The inverse image of a measurable set under a measurable function is measurable?
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I have a confusion with measurable functions.
I just saw that the statement "The inverse image of a measurable set under a measurable function is measurable" is false with counter-example the function on Cantor set but I know the definition of f measurable is:
Let $f:(X,O_X)to (Y,O_Y)$ with $O_X$ $sigma$-algebra of $X$ and $O_Y$ $sigma$-algebra of $Y$.
$f$ said $(O_X-O_Y)$-measurable function if for all $Bin O_Y f^-1(B)in O_X$.
But, What is the difference with
"The inverse image of a measurable set under a measurable function is measurable? "
measure-theory cantor-set measurable-functions
asked 2 hours ago
eraldcoil
15818
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up vote
4
down vote
favorite
I have a confusion with measurable functions.
I just saw that the statement "The inverse image of a measurable set under a measurable function is measurable" is false with counter-example the function on Cantor set but I know the definition of f measurable is:
Let $f:(X,O_X)to (Y,O_Y)$ with $O_X$ $sigma$-algebra of $X$ and $O_Y$ $sigma$-algebra of $Y$.
$f$ said $(O_X-O_Y)$-measurable function if for all $Bin O_Y f^-1(B)in O_X$.
But, What is the difference with
"The inverse image of a measurable set under a measurable function is measurable? "
measure-theory cantor-set measurable-functions
asked 2 hours ago
eraldcoil
15818
3
Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain.
– rubikscube09
2 hours ago
@rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions.
– user10354138
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have a confusion with measurable functions.
I just saw that the statement "The inverse image of a measurable set under a measurable function is measurable" is false with counter-example the function on Cantor set but I know the definition of f measurable is:
Let $f:(X,O_X)to (Y,O_Y)$ with $O_X$ $sigma$-algebra of $X$ and $O_Y$ $sigma$-algebra of $Y$.
$f$ said $(O_X-O_Y)$-measurable function if for all $Bin O_Y f^-1(B)in O_X$.
But, What is the difference with
"The inverse image of a measurable set under a measurable function is measurable? "
measure-theory cantor-set measurable-functions
asked 2 hours ago
eraldcoil
15818
I have a confusion with measurable functions.
I just saw that the statement "The inverse image of a measurable set under a measurable function is measurable" is false with counter-example the function on Cantor set but I know the definition of f measurable is:
Let $f:(X,O_X)to (Y,O_Y)$ with $O_X$ $sigma$-algebra of $X$ and $O_Y$ $sigma$-algebra of $Y$.
$f$ said $(O_X-O_Y)$-measurable function if for all $Bin O_Y f^-1(B)in O_X$.
But, What is the difference with
"The inverse image of a measurable set under a measurable function is measurable? "
measure-theory cantor-set measurable-functions
measure-theory cantor-set measurable-functions
asked 2 hours ago
eraldcoil
15818
asked 2 hours ago
eraldcoil
15818
asked 2 hours ago
eraldcoil
15818
asked 2 hours ago
eraldcoil
15818
asked 2 hours ago
eraldcoil
15818
15818
3
Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain.
– rubikscube09
2 hours ago
@rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions.
– user10354138
1 hour ago
add a comment |Â
3
Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain.
– rubikscube09
2 hours ago
@rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions.
– user10354138
1 hour ago
3
3
Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain.
– rubikscube09
2 hours ago
Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain.
– rubikscube09
2 hours ago
@rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions.
– user10354138
1 hour ago
@rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions.
– user10354138
1 hour ago
add a comment |Â
1 Answer
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It depends what $sigma$ algebra you are considering on the target space.
When everyone talks about measureable functions on $mathbb R$, they mean that $mathcal O_Y$ is the the $sigma$-algebra of Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue measurable set may not be measurable.
answered 2 hours ago
D_S
13.2k51551
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It depends what $sigma$ algebra you are considering on the target space.
When everyone talks about measureable functions on $mathbb R$, they mean that $mathcal O_Y$ is the the $sigma$-algebra of Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue measurable set may not be measurable.
answered 2 hours ago
D_S
13.2k51551
add a comment |Â
up vote
4
down vote
It depends what $sigma$ algebra you are considering on the target space.
When everyone talks about measureable functions on $mathbb R$, they mean that $mathcal O_Y$ is the the $sigma$-algebra of Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue measurable set may not be measurable.
answered 2 hours ago
D_S
13.2k51551
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It depends what $sigma$ algebra you are considering on the target space.
When everyone talks about measureable functions on $mathbb R$, they mean that $mathcal O_Y$ is the the $sigma$-algebra of Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue measurable set may not be measurable.
answered 2 hours ago
D_S
13.2k51551
It depends what $sigma$ algebra you are considering on the target space.
When everyone talks about measureable functions on $mathbb R$, they mean that $mathcal O_Y$ is the the $sigma$-algebra of Borel sets (generated by open intervals). This means that the preimage of an open interval like $(a,b)$ is measurable, but the preimage of a Lebesgue measurable set may not be measurable.
answered 2 hours ago
D_S
13.2k51551
answered 2 hours ago
D_S
13.2k51551
answered 2 hours ago
D_S
13.2k51551
answered 2 hours ago
D_S
13.2k51551
13.2k51551
add a comment |Â
add a comment |Â
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3
Measurable functions from R to R use the borel sigma algebra in the codomain and the lebesgue sigma algebra in the domain.
– rubikscube09
2 hours ago
@rubikscube09: no, that is the definition of Lebesgue measurable functions, not measurable functions.
– user10354138
1 hour ago