Extension of the Lebesgue measure on the extended real line
$begingroup$
Is it possible to extend the Lebesgue measure on $mathbb{R}$ as a measure on $[-infty,infty]$ (defined on the Borel subsets of $[-infty,infty]$), so that $[-infty,infty]backslash mathbb{R}$ is a null set ?
real-analysis measure-theory
asked Jan 1 at 18:26
LCOLCO
464
$endgroup$
add a comment |
$begingroup$
Is it possible to extend the Lebesgue measure on $mathbb{R}$ as a measure on $[-infty,infty]$ (defined on the Borel subsets of $[-infty,infty]$), so that $[-infty,infty]backslash mathbb{R}$ is a null set ?
real-analysis measure-theory
asked Jan 1 at 18:26
LCOLCO
464
$endgroup$
1
$begingroup$
Sure, Just set $mu(A) =lambda(A cap Bbb{R})$.
$endgroup$
– PhoemueX
Jan 1 at 18:29
add a comment |
$begingroup$
Is it possible to extend the Lebesgue measure on $mathbb{R}$ as a measure on $[-infty,infty]$ (defined on the Borel subsets of $[-infty,infty]$), so that $[-infty,infty]backslash mathbb{R}$ is a null set ?
real-analysis measure-theory
asked Jan 1 at 18:26
LCOLCO
464
$endgroup$
Is it possible to extend the Lebesgue measure on $mathbb{R}$ as a measure on $[-infty,infty]$ (defined on the Borel subsets of $[-infty,infty]$), so that $[-infty,infty]backslash mathbb{R}$ is a null set ?
real-analysis measure-theory
real-analysis measure-theory
asked Jan 1 at 18:26
LCOLCO
464
asked Jan 1 at 18:26
LCOLCO
464
asked Jan 1 at 18:26
LCOLCO
464
asked Jan 1 at 18:26
LCOLCO
464
asked Jan 1 at 18:26
LCOLCO
464
464
1
$begingroup$
Sure, Just set $mu(A) =lambda(A cap Bbb{R})$.
$endgroup$
– PhoemueX
Jan 1 at 18:29
add a comment |
1
$begingroup$
Sure, Just set $mu(A) =lambda(A cap Bbb{R})$.
$endgroup$
– PhoemueX
Jan 1 at 18:29
1
1
$begingroup$
Sure, Just set $mu(A) =lambda(A cap Bbb{R})$.
$endgroup$
– PhoemueX
Jan 1 at 18:29
$begingroup$
Sure, Just set $mu(A) =lambda(A cap Bbb{R})$.
$endgroup$
– PhoemueX
Jan 1 at 18:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As proposed in the comment, we can just "extend by zero sets." That is, since we have only added two points it is intuitive to consider them as negligible. So, if $(mathbb{R},Omega, lambda)$ is your initial measure space - where $lambda$ denotes the Lebesgue measure, define $(widehat{mathbb{R}}, widehat{Omega}, mu)$ by declaring a set $A$ to be measurable in $widehat{mathbb{R}}$ if and only if $Acap mathbb{R}$ is Lebesgue Measurable. Further, define
$$ mu(A)=lambda(Acap mathbb{R}).$$
It's not so hard to check that this provides a measure. It solves your problem because $X=[-infty,infty]setminus mathbb{R}={pm infty}$, and $$mu({pminfty})=lambda({pminftycap mathbb{R}})=lambda(varnothing)=0.$$ You might also notice that this technique can be generalized.
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
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add a comment |
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1 Answer
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$begingroup$
As proposed in the comment, we can just "extend by zero sets." That is, since we have only added two points it is intuitive to consider them as negligible. So, if $(mathbb{R},Omega, lambda)$ is your initial measure space - where $lambda$ denotes the Lebesgue measure, define $(widehat{mathbb{R}}, widehat{Omega}, mu)$ by declaring a set $A$ to be measurable in $widehat{mathbb{R}}$ if and only if $Acap mathbb{R}$ is Lebesgue Measurable. Further, define
$$ mu(A)=lambda(Acap mathbb{R}).$$
It's not so hard to check that this provides a measure. It solves your problem because $X=[-infty,infty]setminus mathbb{R}={pm infty}$, and $$mu({pminfty})=lambda({pminftycap mathbb{R}})=lambda(varnothing)=0.$$ You might also notice that this technique can be generalized.
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
As proposed in the comment, we can just "extend by zero sets." That is, since we have only added two points it is intuitive to consider them as negligible. So, if $(mathbb{R},Omega, lambda)$ is your initial measure space - where $lambda$ denotes the Lebesgue measure, define $(widehat{mathbb{R}}, widehat{Omega}, mu)$ by declaring a set $A$ to be measurable in $widehat{mathbb{R}}$ if and only if $Acap mathbb{R}$ is Lebesgue Measurable. Further, define
$$ mu(A)=lambda(Acap mathbb{R}).$$
It's not so hard to check that this provides a measure. It solves your problem because $X=[-infty,infty]setminus mathbb{R}={pm infty}$, and $$mu({pminfty})=lambda({pminftycap mathbb{R}})=lambda(varnothing)=0.$$ You might also notice that this technique can be generalized.
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
As proposed in the comment, we can just "extend by zero sets." That is, since we have only added two points it is intuitive to consider them as negligible. So, if $(mathbb{R},Omega, lambda)$ is your initial measure space - where $lambda$ denotes the Lebesgue measure, define $(widehat{mathbb{R}}, widehat{Omega}, mu)$ by declaring a set $A$ to be measurable in $widehat{mathbb{R}}$ if and only if $Acap mathbb{R}$ is Lebesgue Measurable. Further, define
$$ mu(A)=lambda(Acap mathbb{R}).$$
It's not so hard to check that this provides a measure. It solves your problem because $X=[-infty,infty]setminus mathbb{R}={pm infty}$, and $$mu({pminfty})=lambda({pminftycap mathbb{R}})=lambda(varnothing)=0.$$ You might also notice that this technique can be generalized.
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
As proposed in the comment, we can just "extend by zero sets." That is, since we have only added two points it is intuitive to consider them as negligible. So, if $(mathbb{R},Omega, lambda)$ is your initial measure space - where $lambda$ denotes the Lebesgue measure, define $(widehat{mathbb{R}}, widehat{Omega}, mu)$ by declaring a set $A$ to be measurable in $widehat{mathbb{R}}$ if and only if $Acap mathbb{R}$ is Lebesgue Measurable. Further, define
$$ mu(A)=lambda(Acap mathbb{R}).$$
It's not so hard to check that this provides a measure. It solves your problem because $X=[-infty,infty]setminus mathbb{R}={pm infty}$, and $$mu({pminfty})=lambda({pminftycap mathbb{R}})=lambda(varnothing)=0.$$ You might also notice that this technique can be generalized.
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 1 at 18:49
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
10.6k41741
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Sure, Just set $mu(A) =lambda(A cap Bbb{R})$.
$endgroup$
– PhoemueX
Jan 1 at 18:29