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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,

$$rift(C) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } \mbox{ } \lbrace ( \omega^1, \omega^2 ) \in X \times X \ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace \mbox{ } \mbox{ } \rbrace.$$

Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.

How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$

Definition: the coupling is not necessarily the product between the two measures. Let's consider two measures $\mu^1$ and $\mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure 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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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,

$$rift(C) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in X\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) rbrace \mbox{ } \mbox{ } \rbrace.$$

Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.

How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$

Definition: the coupling is not necessarily the product between the two measures. Let's consider two measures $\mu^1$ and $\mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure 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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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,

$$rift(crift(C) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in X\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) \mbox{ } \mbox{ } \rbrace.$$

Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.

How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$

Definition: the coupling is not necessarily the product between the two measures. Let's consider two measures $\mu^1$ and $\mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure 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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