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$\newcommand{\smin}{\setminus} \newcommand{\sset}{\subseteq}$If $\mu$ is a measure on $X$, $ \nu $ a measure on $Y, \gamma $ a measure on $X \times Y$ s.t. $ \gamma(A \times Y) = \mu (A) $ and $\gamma(X \times B) = \nu(B)$, $ \forall A \sset X $ $\mu$-measurable and $B \sset Y$ $\nu$-measurable then \begin{array}\\ &\gamma ( X \times Y \smin A \times B) \\ & \leq \gamma(((X \smin A) \times Y) \cup (X \times (Y \smin B))) \\ &\leq \gamma((X \smin A) \times Y) + \gamma((X \times Y) \smin B)\\ &= \mu(X \smin A) + \nu (Y \smin B) \end{array}

Does this make sense?

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Just so you know, $X\times Y\setminus A\times B\neq (X\setminus A)\times (Y\setminus B)$ in general. –  Clayton Mar 19 '13 at 0:55

1 Answer 1

up vote 5 down vote accepted

Well, it make sense and actually it's even simpler: note that $$ X\times Y\setminus A\times B \subseteq [(X\setminus A)\times Y]\cup[X\times (Y\setminus B)] $$ which if not obvious, may follow with the help of these pictures: $$ X\times Y\setminus A\times B $$ $$ (X\setminus A)\times Y $$ $$ X\times (Y\setminus B) $$

As a result you get $$ \gamma(X\times Y\setminus A\times B)\leq \mu(X\setminus A) + \nu(Y\setminus B) $$

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