If a measure is $sigma -$finite, why is finite on the compacts set?
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Let $X$ a topological space and $mathcal B$ the Borel set of $X$. Let $mu$ a measure on $mathcal B$. Why if $mu$ is $sigma -$finite, then $mu(K)<infty $ for all compact $Ksubset X$ ?
Attempts
Let $X_i$ a collection of $X$ s.t. $X=bigcup_i=1^infty X_i$ and $mu(X_i)<infty $ for all $i$. Let $U_i_i=1^n$ opens s.t. $$Ksubset bigcup_i=1^nU_isubset bigcup_i=1^infty X_i.$$
I would agree if each $U_i$ is contained in a finite number of $X_j$, but I don't understand why it is really the case.
measure-theory compactness
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
asked Aug 26 at 9:29
Henri
1176
add a comment |Â
up vote
0
down vote
favorite
Let $X$ a topological space and $mathcal B$ the Borel set of $X$. Let $mu$ a measure on $mathcal B$. Why if $mu$ is $sigma -$finite, then $mu(K)<infty $ for all compact $Ksubset X$ ?
Attempts
Let $X_i$ a collection of $X$ s.t. $X=bigcup_i=1^infty X_i$ and $mu(X_i)<infty $ for all $i$. Let $U_i_i=1^n$ opens s.t. $$Ksubset bigcup_i=1^nU_isubset bigcup_i=1^infty X_i.$$
I would agree if each $U_i$ is contained in a finite number of $X_j$, but I don't understand why it is really the case.
measure-theory compactness
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
asked Aug 26 at 9:29
Henri
1176
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ a topological space and $mathcal B$ the Borel set of $X$. Let $mu$ a measure on $mathcal B$. Why if $mu$ is $sigma -$finite, then $mu(K)<infty $ for all compact $Ksubset X$ ?
Attempts
Let $X_i$ a collection of $X$ s.t. $X=bigcup_i=1^infty X_i$ and $mu(X_i)<infty $ for all $i$. Let $U_i_i=1^n$ opens s.t. $$Ksubset bigcup_i=1^nU_isubset bigcup_i=1^infty X_i.$$
I would agree if each $U_i$ is contained in a finite number of $X_j$, but I don't understand why it is really the case.
measure-theory compactness
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
asked Aug 26 at 9:29
Henri
1176
Let $X$ a topological space and $mathcal B$ the Borel set of $X$. Let $mu$ a measure on $mathcal B$. Why if $mu$ is $sigma -$finite, then $mu(K)<infty $ for all compact $Ksubset X$ ?
Attempts
Let $X_i$ a collection of $X$ s.t. $X=bigcup_i=1^infty X_i$ and $mu(X_i)<infty $ for all $i$. Let $U_i_i=1^n$ opens s.t. $$Ksubset bigcup_i=1^nU_isubset bigcup_i=1^infty X_i.$$
I would agree if each $U_i$ is contained in a finite number of $X_j$, but I don't understand why it is really the case.
measure-theory compactness
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
asked Aug 26 at 9:29
Henri
1176
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
edited Aug 26 at 9:36
José Carlos Santos
119k16101182
119k16101182
asked Aug 26 at 9:29
Henri
1176
asked Aug 26 at 9:29
Henri
1176
asked Aug 26 at 9:29
Henri
1176
1176
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
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This is not true. Take $X=mathbb Q$ (with the usual topology) and $mu(A)=#A$. Then $mu$ is $sigma$-finite, but$$muleft(0cupleft,ninmathbb Nrightright)=infty,$$in spite of the fact that $0cupleft,ninmathbb Nright$ is compact.
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is not true. Take $X=mathbb Q$ (with the usual topology) and $mu(A)=#A$. Then $mu$ is $sigma$-finite, but$$muleft(0cupleft,ninmathbb Nrightright)=infty,$$in spite of the fact that $0cupleft,ninmathbb Nright$ is compact.
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
 |Â
show 5 more comments
up vote
2
down vote
accepted
This is not true. Take $X=mathbb Q$ (with the usual topology) and $mu(A)=#A$. Then $mu$ is $sigma$-finite, but$$muleft(0cupleft,ninmathbb Nrightright)=infty,$$in spite of the fact that $0cupleft,ninmathbb Nright$ is compact.
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
 |Â
show 5 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is not true. Take $X=mathbb Q$ (with the usual topology) and $mu(A)=#A$. Then $mu$ is $sigma$-finite, but$$muleft(0cupleft,ninmathbb Nrightright)=infty,$$in spite of the fact that $0cupleft,ninmathbb Nright$ is compact.
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
This is not true. Take $X=mathbb Q$ (with the usual topology) and $mu(A)=#A$. Then $mu$ is $sigma$-finite, but$$muleft(0cupleft,ninmathbb Nrightright)=infty,$$in spite of the fact that $0cupleft,ninmathbb Nright$ is compact.
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
edited Aug 26 at 10:18
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
answered Aug 26 at 9:35
José Carlos Santos
119k16101182
119k16101182
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
 |Â
show 5 more comments
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
Looks to be a very good example, thank you. Something is nevertheless strange : In a book I'm reading, they say that a $mu:mathcal Bto [0,infty ]$ is a Radon measure if $mu(K)<infty $ on compact set. And on wiki they say that a measure is a Radon measure if it's inner regular, outer regular and $sigma -$finite. So both definition should be equivalent no ?
– Henri
Aug 26 at 9:42
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
@Henri I suppose that, yes, they are equivalent. However, the measure that I defined is neither a Radon measuere nor outer regular.
– José Carlos Santos
Aug 26 at 9:43
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
Oh, true. In my mind I was thinking about $mu$ more regular. But anyway, in fact your answer gave me a better idea of "having more regularity". So perfect. Just a last small thing : If $X$ is a topological space and $mathcal B$ its borel $sigma -$algebra, is it true that $X$ is a countable union of compact set ? If no, what condition we have to put on the topology to have the statement true ? (I know for example it's true in $mathbb R^n$ with standard topology).
– Henri
Aug 26 at 9:48
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
@Henri No. If, in $mathbb R$, you consider the discrete topology, then $mathbb R$ cannot be expressed as a countable union of compact subsets (which are the finite subsets, in this context).
– José Carlos Santos
Aug 26 at 9:51
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
I'm confuse on something : you say that $K:=0cupfrac1nmid ninmathbb N^*$ is compact in $mathbb Q$. $mathcal U=0,frac1nmid ninmathbb N^*$ is a covering of open set of $K$. What would be a finite subcovering of $K$ by element of $mathcal U$ ?
– Henri
Aug 26 at 10:14
 |Â
show 5 more comments
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