If $Z$ is $sigma(X,Y)$-measurable, is there a measurable $f$ with $Z=f(X,Y)$?
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Let
$(Omega_i,mathcal A_i)$ be a measurable space- $X:Omega_1toOmega_2$
- $Z:Omega_1tomathbb R$
It's easy to show that $Z$ is $sigma(X)$-measurable if and only if there is a $mathcal A_2$-measurable $f:Omega_2tomathbb R$ with $$Z=f(X).tag1$$ Now, suppose $Y:Omega_1toOmega_3$. Are we able to show that if $Z$ is $sigma(X,Y)$-measurable, then there is a $mathcal A_2otimesmathcal A_3$-measurable $g:Omega_2timesOmega_3tomathbb R$ with $$Z=g(X,Y)tag2?$$
(By the way: Is it possible to generalize $(1)$ to Banach space valued strongly measurable $Z,f$?)
probability-theory measure-theory measurable-functions
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up vote
0
down vote
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Let
$(Omega_i,mathcal A_i)$ be a measurable space- $X:Omega_1toOmega_2$
- $Z:Omega_1tomathbb R$
It's easy to show that $Z$ is $sigma(X)$-measurable if and only if there is a $mathcal A_2$-measurable $f:Omega_2tomathbb R$ with $$Z=f(X).tag1$$ Now, suppose $Y:Omega_1toOmega_3$. Are we able to show that if $Z$ is $sigma(X,Y)$-measurable, then there is a $mathcal A_2otimesmathcal A_3$-measurable $g:Omega_2timesOmega_3tomathbb R$ with $$Z=g(X,Y)tag2?$$
(By the way: Is it possible to generalize $(1)$ to Banach space valued strongly measurable $Z,f$?)
probability-theory measure-theory measurable-functions
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let
$(Omega_i,mathcal A_i)$ be a measurable space- $X:Omega_1toOmega_2$
- $Z:Omega_1tomathbb R$
It's easy to show that $Z$ is $sigma(X)$-measurable if and only if there is a $mathcal A_2$-measurable $f:Omega_2tomathbb R$ with $$Z=f(X).tag1$$ Now, suppose $Y:Omega_1toOmega_3$. Are we able to show that if $Z$ is $sigma(X,Y)$-measurable, then there is a $mathcal A_2otimesmathcal A_3$-measurable $g:Omega_2timesOmega_3tomathbb R$ with $$Z=g(X,Y)tag2?$$
(By the way: Is it possible to generalize $(1)$ to Banach space valued strongly measurable $Z,f$?)
probability-theory measure-theory measurable-functions
Let
$(Omega_i,mathcal A_i)$ be a measurable space- $X:Omega_1toOmega_2$
- $Z:Omega_1tomathbb R$
It's easy to show that $Z$ is $sigma(X)$-measurable if and only if there is a $mathcal A_2$-measurable $f:Omega_2tomathbb R$ with $$Z=f(X).tag1$$ Now, suppose $Y:Omega_1toOmega_3$. Are we able to show that if $Z$ is $sigma(X,Y)$-measurable, then there is a $mathcal A_2otimesmathcal A_3$-measurable $g:Omega_2timesOmega_3tomathbb R$ with $$Z=g(X,Y)tag2?$$
(By the way: Is it possible to generalize $(1)$ to Banach space valued strongly measurable $Z,f$?)
probability-theory measure-theory measurable-functions
probability-theory measure-theory measurable-functions
asked Nov 16 at 22:11
0xbadf00d
1,86041428
1,86041428
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If $Z=I_{X^{-1}(A)cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{Atimes B}$. Now ${Cin mathcal B(mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) text {for some measurable} f:mathbb R^{2} to mathbb R}$ is a sigma algebra which contains measurable rectangles so it contains all Borel sets in $mathbb R^{2}$. It follow now that the result is true for any simple function $Z$ measurable w.r.t. $sigma (X,Y)$. Hence the same holde for non-negative meaurble functions. If $Z_n=f_n(X,Y)$ for all $n$ and $Z_n to Z$ then $Z=limsup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $Z=I_{X^{-1}(A)cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{Atimes B}$. Now ${Cin mathcal B(mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) text {for some measurable} f:mathbb R^{2} to mathbb R}$ is a sigma algebra which contains measurable rectangles so it contains all Borel sets in $mathbb R^{2}$. It follow now that the result is true for any simple function $Z$ measurable w.r.t. $sigma (X,Y)$. Hence the same holde for non-negative meaurble functions. If $Z_n=f_n(X,Y)$ for all $n$ and $Z_n to Z$ then $Z=limsup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
add a comment |
up vote
0
down vote
If $Z=I_{X^{-1}(A)cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{Atimes B}$. Now ${Cin mathcal B(mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) text {for some measurable} f:mathbb R^{2} to mathbb R}$ is a sigma algebra which contains measurable rectangles so it contains all Borel sets in $mathbb R^{2}$. It follow now that the result is true for any simple function $Z$ measurable w.r.t. $sigma (X,Y)$. Hence the same holde for non-negative meaurble functions. If $Z_n=f_n(X,Y)$ for all $n$ and $Z_n to Z$ then $Z=limsup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
add a comment |
up vote
0
down vote
up vote
0
down vote
If $Z=I_{X^{-1}(A)cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{Atimes B}$. Now ${Cin mathcal B(mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) text {for some measurable} f:mathbb R^{2} to mathbb R}$ is a sigma algebra which contains measurable rectangles so it contains all Borel sets in $mathbb R^{2}$. It follow now that the result is true for any simple function $Z$ measurable w.r.t. $sigma (X,Y)$. Hence the same holde for non-negative meaurble functions. If $Z_n=f_n(X,Y)$ for all $n$ and $Z_n to Z$ then $Z=limsup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.
If $Z=I_{X^{-1}(A)cap Y^{-1}(B)}$ then $Z=f(X,Y)$ where $f=I_{Atimes B}$. Now ${Cin mathcal B(mathbb R^{2}):I_{{(X,Y)^{-1}}(C)}=f(X,Y) text {for some measurable} f:mathbb R^{2} to mathbb R}$ is a sigma algebra which contains measurable rectangles so it contains all Borel sets in $mathbb R^{2}$. It follow now that the result is true for any simple function $Z$ measurable w.r.t. $sigma (X,Y)$. Hence the same holde for non-negative meaurble functions. If $Z_n=f_n(X,Y)$ for all $n$ and $Z_n to Z$ then $Z=limsup f_n(X,Y)$. Now write $Z$ as $Z^{+}-Z^{-}$ to complete the proof. The same argument works for strongly measurable Banach valued $Z$.
edited Nov 17 at 11:36
answered Nov 16 at 23:21
Kavi Rama Murthy
42.8k31751
42.8k31751
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
add a comment |
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
My problem with the Banach space case is that there is no $operatorname{lim sup}$.
– 0xbadf00d
Nov 17 at 10:49
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
Something is wrong with your definition of your $sigma$-algebra. Why $Cinmathcal B(mathbb R^2)$?
– 0xbadf00d
Nov 17 at 10:54
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
You are right. I typed the definition wrongly. Please see the revised answer. @Oxbadf00d
– Kavi Rama Murthy
Nov 17 at 11:36
add a comment |
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