Is g measurable?
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Let $(X, A)$ be a measurable space and let $f : X to mathbb R$ be a measurable function.
Let $g(x) = 0$ if $f(x)$ is rational
and $g(x) = 1$ if $f(x)$ is irrational.
Am I correct in saying $g$ is measurable ?
measure-theory
edited Nov 18 at 13:06
amWhy
191k27223439
asked Nov 18 at 12:09
Johnmallu
266
add a comment |
up vote
0
down vote
favorite
Let $(X, A)$ be a measurable space and let $f : X to mathbb R$ be a measurable function.
Let $g(x) = 0$ if $f(x)$ is rational
and $g(x) = 1$ if $f(x)$ is irrational.
Am I correct in saying $g$ is measurable ?
measure-theory
edited Nov 18 at 13:06
amWhy
191k27223439
asked Nov 18 at 12:09
Johnmallu
266
You can write $h(a) = mathbb 1(atext{ is rational})$, then $g(x)=h(f(x))$, $f$ is measurable and clearly $h$ is measurable too, so $hcirc f$ is measurable. Of course you need to justify that $h$ is measurable.
– P. Quinton
Nov 18 at 12:16
How would you prove h to be measurable I know that for h to be measurable it must be a function between two measurable spaces such that the preimage of any measurable set is measurable . @P.Quinton
– Johnmallu
Nov 18 at 12:20
Observe that any subset of $mathbb Q$ is measurable, hence so is $mathbb Rsetminus mathbb Q$. For any measurable set $Xsubseteq mathbb R$, the two sets $Xcap mathbb Q$ and $xsetminus mathbb Q$ are also measurable and the image of $X$ with respect to $h$ is directly a function of those two sets.
– P. Quinton
Nov 18 at 12:32
Ah I see. Thank you for clarifying and the explanation.
– Johnmallu
Nov 18 at 12:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X, A)$ be a measurable space and let $f : X to mathbb R$ be a measurable function.
Let $g(x) = 0$ if $f(x)$ is rational
and $g(x) = 1$ if $f(x)$ is irrational.
Am I correct in saying $g$ is measurable ?
measure-theory
edited Nov 18 at 13:06
amWhy
191k27223439
asked Nov 18 at 12:09
Johnmallu
266
Let $(X, A)$ be a measurable space and let $f : X to mathbb R$ be a measurable function.
Let $g(x) = 0$ if $f(x)$ is rational
and $g(x) = 1$ if $f(x)$ is irrational.
Am I correct in saying $g$ is measurable ?
measure-theory
measure-theory
edited Nov 18 at 13:06
amWhy
191k27223439
asked Nov 18 at 12:09
Johnmallu
266
edited Nov 18 at 13:06
amWhy
191k27223439
asked Nov 18 at 12:09
Johnmallu
266
edited Nov 18 at 13:06
amWhy
191k27223439
edited Nov 18 at 13:06
amWhy
191k27223439
edited Nov 18 at 13:06
amWhy
191k27223439
191k27223439
asked Nov 18 at 12:09
Johnmallu
266
asked Nov 18 at 12:09
Johnmallu
266
asked Nov 18 at 12:09
Johnmallu
266
266
You can write $h(a) = mathbb 1(atext{ is rational})$, then $g(x)=h(f(x))$, $f$ is measurable and clearly $h$ is measurable too, so $hcirc f$ is measurable. Of course you need to justify that $h$ is measurable.
– P. Quinton
Nov 18 at 12:16
How would you prove h to be measurable I know that for h to be measurable it must be a function between two measurable spaces such that the preimage of any measurable set is measurable . @P.Quinton
– Johnmallu
Nov 18 at 12:20
Observe that any subset of $mathbb Q$ is measurable, hence so is $mathbb Rsetminus mathbb Q$. For any measurable set $Xsubseteq mathbb R$, the two sets $Xcap mathbb Q$ and $xsetminus mathbb Q$ are also measurable and the image of $X$ with respect to $h$ is directly a function of those two sets.
– P. Quinton
Nov 18 at 12:32
Ah I see. Thank you for clarifying and the explanation.
– Johnmallu
Nov 18 at 12:39
add a comment |
You can write $h(a) = mathbb 1(atext{ is rational})$, then $g(x)=h(f(x))$, $f$ is measurable and clearly $h$ is measurable too, so $hcirc f$ is measurable. Of course you need to justify that $h$ is measurable.
– P. Quinton
Nov 18 at 12:16
How would you prove h to be measurable I know that for h to be measurable it must be a function between two measurable spaces such that the preimage of any measurable set is measurable . @P.Quinton
– Johnmallu
Nov 18 at 12:20
Observe that any subset of $mathbb Q$ is measurable, hence so is $mathbb Rsetminus mathbb Q$. For any measurable set $Xsubseteq mathbb R$, the two sets $Xcap mathbb Q$ and $xsetminus mathbb Q$ are also measurable and the image of $X$ with respect to $h$ is directly a function of those two sets.
– P. Quinton
Nov 18 at 12:32
Ah I see. Thank you for clarifying and the explanation.
– Johnmallu
Nov 18 at 12:39
You can write $h(a) = mathbb 1(atext{ is rational})$, then $g(x)=h(f(x))$, $f$ is measurable and clearly $h$ is measurable too, so $hcirc f$ is measurable. Of course you need to justify that $h$ is measurable.
– P. Quinton
Nov 18 at 12:16
You can write $h(a) = mathbb 1(atext{ is rational})$, then $g(x)=h(f(x))$, $f$ is measurable and clearly $h$ is measurable too, so $hcirc f$ is measurable. Of course you need to justify that $h$ is measurable.
– P. Quinton
Nov 18 at 12:16
How would you prove h to be measurable I know that for h to be measurable it must be a function between two measurable spaces such that the preimage of any measurable set is measurable . @P.Quinton
– Johnmallu
Nov 18 at 12:20
How would you prove h to be measurable I know that for h to be measurable it must be a function between two measurable spaces such that the preimage of any measurable set is measurable . @P.Quinton
– Johnmallu
Nov 18 at 12:20
Observe that any subset of $mathbb Q$ is measurable, hence so is $mathbb Rsetminus mathbb Q$. For any measurable set $Xsubseteq mathbb R$, the two sets $Xcap mathbb Q$ and $xsetminus mathbb Q$ are also measurable and the image of $X$ with respect to $h$ is directly a function of those two sets.
– P. Quinton
Nov 18 at 12:32
Observe that any subset of $mathbb Q$ is measurable, hence so is $mathbb Rsetminus mathbb Q$. For any measurable set $Xsubseteq mathbb R$, the two sets $Xcap mathbb Q$ and $xsetminus mathbb Q$ are also measurable and the image of $X$ with respect to $h$ is directly a function of those two sets.
– P. Quinton
Nov 18 at 12:32
Ah I see. Thank you for clarifying and the explanation.
– Johnmallu
Nov 18 at 12:39
Ah I see. Thank you for clarifying and the explanation.
– Johnmallu
Nov 18 at 12:39
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The inverse image of any Borel set under $g$ is $emptyset$, $X$, $f^{-1}(mathbb Q)$ or $f^{-1}(mathbb Q^{c})$ and each of these sets is measurable.
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The inverse image of any Borel set under $g$ is $emptyset$, $X$, $f^{-1}(mathbb Q)$ or $f^{-1}(mathbb Q^{c})$ and each of these sets is measurable.
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
add a comment |
up vote
1
down vote
accepted
The inverse image of any Borel set under $g$ is $emptyset$, $X$, $f^{-1}(mathbb Q)$ or $f^{-1}(mathbb Q^{c})$ and each of these sets is measurable.
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The inverse image of any Borel set under $g$ is $emptyset$, $X$, $f^{-1}(mathbb Q)$ or $f^{-1}(mathbb Q^{c})$ and each of these sets is measurable.
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
The inverse image of any Borel set under $g$ is $emptyset$, $X$, $f^{-1}(mathbb Q)$ or $f^{-1}(mathbb Q^{c})$ and each of these sets is measurable.
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
answered Nov 18 at 12:32
Kavi Rama Murthy
44.1k31852
44.1k31852
add a comment |
add a comment |
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You can write $h(a) = mathbb 1(atext{ is rational})$, then $g(x)=h(f(x))$, $f$ is measurable and clearly $h$ is measurable too, so $hcirc f$ is measurable. Of course you need to justify that $h$ is measurable.
– P. Quinton
Nov 18 at 12:16
How would you prove h to be measurable I know that for h to be measurable it must be a function between two measurable spaces such that the preimage of any measurable set is measurable . @P.Quinton
– Johnmallu
Nov 18 at 12:20
Observe that any subset of $mathbb Q$ is measurable, hence so is $mathbb Rsetminus mathbb Q$. For any measurable set $Xsubseteq mathbb R$, the two sets $Xcap mathbb Q$ and $xsetminus mathbb Q$ are also measurable and the image of $X$ with respect to $h$ is directly a function of those two sets.
– P. Quinton
Nov 18 at 12:32
Ah I see. Thank you for clarifying and the explanation.
– Johnmallu
Nov 18 at 12:39