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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    $\begingroup$ Why do you think this should be true? $\endgroup$ Nov 23, 2012 at 20:22
  • $\begingroup$ Can you define the coupling? Then it might be possible for someone who only knew what a measure on a product space to think about this problem. $\endgroup$ Nov 24, 2012 at 9:11
  • $\begingroup$ So the coupling is a measure on the product space with marginals equal to $\mu_i$. Good. Then the next logical question is what is the meaning of $C \{ \omega^1,\omega^2 \in X s.t. \omega^1_s \neq \omega^2_s \}$? From the notation, it seems this is a set of subsets of $X$, but $C$ is a measure on the product space$X\times X$. What is the meaning of this set notation? $\endgroup$ Nov 24, 2012 at 14:03
  • $\begingroup$ Well it is a subset of couples of elements in $X$. What I meant is: $\{ \omega^1, \omega^2 \in X s.t. \omega_s^1 \neq \omega_s^2 \} = \{ (\omega^1, \omega^2) \in (X \times X) s.t. \omega_s^1 \neq \omega_2^2\}$. $\endgroup$ Nov 24, 2012 at 14:41
  • $\begingroup$ I will modify the notation to make it more clear. $\endgroup$ Nov 24, 2012 at 14:42

2 Answers 2


Here is what seems to me a good beginning, if the argument can be carried all the way.

We have that $dist(\mu_1,\mu_2):=\inf_c \sup_s f(s,c)$,

where $f(s,c):= c( \{ (\omega^1,\omega^2) \in X\times X : \omega^1_s \neq \omega^2_s \} )$.

Then for any indices $ \{ s_i \}$ for $i$ from $1$ to $r$, we have

$\sup_s f(s,c) \geq f(s_i,c)$, so that

$r \sup_s f(s,c) \geq \sum_{i=1}^r f(s_i,c)$,

and taking the infimum in $c$ we have

$r dist(\mu_1,\mu_2) \geq \inf_c \sum_{i=1}^r f(s_i,c)$.

But $c$ is a measure, and so we should be able to arrange the right hand side as $c$ of some set where the components differ in at most $r$ spots, the infimum of which can be bounded below by the difference of the measures.

For this to work, maybe the set in the definition of $rift$ should have ONLY $\omega^1_s \neq \omega^2_s$?


For any specific coupling $C$, the left hand side can be bounded by the probability that $\omega^1_s \ne \omega^2_s$ for some $s\in I$, where $I$ is the set of indices on which the cylinder $A$ is defined. This is a most $r$ times $rift(C)$.


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