Doob Dynkin lemma, uniqueness of the Borel function
For $X$ and $Y$ defined on some probability space $(Omega,mathcal{F},mathbf{P})$, $Y$ is $sigma(X)$-measurable iff there is a Borel function $f$ such that $Y=f(X)$.
My question is, given $X$ and $Y$ and knowing that $Y$ being $sigma(X)$-measurable, is the Borel function $f$ unique? Can there be more than one such Borel functions fulfill the condition?
probability-theory measure-theory
asked Nov 19 '18 at 1:10
lychtalent
619
add a comment |
For $X$ and $Y$ defined on some probability space $(Omega,mathcal{F},mathbf{P})$, $Y$ is $sigma(X)$-measurable iff there is a Borel function $f$ such that $Y=f(X)$.
My question is, given $X$ and $Y$ and knowing that $Y$ being $sigma(X)$-measurable, is the Borel function $f$ unique? Can there be more than one such Borel functions fulfill the condition?
probability-theory measure-theory
asked Nov 19 '18 at 1:10
lychtalent
619
add a comment |
For $X$ and $Y$ defined on some probability space $(Omega,mathcal{F},mathbf{P})$, $Y$ is $sigma(X)$-measurable iff there is a Borel function $f$ such that $Y=f(X)$.
My question is, given $X$ and $Y$ and knowing that $Y$ being $sigma(X)$-measurable, is the Borel function $f$ unique? Can there be more than one such Borel functions fulfill the condition?
probability-theory measure-theory
asked Nov 19 '18 at 1:10
lychtalent
619
For $X$ and $Y$ defined on some probability space $(Omega,mathcal{F},mathbf{P})$, $Y$ is $sigma(X)$-measurable iff there is a Borel function $f$ such that $Y=f(X)$.
My question is, given $X$ and $Y$ and knowing that $Y$ being $sigma(X)$-measurable, is the Borel function $f$ unique? Can there be more than one such Borel functions fulfill the condition?
probability-theory measure-theory
probability-theory measure-theory
asked Nov 19 '18 at 1:10
lychtalent
619
asked Nov 19 '18 at 1:10
lychtalent
619
edited Nov 19 '18 at 10:23
asked Nov 19 '18 at 1:10
lychtalent
619
asked Nov 19 '18 at 1:10
lychtalent
619
asked Nov 19 '18 at 1:10
lychtalent
619
619
add a comment |
add a comment |
1 Answer
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oldest
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It helps to rephrase this question in terms of pure measure theory: let measurable $X,YinOmegatoPsi$ be such that $Y=fcirc X$ for some Borel $f$. Is $f$ unique?
This immediately makes it clear that Borelness is not relevant to this problem. The answer is the well-known theory of lifts in the category of sets.
If $X$ is not surjective, the answer is clearly no; we can make arbitrary changes to $f$ on the complement of $text{im}{(X)}$.
If $X$ is surjective, then $f$ is unique by linearity. For, suppose $Y=fcirc X=gcirc X$; then, by subtracting, we have $(f-g)circ X=0$. As $X$ is surjective, $f-g=0$.
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It helps to rephrase this question in terms of pure measure theory: let measurable $X,YinOmegatoPsi$ be such that $Y=fcirc X$ for some Borel $f$. Is $f$ unique?
This immediately makes it clear that Borelness is not relevant to this problem. The answer is the well-known theory of lifts in the category of sets.
If $X$ is not surjective, the answer is clearly no; we can make arbitrary changes to $f$ on the complement of $text{im}{(X)}$.
If $X$ is surjective, then $f$ is unique by linearity. For, suppose $Y=fcirc X=gcirc X$; then, by subtracting, we have $(f-g)circ X=0$. As $X$ is surjective, $f-g=0$.
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
add a comment |
It helps to rephrase this question in terms of pure measure theory: let measurable $X,YinOmegatoPsi$ be such that $Y=fcirc X$ for some Borel $f$. Is $f$ unique?
This immediately makes it clear that Borelness is not relevant to this problem. The answer is the well-known theory of lifts in the category of sets.
If $X$ is not surjective, the answer is clearly no; we can make arbitrary changes to $f$ on the complement of $text{im}{(X)}$.
If $X$ is surjective, then $f$ is unique by linearity. For, suppose $Y=fcirc X=gcirc X$; then, by subtracting, we have $(f-g)circ X=0$. As $X$ is surjective, $f-g=0$.
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
add a comment |
It helps to rephrase this question in terms of pure measure theory: let measurable $X,YinOmegatoPsi$ be such that $Y=fcirc X$ for some Borel $f$. Is $f$ unique?
This immediately makes it clear that Borelness is not relevant to this problem. The answer is the well-known theory of lifts in the category of sets.
If $X$ is not surjective, the answer is clearly no; we can make arbitrary changes to $f$ on the complement of $text{im}{(X)}$.
If $X$ is surjective, then $f$ is unique by linearity. For, suppose $Y=fcirc X=gcirc X$; then, by subtracting, we have $(f-g)circ X=0$. As $X$ is surjective, $f-g=0$.
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
It helps to rephrase this question in terms of pure measure theory: let measurable $X,YinOmegatoPsi$ be such that $Y=fcirc X$ for some Borel $f$. Is $f$ unique?
This immediately makes it clear that Borelness is not relevant to this problem. The answer is the well-known theory of lifts in the category of sets.
If $X$ is not surjective, the answer is clearly no; we can make arbitrary changes to $f$ on the complement of $text{im}{(X)}$.
If $X$ is surjective, then $f$ is unique by linearity. For, suppose $Y=fcirc X=gcirc X$; then, by subtracting, we have $(f-g)circ X=0$. As $X$ is surjective, $f-g=0$.
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
answered Nov 19 '18 at 1:22
Jacob Manaker
1,119414
1,119414
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
add a comment |
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
Thanks! This is an excellent answer!
– lychtalent
Nov 19 '18 at 10:23
add a comment |
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