Existence of a function that generates a Borel set
How to prove that there exists $xi: Omega to mathbb R$, which is not a random variable such that for all $x in mathbb R$ $xi^{-1}(x) = { omega | xi(omega) = x }$ is a Borel set?
probability-theory random-variables borel-sets
asked Dec 11 '18 at 5:46
Lisa
255
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How to prove that there exists $xi: Omega to mathbb R$, which is not a random variable such that for all $x in mathbb R$ $xi^{-1}(x) = { omega | xi(omega) = x }$ is a Borel set?
probability-theory random-variables borel-sets
asked Dec 11 '18 at 5:46
Lisa
255
What is $Omega$ and what do you mean by a Borel set in $Omega$?
– Kavi Rama Murthy
Dec 11 '18 at 6:06
$Omega$ is $Omega$ from some $probability space$ and $(Omega, mathcal F)$ is measurable. @KaviRamaMurthy
– Lisa
Dec 11 '18 at 6:21
add a comment |
How to prove that there exists $xi: Omega to mathbb R$, which is not a random variable such that for all $x in mathbb R$ $xi^{-1}(x) = { omega | xi(omega) = x }$ is a Borel set?
probability-theory random-variables borel-sets
asked Dec 11 '18 at 5:46
Lisa
255
How to prove that there exists $xi: Omega to mathbb R$, which is not a random variable such that for all $x in mathbb R$ $xi^{-1}(x) = { omega | xi(omega) = x }$ is a Borel set?
probability-theory random-variables borel-sets
probability-theory random-variables borel-sets
asked Dec 11 '18 at 5:46
Lisa
255
asked Dec 11 '18 at 5:46
Lisa
255
asked Dec 11 '18 at 5:46
Lisa
255
asked Dec 11 '18 at 5:46
Lisa
255
asked Dec 11 '18 at 5:46
Lisa
255
255
What is $Omega$ and what do you mean by a Borel set in $Omega$?
– Kavi Rama Murthy
Dec 11 '18 at 6:06
$Omega$ is $Omega$ from some $probability space$ and $(Omega, mathcal F)$ is measurable. @KaviRamaMurthy
– Lisa
Dec 11 '18 at 6:21
add a comment |
What is $Omega$ and what do you mean by a Borel set in $Omega$?
– Kavi Rama Murthy
Dec 11 '18 at 6:06
$Omega$ is $Omega$ from some $probability space$ and $(Omega, mathcal F)$ is measurable. @KaviRamaMurthy
– Lisa
Dec 11 '18 at 6:21
What is $Omega$ and what do you mean by a Borel set in $Omega$?
– Kavi Rama Murthy
Dec 11 '18 at 6:06
What is $Omega$ and what do you mean by a Borel set in $Omega$?
– Kavi Rama Murthy
Dec 11 '18 at 6:06
$Omega$ is $Omega$ from some $probability space$ and $(Omega, mathcal F)$ is measurable. @KaviRamaMurthy
– Lisa
Dec 11 '18 at 6:21
$Omega$ is $Omega$ from some $probability space$ and $(Omega, mathcal F)$ is measurable. @KaviRamaMurthy
– Lisa
Dec 11 '18 at 6:21
add a comment |
1 Answer
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Let $Omega=mathbb{R}$, $mathcal{F}=sigma({omega})$, and $xi(omega):=omega$. Then $xi^{-1}({x})={x}inmathcal{F}$. However, for any Borel set $B$ that is not countable or co-countable $xi^{-1}(B)notinmathcal{F}$ (e.g. $B=(-infty,0]$).
Let $Omega=[0,1]$, $mathcal{F}=mathcal{B}([0,1])$, and $xi(omega):=omega+2cdot1{omeganotin N}$, where $Nsubset [0,1]$ is a non-Borel set. Then $xi^{-1}({x})inmathcal{F}$. However, $xi^{-1}([0,1])=Vnotin mathcal{F}$.
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Let $Omega=mathbb{R}$, $mathcal{F}=sigma({omega})$, and $xi(omega):=omega$. Then $xi^{-1}({x})={x}inmathcal{F}$. However, for any Borel set $B$ that is not countable or co-countable $xi^{-1}(B)notinmathcal{F}$ (e.g. $B=(-infty,0]$).
Let $Omega=[0,1]$, $mathcal{F}=mathcal{B}([0,1])$, and $xi(omega):=omega+2cdot1{omeganotin N}$, where $Nsubset [0,1]$ is a non-Borel set. Then $xi^{-1}({x})inmathcal{F}$. However, $xi^{-1}([0,1])=Vnotin mathcal{F}$.
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
add a comment |
Let $Omega=mathbb{R}$, $mathcal{F}=sigma({omega})$, and $xi(omega):=omega$. Then $xi^{-1}({x})={x}inmathcal{F}$. However, for any Borel set $B$ that is not countable or co-countable $xi^{-1}(B)notinmathcal{F}$ (e.g. $B=(-infty,0]$).
Let $Omega=[0,1]$, $mathcal{F}=mathcal{B}([0,1])$, and $xi(omega):=omega+2cdot1{omeganotin N}$, where $Nsubset [0,1]$ is a non-Borel set. Then $xi^{-1}({x})inmathcal{F}$. However, $xi^{-1}([0,1])=Vnotin mathcal{F}$.
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
add a comment |
Let $Omega=mathbb{R}$, $mathcal{F}=sigma({omega})$, and $xi(omega):=omega$. Then $xi^{-1}({x})={x}inmathcal{F}$. However, for any Borel set $B$ that is not countable or co-countable $xi^{-1}(B)notinmathcal{F}$ (e.g. $B=(-infty,0]$).
Let $Omega=[0,1]$, $mathcal{F}=mathcal{B}([0,1])$, and $xi(omega):=omega+2cdot1{omeganotin N}$, where $Nsubset [0,1]$ is a non-Borel set. Then $xi^{-1}({x})inmathcal{F}$. However, $xi^{-1}([0,1])=Vnotin mathcal{F}$.
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
Let $Omega=mathbb{R}$, $mathcal{F}=sigma({omega})$, and $xi(omega):=omega$. Then $xi^{-1}({x})={x}inmathcal{F}$. However, for any Borel set $B$ that is not countable or co-countable $xi^{-1}(B)notinmathcal{F}$ (e.g. $B=(-infty,0]$).
Let $Omega=[0,1]$, $mathcal{F}=mathcal{B}([0,1])$, and $xi(omega):=omega+2cdot1{omeganotin N}$, where $Nsubset [0,1]$ is a non-Borel set. Then $xi^{-1}({x})inmathcal{F}$. However, $xi^{-1}([0,1])=Vnotin mathcal{F}$.
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
edited Dec 12 '18 at 0:19
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
answered Dec 11 '18 at 20:09
d.k.o.
8,612528
8,612528
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What is $Omega$ and what do you mean by a Borel set in $Omega$?
– Kavi Rama Murthy
Dec 11 '18 at 6:06
$Omega$ is $Omega$ from some $probability space$ and $(Omega, mathcal F)$ is measurable. @KaviRamaMurthy
– Lisa
Dec 11 '18 at 6:21