- #1
Sarcasticus
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Let (X, [tex]\mathcal{A}[/tex]), (Y, [tex]\mathcal{B}[/tex]) be measurable spaces. A function [tex]K: X \times \mathcal{B} \rightarrow [0, +\infty[/tex]] is called a kernel from (X, [tex]\mathcal{A}[/tex]) to (Y, [tex]\mathcal{B}[/tex]) if
i) for each x in X, the function B [tex]\mapsto[/tex] K(x,B) is a measure on (Y, [tex]\mathcal{B}[/tex]), and
ii) for each B in [tex]\mathcal{B}[/tex], the function x [tex]\mapsto[/tex] K(x,B) is [tex]\mathcal{A}[/tex]-measurable.
Suppose that K is a kernel from (X, [tex]\mathcal{A}[/tex]) to (Y, [tex]\mathcal{B}[/tex]), that [tex]\mu[/tex] is a measure on (X, [tex]\mathcal{A}[/tex]) and that f is a [0, +[tex]\infty[/tex]]-valued [tex]\mathcal{B}[/tex]-measurable function on Y. Show that
a) B [tex]\mapsto \int K(x,B) \mu(dx)[/tex] is a measure on (Y, [tex]\mathcal{B}[/tex])
b) x [tex]\mapsto \int f(y) K(x,dy)[/tex] is an [tex]\mathcal{A}[/tex]-measurable function on X, and
c) if [tex]\nu[/tex] is the measure on (Y, [tex]\mathcal{B}[/tex]) defined in part (a), then
[tex]\int f(y) \nu(dy) = \int [\int f(y) K(x,dy)] \mu (dx)[/tex]
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 [tex]\mathcal{A}[/tex]-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
[tex] \int f(y)\nu(dy) = \int f(y) ( \int K(x,dy) \mu(dx) )[/tex]
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 [tex]\mathcal{A}[/tex]-measurable, then
[tex] f(y) \int K(x,dy) \mu(dx) = \int f(y) K(x,dy) \mu(dx) [/tex]
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!
i) for each x in X, the function B [tex]\mapsto[/tex] K(x,B) is a measure on (Y, [tex]\mathcal{B}[/tex]), and
ii) for each B in [tex]\mathcal{B}[/tex], the function x [tex]\mapsto[/tex] K(x,B) is [tex]\mathcal{A}[/tex]-measurable.
Suppose that K is a kernel from (X, [tex]\mathcal{A}[/tex]) to (Y, [tex]\mathcal{B}[/tex]), that [tex]\mu[/tex] is a measure on (X, [tex]\mathcal{A}[/tex]) and that f is a [0, +[tex]\infty[/tex]]-valued [tex]\mathcal{B}[/tex]-measurable function on Y. Show that
a) B [tex]\mapsto \int K(x,B) \mu(dx)[/tex] is a measure on (Y, [tex]\mathcal{B}[/tex])
b) x [tex]\mapsto \int f(y) K(x,dy)[/tex] is an [tex]\mathcal{A}[/tex]-measurable function on X, and
c) if [tex]\nu[/tex] is the measure on (Y, [tex]\mathcal{B}[/tex]) defined in part (a), then
[tex]\int f(y) \nu(dy) = \int [\int f(y) K(x,dy)] \mu (dx)[/tex]
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 [tex]\mathcal{A}[/tex]-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
[tex] \int f(y)\nu(dy) = \int f(y) ( \int K(x,dy) \mu(dx) )[/tex]
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 [tex]\mathcal{A}[/tex]-measurable, then
[tex] f(y) \int K(x,dy) \mu(dx) = \int f(y) K(x,dy) \mu(dx) [/tex]
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!