Difference between two measures 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.
 A: 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 $A, B$.
A: 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 $\widetilde \nu(F)=0$.
