Measurable set on Manifold.
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In my textbook, we have the following definition,
If $M$ is a smooth manifold of dimension $n$. We say that a set $Asubset M$ is measurable if, for any chart $U$, the intersection $Acap U$ is a Lebesgue measurable set in $U$.
What really means this? A chart $U$ on $M$ is any couple $(U,varphi)$ where $U$ is an open subset of $M$ and $varphi$ is a homeomorphism of $U$ onto an open subset of $mathbb{R}^{n}$. So, if for any chart $U$, the intersection $Acap Uimplies varphi(Acap U)subsetmathbb{R}^{n}$ is Lesbesgue measurable set in $U$?
And, also says that
The family of all measurable sets in $M$ forms a $sigma-$álgebra.
But I don't see how this form an $sigma-$algebra, i.e., is the union of measurable set measurable?. Thanks!
manifolds smooth-manifolds
$endgroup$
add a comment |
$begingroup$
In my textbook, we have the following definition,
If $M$ is a smooth manifold of dimension $n$. We say that a set $Asubset M$ is measurable if, for any chart $U$, the intersection $Acap U$ is a Lebesgue measurable set in $U$.
What really means this? A chart $U$ on $M$ is any couple $(U,varphi)$ where $U$ is an open subset of $M$ and $varphi$ is a homeomorphism of $U$ onto an open subset of $mathbb{R}^{n}$. So, if for any chart $U$, the intersection $Acap Uimplies varphi(Acap U)subsetmathbb{R}^{n}$ is Lesbesgue measurable set in $U$?
And, also says that
The family of all measurable sets in $M$ forms a $sigma-$álgebra.
But I don't see how this form an $sigma-$algebra, i.e., is the union of measurable set measurable?. Thanks!
manifolds smooth-manifolds
$endgroup$
$begingroup$
For a sigma-algebra one needs closure under countable unions.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:11
1
$begingroup$
I write a chart as $c = (mathrm{U}, varphi, n)$ where $mathrm{U}$ is an open subset of the manifold and $n$ is the dimension of the chart. Then, what measurability means is that $varphi(mathrm{A} cap mathrm{U})$ is a Lebuesgue measurable set in $mathbf{R}^n.$
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– Will M.
Dec 21 '18 at 5:13
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@WillM. and these open $U$ (their collection) form a $sigma$-algebra correct? closed under complement and (countable) union.
$endgroup$
– Hossien Sahebjame
Dec 21 '18 at 5:18
1
$begingroup$
No. Because, in general, the complement of an open set is not open.
$endgroup$
– Will M.
Dec 21 '18 at 5:20
add a comment |
$begingroup$
In my textbook, we have the following definition,
If $M$ is a smooth manifold of dimension $n$. We say that a set $Asubset M$ is measurable if, for any chart $U$, the intersection $Acap U$ is a Lebesgue measurable set in $U$.
What really means this? A chart $U$ on $M$ is any couple $(U,varphi)$ where $U$ is an open subset of $M$ and $varphi$ is a homeomorphism of $U$ onto an open subset of $mathbb{R}^{n}$. So, if for any chart $U$, the intersection $Acap Uimplies varphi(Acap U)subsetmathbb{R}^{n}$ is Lesbesgue measurable set in $U$?
And, also says that
The family of all measurable sets in $M$ forms a $sigma-$álgebra.
But I don't see how this form an $sigma-$algebra, i.e., is the union of measurable set measurable?. Thanks!
manifolds smooth-manifolds
$endgroup$
In my textbook, we have the following definition,
If $M$ is a smooth manifold of dimension $n$. We say that a set $Asubset M$ is measurable if, for any chart $U$, the intersection $Acap U$ is a Lebesgue measurable set in $U$.
What really means this? A chart $U$ on $M$ is any couple $(U,varphi)$ where $U$ is an open subset of $M$ and $varphi$ is a homeomorphism of $U$ onto an open subset of $mathbb{R}^{n}$. So, if for any chart $U$, the intersection $Acap Uimplies varphi(Acap U)subsetmathbb{R}^{n}$ is Lesbesgue measurable set in $U$?
And, also says that
The family of all measurable sets in $M$ forms a $sigma-$álgebra.
But I don't see how this form an $sigma-$algebra, i.e., is the union of measurable set measurable?. Thanks!
manifolds smooth-manifolds
manifolds smooth-manifolds
edited Dec 21 '18 at 6:42
asked Dec 21 '18 at 5:05
user570343
$begingroup$
For a sigma-algebra one needs closure under countable unions.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:11
1
$begingroup$
I write a chart as $c = (mathrm{U}, varphi, n)$ where $mathrm{U}$ is an open subset of the manifold and $n$ is the dimension of the chart. Then, what measurability means is that $varphi(mathrm{A} cap mathrm{U})$ is a Lebuesgue measurable set in $mathbf{R}^n.$
$endgroup$
– Will M.
Dec 21 '18 at 5:13
$begingroup$
@WillM. and these open $U$ (their collection) form a $sigma$-algebra correct? closed under complement and (countable) union.
$endgroup$
– Hossien Sahebjame
Dec 21 '18 at 5:18
1
$begingroup$
No. Because, in general, the complement of an open set is not open.
$endgroup$
– Will M.
Dec 21 '18 at 5:20
add a comment |
$begingroup$
For a sigma-algebra one needs closure under countable unions.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:11
1
$begingroup$
I write a chart as $c = (mathrm{U}, varphi, n)$ where $mathrm{U}$ is an open subset of the manifold and $n$ is the dimension of the chart. Then, what measurability means is that $varphi(mathrm{A} cap mathrm{U})$ is a Lebuesgue measurable set in $mathbf{R}^n.$
$endgroup$
– Will M.
Dec 21 '18 at 5:13
$begingroup$
@WillM. and these open $U$ (their collection) form a $sigma$-algebra correct? closed under complement and (countable) union.
$endgroup$
– Hossien Sahebjame
Dec 21 '18 at 5:18
1
$begingroup$
No. Because, in general, the complement of an open set is not open.
$endgroup$
– Will M.
Dec 21 '18 at 5:20
$begingroup$
For a sigma-algebra one needs closure under countable unions.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:11
$begingroup$
For a sigma-algebra one needs closure under countable unions.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:11
1
1
$begingroup$
I write a chart as $c = (mathrm{U}, varphi, n)$ where $mathrm{U}$ is an open subset of the manifold and $n$ is the dimension of the chart. Then, what measurability means is that $varphi(mathrm{A} cap mathrm{U})$ is a Lebuesgue measurable set in $mathbf{R}^n.$
$endgroup$
– Will M.
Dec 21 '18 at 5:13
$begingroup$
I write a chart as $c = (mathrm{U}, varphi, n)$ where $mathrm{U}$ is an open subset of the manifold and $n$ is the dimension of the chart. Then, what measurability means is that $varphi(mathrm{A} cap mathrm{U})$ is a Lebuesgue measurable set in $mathbf{R}^n.$
$endgroup$
– Will M.
Dec 21 '18 at 5:13
$begingroup$
@WillM. and these open $U$ (their collection) form a $sigma$-algebra correct? closed under complement and (countable) union.
$endgroup$
– Hossien Sahebjame
Dec 21 '18 at 5:18
$begingroup$
@WillM. and these open $U$ (their collection) form a $sigma$-algebra correct? closed under complement and (countable) union.
$endgroup$
– Hossien Sahebjame
Dec 21 '18 at 5:18
1
1
$begingroup$
No. Because, in general, the complement of an open set is not open.
$endgroup$
– Will M.
Dec 21 '18 at 5:20
$begingroup$
No. Because, in general, the complement of an open set is not open.
$endgroup$
– Will M.
Dec 21 '18 at 5:20
add a comment |
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$begingroup$
For a sigma-algebra one needs closure under countable unions.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:11
1
$begingroup$
I write a chart as $c = (mathrm{U}, varphi, n)$ where $mathrm{U}$ is an open subset of the manifold and $n$ is the dimension of the chart. Then, what measurability means is that $varphi(mathrm{A} cap mathrm{U})$ is a Lebuesgue measurable set in $mathbf{R}^n.$
$endgroup$
– Will M.
Dec 21 '18 at 5:13
$begingroup$
@WillM. and these open $U$ (their collection) form a $sigma$-algebra correct? closed under complement and (countable) union.
$endgroup$
– Hossien Sahebjame
Dec 21 '18 at 5:18
1
$begingroup$
No. Because, in general, the complement of an open set is not open.
$endgroup$
– Will M.
Dec 21 '18 at 5:20