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Measure space and measurable set : do we need a measure on a space to have a measurable set?
Clash Royale CLAN TAG#URR8PPP
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I'm a little bit in truble with definition of measurable set. In one definition, I have :
Let $X$ a set $mathcal F$ a $sigma -$algebra on $X$. Then a set $Asubset X$ is measurable if $Ain mathcal F$.
I guess that in "measurable set", we can measure them, no ? So in such a space, what could be a measure ?
For example, $mathcal P(mathbb R)$ (the power set of $mathbb R$) is a a $sigma -$algebra on $mathbb R$. And with this $sigma -$algebra, all sets are measurables. But in such space, what could be a measure ? I sa somewhere (If my memory is good), that there are no measure on $(mathbb R,mathcal P(mathbb R))$ because for all function $mu:mathcal P(mathbb R)to mathbb R^+$ s.t.
1) $mu(varnothing )=0$
2) $Asubset Bimplies mu(A)leq mu(B)$, $A,Bin mathcal P(mathbb R)$
3) $muleft(bigcup_ninmathbb N A_iright)leqsum_ninmathbb Nmu(A_i)$, $A_iin mathcal P(mathbb R)$
then there are always sets $C,Din mathcal P(mathbb R)$ s.t. $$mu(Acup B)<mu(A)+mu(B),$$
and thus there is no measure on $(mathbb R, mathcal P(mathbb R))$.
- So how can we prove that a space $(X,mathcal F)$ has a measure or not, and if there is no measure, can we talk about measurable space ?
measure-theory
asked 1 hour ago
seb
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up vote
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favorite
I'm a little bit in truble with definition of measurable set. In one definition, I have :
Let $X$ a set $mathcal F$ a $sigma -$algebra on $X$. Then a set $Asubset X$ is measurable if $Ain mathcal F$.
I guess that in "measurable set", we can measure them, no ? So in such a space, what could be a measure ?
For example, $mathcal P(mathbb R)$ (the power set of $mathbb R$) is a a $sigma -$algebra on $mathbb R$. And with this $sigma -$algebra, all sets are measurables. But in such space, what could be a measure ? I sa somewhere (If my memory is good), that there are no measure on $(mathbb R,mathcal P(mathbb R))$ because for all function $mu:mathcal P(mathbb R)to mathbb R^+$ s.t.
1) $mu(varnothing )=0$
2) $Asubset Bimplies mu(A)leq mu(B)$, $A,Bin mathcal P(mathbb R)$
3) $muleft(bigcup_ninmathbb N A_iright)leqsum_ninmathbb Nmu(A_i)$, $A_iin mathcal P(mathbb R)$
then there are always sets $C,Din mathcal P(mathbb R)$ s.t. $$mu(Acup B)<mu(A)+mu(B),$$
and thus there is no measure on $(mathbb R, mathcal P(mathbb R))$.
- So how can we prove that a space $(X,mathcal F)$ has a measure or not, and if there is no measure, can we talk about measurable space ?
measure-theory
asked 1 hour ago
seb
102
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The notion "measurable set" doesn't require a measure on your set. Though, the 0-map is always a measure on any $sigma$-algebra on any set.
– Dominique MATTEI
58 mins ago
There are measures on $mathcal P(Bbb R)$. Namely, the counting measure $mu(A)=begincasesinfty&textif lvert Arvertgealeph_0\ lvert Arvert&textif lvert Arvert<aleph_0endcases$. What you cannot have (under axiom of choice) is a translation-invariant measure such that $mu((0,1))=1$.
– Saucy O'Path
56 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm a little bit in truble with definition of measurable set. In one definition, I have :
Let $X$ a set $mathcal F$ a $sigma -$algebra on $X$. Then a set $Asubset X$ is measurable if $Ain mathcal F$.
I guess that in "measurable set", we can measure them, no ? So in such a space, what could be a measure ?
For example, $mathcal P(mathbb R)$ (the power set of $mathbb R$) is a a $sigma -$algebra on $mathbb R$. And with this $sigma -$algebra, all sets are measurables. But in such space, what could be a measure ? I sa somewhere (If my memory is good), that there are no measure on $(mathbb R,mathcal P(mathbb R))$ because for all function $mu:mathcal P(mathbb R)to mathbb R^+$ s.t.
1) $mu(varnothing )=0$
2) $Asubset Bimplies mu(A)leq mu(B)$, $A,Bin mathcal P(mathbb R)$
3) $muleft(bigcup_ninmathbb N A_iright)leqsum_ninmathbb Nmu(A_i)$, $A_iin mathcal P(mathbb R)$
then there are always sets $C,Din mathcal P(mathbb R)$ s.t. $$mu(Acup B)<mu(A)+mu(B),$$
and thus there is no measure on $(mathbb R, mathcal P(mathbb R))$.
- So how can we prove that a space $(X,mathcal F)$ has a measure or not, and if there is no measure, can we talk about measurable space ?
measure-theory
asked 1 hour ago
seb
102
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm a little bit in truble with definition of measurable set. In one definition, I have :
Let $X$ a set $mathcal F$ a $sigma -$algebra on $X$. Then a set $Asubset X$ is measurable if $Ain mathcal F$.
I guess that in "measurable set", we can measure them, no ? So in such a space, what could be a measure ?
For example, $mathcal P(mathbb R)$ (the power set of $mathbb R$) is a a $sigma -$algebra on $mathbb R$. And with this $sigma -$algebra, all sets are measurables. But in such space, what could be a measure ? I sa somewhere (If my memory is good), that there are no measure on $(mathbb R,mathcal P(mathbb R))$ because for all function $mu:mathcal P(mathbb R)to mathbb R^+$ s.t.
1) $mu(varnothing )=0$
2) $Asubset Bimplies mu(A)leq mu(B)$, $A,Bin mathcal P(mathbb R)$
3) $muleft(bigcup_ninmathbb N A_iright)leqsum_ninmathbb Nmu(A_i)$, $A_iin mathcal P(mathbb R)$
then there are always sets $C,Din mathcal P(mathbb R)$ s.t. $$mu(Acup B)<mu(A)+mu(B),$$
and thus there is no measure on $(mathbb R, mathcal P(mathbb R))$.
- So how can we prove that a space $(X,mathcal F)$ has a measure or not, and if there is no measure, can we talk about measurable space ?
measure-theory
measure-theory
asked 1 hour ago
seb
102
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
seb
102
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
seb
102
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
seb
102
asked 1 hour ago
seb
102
102
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
seb is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
1
The notion "measurable set" doesn't require a measure on your set. Though, the 0-map is always a measure on any $sigma$-algebra on any set.
– Dominique MATTEI
58 mins ago
There are measures on $mathcal P(Bbb R)$. Namely, the counting measure $mu(A)=begincasesinfty&textif lvert Arvertgealeph_0\ lvert Arvert&textif lvert Arvert<aleph_0endcases$. What you cannot have (under axiom of choice) is a translation-invariant measure such that $mu((0,1))=1$.
– Saucy O'Path
56 mins ago
add a comment |Â
1
The notion "measurable set" doesn't require a measure on your set. Though, the 0-map is always a measure on any $sigma$-algebra on any set.
– Dominique MATTEI
58 mins ago
There are measures on $mathcal P(Bbb R)$. Namely, the counting measure $mu(A)=begincasesinfty&textif lvert Arvertgealeph_0\ lvert Arvert&textif lvert Arvert<aleph_0endcases$. What you cannot have (under axiom of choice) is a translation-invariant measure such that $mu((0,1))=1$.
– Saucy O'Path
56 mins ago
1
1
The notion "measurable set" doesn't require a measure on your set. Though, the 0-map is always a measure on any $sigma$-algebra on any set.
– Dominique MATTEI
58 mins ago
The notion "measurable set" doesn't require a measure on your set. Though, the 0-map is always a measure on any $sigma$-algebra on any set.
– Dominique MATTEI
58 mins ago
There are measures on $mathcal P(Bbb R)$. Namely, the counting measure $mu(A)=begincasesinfty&textif lvert Arvertgealeph_0\ lvert Arvert&textif lvert Arvert<aleph_0endcases$. What you cannot have (under axiom of choice) is a translation-invariant measure such that $mu((0,1))=1$.
– Saucy O'Path
56 mins ago
There are measures on $mathcal P(Bbb R)$. Namely, the counting measure $mu(A)=begincasesinfty&textif lvert Arvertgealeph_0\ lvert Arvert&textif lvert Arvert<aleph_0endcases$. What you cannot have (under axiom of choice) is a translation-invariant measure such that $mu((0,1))=1$.
– Saucy O'Path
56 mins ago
add a comment |Â
3 Answers
3
active
oldest
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up vote
3
down vote
accepted
Yes, a measure space has a measure function that is defined on a $sigma$-algebra. So having a $sigma$-algebra is a prerequisite for having a measure, and hence a set with just a $sigma$-algebra is "measurable" (we could define a measure on it, but in some cases we cannot..) The $sigma$-algebra is the "wish list": all sets we would like to be able to measure using a measure function. We already anticipate on our wish and call all sets in the $sigma$-algebra "measurable").
It's analogous to having a set with a topology and calling its members "open". We have a set and a $sigma$-algebra and call its members "measurable", or we have a set with a bornology and call its members "bounded", or convexity and "convex" etc.
On the powerset of a set $X$ we can always define a measure pick $p in X$ and define: $mu(A) = 1$ iff $p in A$, $mu(A) = 0$ if $p notin A$. Pretty boring, but a valid measure.
On the power set of the reals we cannot define a measure that gives all intervals the Lebesgue measure (so $mu([a,b]) = b-a$ for $a < b$) and also is translation-invariant (so that $mu(A+x) = mu(A)$ for all $A$ and all $x in mathbbR$). We always have measures we can define, but not always "nice" measures with extra good properties.
answered 53 mins ago
Henno Brandsma
95.2k343104
add a comment |Â
up vote
2
down vote
What makes you think there is no measure on $(mathbb R, mathcal P(mathbb R)$? $mu (A)=1$ if $0 in A$ and $0$ otherwise defines a measure on this space. This construction works in any measurable space.
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
1
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
add a comment |Â
up vote
1
down vote
If $ mathcal F$ is a $ sigma$ - algebra on $X$ we have, for example, the following measures on $mathcal F$:
Example 1: $ mu(A)=0$ for all $A in mathcal F$.
Example 2: $mu(A)= infty$ for all $A in mathcal F$ with $ A ne emptyset$ and $mu(emptyset)=0$.
answered 53 mins ago
Fred
39.3k1238
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes, a measure space has a measure function that is defined on a $sigma$-algebra. So having a $sigma$-algebra is a prerequisite for having a measure, and hence a set with just a $sigma$-algebra is "measurable" (we could define a measure on it, but in some cases we cannot..) The $sigma$-algebra is the "wish list": all sets we would like to be able to measure using a measure function. We already anticipate on our wish and call all sets in the $sigma$-algebra "measurable").
It's analogous to having a set with a topology and calling its members "open". We have a set and a $sigma$-algebra and call its members "measurable", or we have a set with a bornology and call its members "bounded", or convexity and "convex" etc.
On the powerset of a set $X$ we can always define a measure pick $p in X$ and define: $mu(A) = 1$ iff $p in A$, $mu(A) = 0$ if $p notin A$. Pretty boring, but a valid measure.
On the power set of the reals we cannot define a measure that gives all intervals the Lebesgue measure (so $mu([a,b]) = b-a$ for $a < b$) and also is translation-invariant (so that $mu(A+x) = mu(A)$ for all $A$ and all $x in mathbbR$). We always have measures we can define, but not always "nice" measures with extra good properties.
answered 53 mins ago
Henno Brandsma
95.2k343104
add a comment |Â
up vote
3
down vote
accepted
Yes, a measure space has a measure function that is defined on a $sigma$-algebra. So having a $sigma$-algebra is a prerequisite for having a measure, and hence a set with just a $sigma$-algebra is "measurable" (we could define a measure on it, but in some cases we cannot..) The $sigma$-algebra is the "wish list": all sets we would like to be able to measure using a measure function. We already anticipate on our wish and call all sets in the $sigma$-algebra "measurable").
It's analogous to having a set with a topology and calling its members "open". We have a set and a $sigma$-algebra and call its members "measurable", or we have a set with a bornology and call its members "bounded", or convexity and "convex" etc.
On the powerset of a set $X$ we can always define a measure pick $p in X$ and define: $mu(A) = 1$ iff $p in A$, $mu(A) = 0$ if $p notin A$. Pretty boring, but a valid measure.
On the power set of the reals we cannot define a measure that gives all intervals the Lebesgue measure (so $mu([a,b]) = b-a$ for $a < b$) and also is translation-invariant (so that $mu(A+x) = mu(A)$ for all $A$ and all $x in mathbbR$). We always have measures we can define, but not always "nice" measures with extra good properties.
answered 53 mins ago
Henno Brandsma
95.2k343104
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes, a measure space has a measure function that is defined on a $sigma$-algebra. So having a $sigma$-algebra is a prerequisite for having a measure, and hence a set with just a $sigma$-algebra is "measurable" (we could define a measure on it, but in some cases we cannot..) The $sigma$-algebra is the "wish list": all sets we would like to be able to measure using a measure function. We already anticipate on our wish and call all sets in the $sigma$-algebra "measurable").
It's analogous to having a set with a topology and calling its members "open". We have a set and a $sigma$-algebra and call its members "measurable", or we have a set with a bornology and call its members "bounded", or convexity and "convex" etc.
On the powerset of a set $X$ we can always define a measure pick $p in X$ and define: $mu(A) = 1$ iff $p in A$, $mu(A) = 0$ if $p notin A$. Pretty boring, but a valid measure.
On the power set of the reals we cannot define a measure that gives all intervals the Lebesgue measure (so $mu([a,b]) = b-a$ for $a < b$) and also is translation-invariant (so that $mu(A+x) = mu(A)$ for all $A$ and all $x in mathbbR$). We always have measures we can define, but not always "nice" measures with extra good properties.
answered 53 mins ago
Henno Brandsma
95.2k343104
Yes, a measure space has a measure function that is defined on a $sigma$-algebra. So having a $sigma$-algebra is a prerequisite for having a measure, and hence a set with just a $sigma$-algebra is "measurable" (we could define a measure on it, but in some cases we cannot..) The $sigma$-algebra is the "wish list": all sets we would like to be able to measure using a measure function. We already anticipate on our wish and call all sets in the $sigma$-algebra "measurable").
It's analogous to having a set with a topology and calling its members "open". We have a set and a $sigma$-algebra and call its members "measurable", or we have a set with a bornology and call its members "bounded", or convexity and "convex" etc.
On the powerset of a set $X$ we can always define a measure pick $p in X$ and define: $mu(A) = 1$ iff $p in A$, $mu(A) = 0$ if $p notin A$. Pretty boring, but a valid measure.
On the power set of the reals we cannot define a measure that gives all intervals the Lebesgue measure (so $mu([a,b]) = b-a$ for $a < b$) and also is translation-invariant (so that $mu(A+x) = mu(A)$ for all $A$ and all $x in mathbbR$). We always have measures we can define, but not always "nice" measures with extra good properties.
answered 53 mins ago
Henno Brandsma
95.2k343104
answered 53 mins ago
Henno Brandsma
95.2k343104
answered 53 mins ago
Henno Brandsma
95.2k343104
answered 53 mins ago
Henno Brandsma
95.2k343104
95.2k343104
add a comment |Â
add a comment |Â
up vote
2
down vote
What makes you think there is no measure on $(mathbb R, mathcal P(mathbb R)$? $mu (A)=1$ if $0 in A$ and $0$ otherwise defines a measure on this space. This construction works in any measurable space.
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
1
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
add a comment |Â
up vote
2
down vote
What makes you think there is no measure on $(mathbb R, mathcal P(mathbb R)$? $mu (A)=1$ if $0 in A$ and $0$ otherwise defines a measure on this space. This construction works in any measurable space.
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
1
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
What makes you think there is no measure on $(mathbb R, mathcal P(mathbb R)$? $mu (A)=1$ if $0 in A$ and $0$ otherwise defines a measure on this space. This construction works in any measurable space.
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
What makes you think there is no measure on $(mathbb R, mathcal P(mathbb R)$? $mu (A)=1$ if $0 in A$ and $0$ otherwise defines a measure on this space. This construction works in any measurable space.
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
answered 1 hour ago
Kavi Rama Murthy
28.5k31439
28.5k31439
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
1
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
add a comment |Â
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
1
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
Oh, yes, true. But to talk about measurable set, do we need a measure, or only a $sigma -$algebra ?
– seb
56 mins ago
1
1
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
Only a sigma algebra.
– Kavi Rama Murthy
54 mins ago
add a comment |Â
up vote
1
down vote
If $ mathcal F$ is a $ sigma$ - algebra on $X$ we have, for example, the following measures on $mathcal F$:
Example 1: $ mu(A)=0$ for all $A in mathcal F$.
Example 2: $mu(A)= infty$ for all $A in mathcal F$ with $ A ne emptyset$ and $mu(emptyset)=0$.
answered 53 mins ago
Fred
39.3k1238
add a comment |Â
up vote
1
down vote
If $ mathcal F$ is a $ sigma$ - algebra on $X$ we have, for example, the following measures on $mathcal F$:
Example 1: $ mu(A)=0$ for all $A in mathcal F$.
Example 2: $mu(A)= infty$ for all $A in mathcal F$ with $ A ne emptyset$ and $mu(emptyset)=0$.
answered 53 mins ago
Fred
39.3k1238
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $ mathcal F$ is a $ sigma$ - algebra on $X$ we have, for example, the following measures on $mathcal F$:
Example 1: $ mu(A)=0$ for all $A in mathcal F$.
Example 2: $mu(A)= infty$ for all $A in mathcal F$ with $ A ne emptyset$ and $mu(emptyset)=0$.
answered 53 mins ago
Fred
39.3k1238
If $ mathcal F$ is a $ sigma$ - algebra on $X$ we have, for example, the following measures on $mathcal F$:
Example 1: $ mu(A)=0$ for all $A in mathcal F$.
Example 2: $mu(A)= infty$ for all $A in mathcal F$ with $ A ne emptyset$ and $mu(emptyset)=0$.
answered 53 mins ago
Fred
39.3k1238
edited 48 mins ago
answered 53 mins ago
Fred
39.3k1238
answered 53 mins ago
Fred
39.3k1238
answered 53 mins ago
Fred
39.3k1238
39.3k1238
add a comment |Â
add a comment |Â
seb is a new contributor. Be nice, and check out our Code of Conduct.
seb is a new contributor. Be nice, and check out our Code of Conduct.
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1
The notion "measurable set" doesn't require a measure on your set. Though, the 0-map is always a measure on any $sigma$-algebra on any set.
– Dominique MATTEI
58 mins ago
There are measures on $mathcal P(Bbb R)$. Namely, the counting measure $mu(A)=begincasesinfty&textif lvert Arvertgealeph_0\ lvert Arvert&textif lvert Arvert<aleph_0endcases$. What you cannot have (under axiom of choice) is a translation-invariant measure such that $mu((0,1))=1$.
– Saucy O'Path
56 mins ago