# Inequality between product measure and its projection

$\newcommand{\smin}{\setminus} \newcommand{\sset}{\subseteq}$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

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 $$\gamma(X\times Y\setminus A\times B)\leq \mu(X\setminus A) + \nu(Y\setminus B)$$