# Equality of certain integrals over product space

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 $$X \times Y$$?

• Can you define that "dy" properly? – Yanko Dec 20 '19 at 12:26
• For every $x$, $k(x, -)$ is a (finite) measure on $Y$. – 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.
• Is there an example where $f$ and $g$ are strictly positive and have finite integral over the whole of $X \times Y$? – daon Dec 21 '19 at 10:46
• @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. – Kavi Rama Murthy Dec 21 '19 at 11:31