Measure theory: kernel mapping

[Sarcasticus]

Let (X, $$\mathcal{A}$$), (Y, $$\mathcal{B}$$) be measurable spaces. A function $$K: X \times \mathcal{B} \rightarrow [0, +\infty$$] is called a kernel from (X, $$\mathcal{A}$$) to (Y, $$\mathcal{B}$$) if

i) for each x in X, the function B $$\mapsto$$ K(x,B) is a measure on (Y, $$\mathcal{B}$$), and
ii) for each B in $$\mathcal{B}$$, the function x $$\mapsto$$ K(x,B) is $$\mathcal{A}$$-measurable.

Suppose that K is a kernel from (X, $$\mathcal{A}$$) to (Y, $$\mathcal{B}$$), that $$\mu$$ is a measure on (X, $$\mathcal{A}$$) and that f is a [0, +$$\infty$$]-valued $$\mathcal{B}$$-measurable function on Y. Show that

a) B $$\mapsto \int K(x,B) \mu(dx)$$ is a measure on (Y, $$\mathcal{B}$$)

b) x $$\mapsto \int f(y) K(x,dy)$$ is an $$\mathcal{A}$$-measurable function on X, and

c) if $$\nu$$ is the measure on (Y, $$\mathcal{B}$$) defined in part (a), then
$$\int f(y) \nu(dy) = \int [\int f(y) K(x,dy)] \mu (dx)$$

Solution:
I can show part (a) quite easily, just unwind the definition of measure (countable additive follows from the properties of the kernel.)
I can show part (b) as well; it will end up being a finite sum of $$\mathcal{A}$$-measurable functions (again, from the dual nature of the kernel.)

But I am having the darndest time with part (c). Essentially, I can show that

$$\int f(y)\nu(dy) = \int f(y) ( \int K(x,dy) \mu(dx) )$$

just by definition, but I can't get past that. I was trying to isolate the kernel K(x,dy) within the bracketed integral; and say that since it's $$\mathcal{A}$$-measurable, then

$$f(y) \int K(x,dy) \mu(dx) = \int f(y) K(x,dy) \mu(dx)$$

as f(y) is a real number. But I don't think that's quite right; and definitely not rigorous enough for my tastes.

Any help is greatly appreciated!

 

[lippyka]

Solution:We can prove (c) by showing that \int f(y) \nu(dy) \geq \int [\int f(y) K(x,dy)] \mu (dx) and \int f(y) \nu(dy) \leq \int [\int f(y) K(x,dy)] \mu (dx). For the first inequality, we have \begin{align*}\int f(y) \nu(dy) &= \int \left[ \int K(x,dy) \mu(dx) \right] f(y) dy \\&\leq \int \left[ \int K(x,dy) \mu(dx)f(y) \right] dy \\&= \int \left[ \int f(y)K(x,dy) \mu(dx) \right] dx\end{align*}where we used the fact that $K(x,dy)$ is a measure for each $x$ and $K(x,dy) \mu(dx)$ is a measure over $dy$ for each fixed $x$, so that it can be pulled out of the integral with $f$. For the second inequality, we have \begin{align*}\int f(y) \nu(dy) &= \int \left[ \int K(x,dy) \mu(dx) \right] f(y) dy \\&\geq \int \left[ \int \min(f(y),K(x,dy)) \mu(dx) \right] dy \\&= \int \left[ \int \min(f(y),K(x,dy)) \mu(dx)f(y) \right] dy \\&= \int \left[ \int f(y) \min(f(y),K(x,dy)) \mu(dx) \right] dx \\&= \int \left[ \int f(y)K(x,dy) \mu(dx) \right] dx\end{align*