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Let $C$ be a coupling between two measures, $C= \mu^1 \mbox{ } t \mbox{ } \mu^2$ ($t$ is the symbol of binary operator of the coupling (I can't find a more proper symbol here)). The measures are both defined on a product space $X = \prod_{s \in S} \Omega_s$. How can I prove that,

$$(\mu^1 \circ \pi^{-1}_s) \mbox{ } t \mbox{ } (\mu^2 \circ \pi_s^{-1} ) = (\mu^1 \mbox{ } t \mbox{ } \mu^2) \circ \pi_s^{-1},$$ that means that the projection on a coordinate $s$ of the coupling between two measures is the coupling of the projections of those measures on the same coordinate?

P.S. the coupling is not necessarily the product between the two measures. The coupling $C = \mu^1 \mbox{ } t \mbox{ } \mu^2$ is any measure $C$ acting on the $\sigma$-algebra of subsets of $X^1 \times X^2$, with $X^i = X$, which has as marginals $\mu^1$ and $\mu^2$, i.e. $\mu^1 = C \circ \pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : X^1 \times X^2 \longrightarrow X^i$.

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Not sure what you mean by "the projection of the coupling is the coupling of the projections"? Do you mean is a coupling of the projections? –  Anthony Quas Nov 23 '12 at 20:10
I want to prove that the projection of a certain coupling between two measures on a certain coordinate of the product space is equal to the coupling between the projection of the two measures on the same coordinate. The coupling operator is the same in the two cases, but in the first case it acts on the two measures and later one considers the projection, while in the second case it acts directly on the projection of the two measures. –  QuantumLogarithm Nov 24 '12 at 16:36
What you call "coupling" is known in ergodic theory as "joining". As far as I understand, in probability coupling is defined for measures on product spaces and requires an additional property: when projected onto the coordinate spaces $X_n^\infty$ (only the coordinates with indices $\ge n$ are retained), the images of the measure $C$ converge to the diagonal measure in total variation. Could you please clarify your definition of coupling. –  R W Nov 24 '12 at 19:39

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