# coupling of the projection and projection of the coupling

Let $S$ be a countable space, $\Omega_s$ be a finite space and $\Omega_S = \prod_{s \in S} \Omega_s$ be the product space, equiped with the product topology. Let $\mu^1$ and $\mu^2$ be two probability measures defined in the product space. Let's define $\mu^1|_s$, $\mu^2|_s$ their projection on the element $s \in S$, $c = \mu^1 \, \, t \, \, \mu^2$ a coupling between the two measures, $c_s$ the same coupling, but between the projections $\mu^1|_s$ and $\mu^2|_s$, $c|_s$ the projection of the coupling $c$ on $s$.

How to prove that $$c|_s = c_s ?$$

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We have to show that $c\mid_s(A\times B)=c_s(A\times B)$ for all $A$ and $B$ subsets of $\Omega_s$. We have $$c_s(A\times B)=\mu^1(\pi_s^{-1}A)\cdot \mu^2(\pi_s^{-1}B)$$ and $$c\mid_s(A\times B)=\mu^1\times \mu^2(\{(x,y)\in \Omega_S^2, (\pi_s(x),\pi_s(y))\in A\times B),$$ which is what we want.
Then we conclude using the fact that finite unions of products of measurable sets form a generating algebra (which gives the result when $\Omega_s$ is not necessarily finite).
Sorry I wrote wrong because I wanted to mean that the coupling is not just the product measure, but it is a general coupling between the two measures. In this case how can I cannot write any more that $c_s(A \times B) = \mu^1 (\pi_s ^{-1} A) \mu^2 (\pi_s^{-1}B)$, right? How can I prove it, then? – QuantumLogarithm Nov 23 '12 at 10:56
So my question is: what is the mathematical justification of writing $$( \mu^1 \, \, t \, \,\mu^2 ) \circ \pi^{-1}_s = (\mu^1 \circ \pi^{-1}_s)\,\, t \, \, (\mu^2 \circ \pi^{-1}_s),$$ where t is the operator of a certain coupling (the product is only a special one). – QuantumLogarithm Nov 23 '12 at 11:10