13:59

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