Let $(X, \Sigma_X)$ and $(Y, \Sigma_Y)$ be measurable spaces. (I'm happy to restrict this to standard Borel spaces.)

Suppose there are finite measures $\mu$ and $\nu$ on $X$ and a finite kernel $k$ from $X$ to $Y$ (i.e. $k : X \times \Sigma_Y \to \mathbb{R}$).

Suppose $f, g : X \times Y \to \mathbb{R}$ are such that for all $y \in Y$, $$ \int_X f(x, y) \mu(dx) = \int_X g(x, y)\nu(dx)$$

Does the following equality hold? $$ \int_X \int_Y f(x, y) k(x, dy) \mu(dx) = \int_X \int_Y g(x, y) k(x, dy) \nu(dx)$$

EDIT: This is not true in general (see Kavi Rama Murthy's answer). What about when $f$ and $g$ are strictly non-negative and have finite integral over $X \times Y$?

  • $\begingroup$ Can you define that "dy" properly? $\endgroup$ – Yanko Dec 20 '19 at 12:26
  • $\begingroup$ For every $x$, $k(x, -)$ is a (finite) measure on $Y$. $\endgroup$ – daon Dec 20 '19 at 12:27

Not true. Let $X=Y=(-1,1)$ with the Borel sigma algebra. Let $k(x,A)=(1+x) \lambda (A)$, (where $\lambda$ is the Lebesgue measure), $\mu=\lambda$ and $\nu =2\mu$. Let $f(x,y)=x$ and $g(x,y)=xy$. Then $\int f(x,y)d\mu (x)=0=\int g(x,y) d\nu (x)$. But the conclusion does not hold.

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  • $\begingroup$ Is there an example where $f$ and $g$ are strictly positive and have finite integral over the whole of $X \times Y$? $\endgroup$ – daon Dec 21 '19 at 10:46
  • $\begingroup$ @daon In my example change $f(x,y)$ to $x+4$ and change $g(x,y)$ to $xy+2$. Then all your coniditions are satisfied and the conclusion is still false. $\endgroup$ – Kavi Rama Murthy Dec 21 '19 at 11:31

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