Let $X$ and $Y$ spaces endowed with measures $\mu_x$ and $\mu_y$ defined on set semirings $\mathfrak{S}_x$ and $\mathfrak{S}_y$ and let $A\subset X\times Y$ be a subset of $X\times Y$ such that $(\mu_x\otimes\mu_y)(A)$, where $\mu_x\otimes\mu_y$ is the Lebesgue extension of measure $\mu_x\times\mu_y$ definined by $(\mu_x\times\mu_y)(S\times S')=\mu_x(S)\mu_y(S')$ for $S\in\mathfrak{S}_x$, $S'\in\mathfrak{S}_y$. Let $B=\{(y,x)\in Y\times X:(x,y)\in A\}$.

I understand, by using the definition of Lebesgue measure and the obvious fact that $\mu_x(S)\mu_y(S')=\mu_y(S')\mu_x(S)$, that $$(\mu_x\otimes\mu_y)(A)=(\mu_y\otimes\mu_x)(B)$$where $\mu_y\otimes\mu_x$ is defined alalogously to $\mu_x\otimes\mu_y$. That would also have the consequence that $\int_A f(x,y)d\mu_x\otimes\mu_y=\int_B f(x,y)d\mu_y\otimes\mu_x$.

Is what I am saying correct? I apologise if the issue is trivial, but I consider it worth being clear and I do not find it explicitly stated in the texts I have read. Many thanks for any answer!!!

  • 2
    $\begingroup$ This looks like a change of variables (transformation of measure) formula. Note that $1_A(x,y)=1_B(y,x)$. It is true if you interpret the right hand side as $$\int _B f(x,y)d(\mu_y\otimes\mu_x) (y,x).$$ $\endgroup$ – Jonas Dahlbæk Oct 24 '14 at 8:48

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