Show that the composition of a sub-invariant measure with a sub-Markov kernel is a contraction on $L^p$
$begingroup$
Let
$(Omega,mathcal A)$ be a measurable space
$mu$ be a finite measure on $(Omega,mathcal A)$
$kappa$ be a sub-Markov kernel on $(Omega,mathcal A)$
- $pge1$
I'll denote the composition of $mu$ and $kappa$ by $mukappa$. Moreover, $$kappa f:=intkappa(x,{rm d}y)f(y)$$ and $$mu f:=int f:{rm d}mu$$ whenever the integrals are well-defined. It's easy to see that if $f:Omegatooverline{mathbb R}$ is $mathcal A$-measurable and either nonnegative or bounded, then $kappa f$ is $mathcal A$-measurable.
Now, assume $$mukappalemutag1$$ and let $ginmathcal L^p(mu)$. I want to show that $kappa gin L^p(mu)$ and $$left|kappa gright|_{L^p(mu)}leleft|gright|_{L^p(mu)}tag2.$$
By the aforementioned property, $kappa|g|$ is well-defined and $mathcal A$-measurable. Moreover, by a Fubini like result $$mu(kappa f)=(mukappa)ftag3$$ for all $mathcal A$-measurable $f:Omegato[0,infty]$ and hence $$mu(kappa|g|)=(mukappa)|g|<infty.$$ Thus, $$kappa|g|<infty;;;mutext{-almost everywhere}tag4$$ and hence $kappa g$ is $mu$-almost everywhere well-defined. We can find a sequence $(g_n)_{ninmathbb N}$ of (elementary, simple, step - call it however you like) $mathcal A$-measurable functions with $$g_nxrightarrow{ntoinfty}gtag5$$ and $$left|g_nright|leleft|gright|;;;text{for all }ninmathbb Ntag6.$$ Thus, by dominated convergence, $$kappa g_nxrightarrow{ntoinfty}kappa g;;;mutext{-almost everywhere}tag7.$$ So, there is a $mathcal A$-measurable $f:Omegatomathbb R$ (e.g. $f:=limsup_{ntoinfty}kappa g_n$) with $$f=kappa g;;;mutext{-almost everywhere}tag8.$$ By Jensen's inequality $$left|kappa gright|^plekappa|g|^p$$ and hence $$left|fright|_{L^p(mu)}^plemu(kappa|g|^p)=(mukappa)|g|^p=left|gright|_{L^p(mu)}^ptag9.$$
Did I overcomplicate anything? Please let me know. In any case, my actual question is: Is it correct that we're not able to show that $kappa g$ is well-defined and $mathcal A$-measurable on all of $Omega$? As shown in my proof, I guess that it's only possible to show that there is a $mathcal A$-measurable $mu$-version of $kappa g$. Or am I missing something?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
$endgroup$
add a comment |
$begingroup$
Let
$(Omega,mathcal A)$ be a measurable space
$mu$ be a finite measure on $(Omega,mathcal A)$
$kappa$ be a sub-Markov kernel on $(Omega,mathcal A)$
- $pge1$
I'll denote the composition of $mu$ and $kappa$ by $mukappa$. Moreover, $$kappa f:=intkappa(x,{rm d}y)f(y)$$ and $$mu f:=int f:{rm d}mu$$ whenever the integrals are well-defined. It's easy to see that if $f:Omegatooverline{mathbb R}$ is $mathcal A$-measurable and either nonnegative or bounded, then $kappa f$ is $mathcal A$-measurable.
Now, assume $$mukappalemutag1$$ and let $ginmathcal L^p(mu)$. I want to show that $kappa gin L^p(mu)$ and $$left|kappa gright|_{L^p(mu)}leleft|gright|_{L^p(mu)}tag2.$$
By the aforementioned property, $kappa|g|$ is well-defined and $mathcal A$-measurable. Moreover, by a Fubini like result $$mu(kappa f)=(mukappa)ftag3$$ for all $mathcal A$-measurable $f:Omegato[0,infty]$ and hence $$mu(kappa|g|)=(mukappa)|g|<infty.$$ Thus, $$kappa|g|<infty;;;mutext{-almost everywhere}tag4$$ and hence $kappa g$ is $mu$-almost everywhere well-defined. We can find a sequence $(g_n)_{ninmathbb N}$ of (elementary, simple, step - call it however you like) $mathcal A$-measurable functions with $$g_nxrightarrow{ntoinfty}gtag5$$ and $$left|g_nright|leleft|gright|;;;text{for all }ninmathbb Ntag6.$$ Thus, by dominated convergence, $$kappa g_nxrightarrow{ntoinfty}kappa g;;;mutext{-almost everywhere}tag7.$$ So, there is a $mathcal A$-measurable $f:Omegatomathbb R$ (e.g. $f:=limsup_{ntoinfty}kappa g_n$) with $$f=kappa g;;;mutext{-almost everywhere}tag8.$$ By Jensen's inequality $$left|kappa gright|^plekappa|g|^p$$ and hence $$left|fright|_{L^p(mu)}^plemu(kappa|g|^p)=(mukappa)|g|^p=left|gright|_{L^p(mu)}^ptag9.$$
Did I overcomplicate anything? Please let me know. In any case, my actual question is: Is it correct that we're not able to show that $kappa g$ is well-defined and $mathcal A$-measurable on all of $Omega$? As shown in my proof, I guess that it's only possible to show that there is a $mathcal A$-measurable $mu$-version of $kappa g$. Or am I missing something?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
$endgroup$
add a comment |
$begingroup$
Let
$(Omega,mathcal A)$ be a measurable space
$mu$ be a finite measure on $(Omega,mathcal A)$
$kappa$ be a sub-Markov kernel on $(Omega,mathcal A)$
- $pge1$
I'll denote the composition of $mu$ and $kappa$ by $mukappa$. Moreover, $$kappa f:=intkappa(x,{rm d}y)f(y)$$ and $$mu f:=int f:{rm d}mu$$ whenever the integrals are well-defined. It's easy to see that if $f:Omegatooverline{mathbb R}$ is $mathcal A$-measurable and either nonnegative or bounded, then $kappa f$ is $mathcal A$-measurable.
Now, assume $$mukappalemutag1$$ and let $ginmathcal L^p(mu)$. I want to show that $kappa gin L^p(mu)$ and $$left|kappa gright|_{L^p(mu)}leleft|gright|_{L^p(mu)}tag2.$$
By the aforementioned property, $kappa|g|$ is well-defined and $mathcal A$-measurable. Moreover, by a Fubini like result $$mu(kappa f)=(mukappa)ftag3$$ for all $mathcal A$-measurable $f:Omegato[0,infty]$ and hence $$mu(kappa|g|)=(mukappa)|g|<infty.$$ Thus, $$kappa|g|<infty;;;mutext{-almost everywhere}tag4$$ and hence $kappa g$ is $mu$-almost everywhere well-defined. We can find a sequence $(g_n)_{ninmathbb N}$ of (elementary, simple, step - call it however you like) $mathcal A$-measurable functions with $$g_nxrightarrow{ntoinfty}gtag5$$ and $$left|g_nright|leleft|gright|;;;text{for all }ninmathbb Ntag6.$$ Thus, by dominated convergence, $$kappa g_nxrightarrow{ntoinfty}kappa g;;;mutext{-almost everywhere}tag7.$$ So, there is a $mathcal A$-measurable $f:Omegatomathbb R$ (e.g. $f:=limsup_{ntoinfty}kappa g_n$) with $$f=kappa g;;;mutext{-almost everywhere}tag8.$$ By Jensen's inequality $$left|kappa gright|^plekappa|g|^p$$ and hence $$left|fright|_{L^p(mu)}^plemu(kappa|g|^p)=(mukappa)|g|^p=left|gright|_{L^p(mu)}^ptag9.$$
Did I overcomplicate anything? Please let me know. In any case, my actual question is: Is it correct that we're not able to show that $kappa g$ is well-defined and $mathcal A$-measurable on all of $Omega$? As shown in my proof, I guess that it's only possible to show that there is a $mathcal A$-measurable $mu$-version of $kappa g$. Or am I missing something?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
$endgroup$
Let
$(Omega,mathcal A)$ be a measurable space
$mu$ be a finite measure on $(Omega,mathcal A)$
$kappa$ be a sub-Markov kernel on $(Omega,mathcal A)$
- $pge1$
I'll denote the composition of $mu$ and $kappa$ by $mukappa$. Moreover, $$kappa f:=intkappa(x,{rm d}y)f(y)$$ and $$mu f:=int f:{rm d}mu$$ whenever the integrals are well-defined. It's easy to see that if $f:Omegatooverline{mathbb R}$ is $mathcal A$-measurable and either nonnegative or bounded, then $kappa f$ is $mathcal A$-measurable.
Now, assume $$mukappalemutag1$$ and let $ginmathcal L^p(mu)$. I want to show that $kappa gin L^p(mu)$ and $$left|kappa gright|_{L^p(mu)}leleft|gright|_{L^p(mu)}tag2.$$
By the aforementioned property, $kappa|g|$ is well-defined and $mathcal A$-measurable. Moreover, by a Fubini like result $$mu(kappa f)=(mukappa)ftag3$$ for all $mathcal A$-measurable $f:Omegato[0,infty]$ and hence $$mu(kappa|g|)=(mukappa)|g|<infty.$$ Thus, $$kappa|g|<infty;;;mutext{-almost everywhere}tag4$$ and hence $kappa g$ is $mu$-almost everywhere well-defined. We can find a sequence $(g_n)_{ninmathbb N}$ of (elementary, simple, step - call it however you like) $mathcal A$-measurable functions with $$g_nxrightarrow{ntoinfty}gtag5$$ and $$left|g_nright|leleft|gright|;;;text{for all }ninmathbb Ntag6.$$ Thus, by dominated convergence, $$kappa g_nxrightarrow{ntoinfty}kappa g;;;mutext{-almost everywhere}tag7.$$ So, there is a $mathcal A$-measurable $f:Omegatomathbb R$ (e.g. $f:=limsup_{ntoinfty}kappa g_n$) with $$f=kappa g;;;mutext{-almost everywhere}tag8.$$ By Jensen's inequality $$left|kappa gright|^plekappa|g|^p$$ and hence $$left|fright|_{L^p(mu)}^plemu(kappa|g|^p)=(mukappa)|g|^p=left|gright|_{L^p(mu)}^ptag9.$$
Did I overcomplicate anything? Please let me know. In any case, my actual question is: Is it correct that we're not able to show that $kappa g$ is well-defined and $mathcal A$-measurable on all of $Omega$? As shown in my proof, I guess that it's only possible to show that there is a $mathcal A$-measurable $mu$-version of $kappa g$. Or am I missing something?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
asked Dec 30 '18 at 0:36
0xbadf00d0xbadf00d
1,68041534
1,68041534
add a comment |
add a comment |
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