Doob Dynkin lemma, uniqueness of the Borel function

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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?

edited Nov 19 at 10:23

asked Nov 19 at 1:10

lychtalent

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up vote
0
down vote

favorite

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?

edited Nov 19 at 10:23

asked Nov 19 at 1:10

lychtalent

699

add a comment |

up vote
0
down vote

favorite

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?

edited Nov 19 at 10:23

asked Nov 19 at 1:10

lychtalent

699

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

edited Nov 19 at 10:23

asked Nov 19 at 1:10

lychtalent

699

edited Nov 19 at 10:23

asked Nov 19 at 1:10

lychtalent

699

edited Nov 19 at 10:23

asked Nov 19 at 1:10

lychtalent

699

asked Nov 19 at 1:10

lychtalent

699

asked Nov 19 at 1:10

lychtalent

699

add a comment |

1 Answer
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oldest

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2
down vote

accepted

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 at 1:22

Jacob Manaker

1,099414

Thanks! This is an excellent answer!
– lychtalent
Nov 19 at 10:23

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

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 at 1:22

Jacob Manaker

1,099414

Thanks! This is an excellent answer!
– lychtalent
Nov 19 at 10:23

add a comment |

up vote
2
down vote

accepted

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 at 1:22

Jacob Manaker

1,099414

Thanks! This is an excellent answer!
– lychtalent
Nov 19 at 10:23

add a comment |

up vote
2
down vote

accepted

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 at 1:22

Jacob Manaker

1,099414

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 at 1:22

Jacob Manaker

1,099414

answered Nov 19 at 1:22

Jacob Manaker

1,099414

answered Nov 19 at 1:22

Jacob Manaker

1,099414

answered Nov 19 at 1:22

Jacob Manaker

1,099414

Thanks! This is an excellent answer!
– lychtalent
Nov 19 at 10:23

add a comment |

Thanks! This is an excellent answer!
– lychtalent
Nov 19 at 10:23

Thanks! This is an excellent answer!
– lychtalent
Nov 19 at 10:23

add a comment |

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