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Suppose that $(\mathscr X, \mathscr B)$ is a measurable space and denote $\Omega = \mathscr X^n$ and let $\mathscr F$ be its product $\sigma$-algebra. Consider two measures $\nu$ and $\tilde \nu$ on $\Omega$ with a condition that $$ |\tilde \nu(B_0\times\dots\times B_n) - \nu(B_0\times\dots\times B_n)|\leq \epsilon $$ for any collection $B_0,\dots,B_n\in \mathscr B$. I wonder how to prove (if it is indeed true) that $$ |\tilde\nu(F) - \nu(F)|\leq \epsilon $$ where $F\in \mathscr F$ is arbitrary. I appreciate any hint.

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The problem has been already solved, but now there is a question: if $|\mu_1(B_1\times B_2)-\mu_2(B_1\times B_2)|\leq 1$ for all $B_1,B_2$ measurable, what can be $\max |\mu_1(F)-\mu_2(F)|$ for $F\in \mathcal B\otimes \mathcal B$? –  Davide Giraudo Mar 29 '12 at 20:52
    
@DavideGiraudo this is indeed an interesting question - but there is another way to think about it: in none of examples measures $\mu,\nu$ are product measures. I will ask a bit more specialized question now –  Ilya Mar 30 '12 at 8:44
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2 Answers

up vote 4 down vote accepted

It is not true. Consider the case $n=2$, ${\mathscr X} = \{0,1\}$, where $\nu$ gives mass $1$ to the points $(0,0)$ and $(1,1)$ and $0$ to the others while $\tilde{\nu}$ gives mass $1$ to $(0,1)$ and $(1,0)$ and $0$ to the others. Then for $F = \{(0,0),(1,1)\}$ we have $\nu(F) - \tilde{\nu}(F) = 2$, but $|\nu(A \times B) - \tilde{\nu}(A \times B)| \le 1$ for all $A, B$.

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Take $X=\{0,1\}$, $n=2$, $\nu=\frac 25\delta_{0,0}+\frac 25\delta(1,0)+\frac 25\delta (1,1)$ and $\widetilde \nu=\delta(0,1)$. We have $|\nu\{(i,j)\}-\widetilde \nu \{(i,j)\}|\leq 1$ for $i,j\in \{0,1\}$, and $|\mu(\{1\}\times X)-\widetilde\mu(\{1\}\times X)|=|\frac 25-1|=3/5\leq 1$, $|\mu(\{0\}\times X)-\widetilde\mu(\{0\}\times X)|=|\frac 45-1|=1/5\leq 1$, $|\mu(X\times \{0\})-\widetilde\mu(X\times \{0\})|=|\frac 25-1|=3/5\leq 1$, $|\mu(X\times \{1\})-\widetilde\mu(X\times \{1\})|=|\frac 45-0|=4/5\leq 1$ and $|\mu(X\times X)-\widetilde\mu(X\times X)|=|\frac 65-1|=1/5\leq 1$.

But if $F=\{(0,0)\}\cup\{(0,1)\}\cup\{(1,1)\}$, we have $\nu(F)=6/5$ but $\widetilde \nu(F)=0$.

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