How can a function(measure) be countably additive on a non sigma-ringLargest $sigma$-algebra on which an outer measure is countably additiveCountably additive map on an algebra.Existence of the Lebesgue measure, proof thereof.Extension of $sigma$-additive measure beyond Lebesgue-measurable sets.right continuous implies countably additiveCountably additive property of Lebesgue measuresigma-ring of sets generated by semiring, semiring closed under countable intersectionsAre measures always sigma-additive?Constructing a $sigma$-finite measure using measures of its subsetsCan Nedoma's pathology and other pathologies of larger measure spaces be avoided by changing the definition of sigma algebra?
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How can a function(measure) be countably additive on a non sigma-ring
Largest $sigma$-algebra on which an outer measure is countably additiveCountably additive map on an algebra.Existence of the Lebesgue measure, proof thereof.Extension of $sigma$-additive measure beyond Lebesgue-measurable sets.right continuous implies countably additiveCountably additive property of Lebesgue measuresigma-ring of sets generated by semiring, semiring closed under countable intersectionsAre measures always sigma-additive?Constructing a $sigma$-finite measure using measures of its subsetsCan Nedoma's pathology and other pathologies of larger measure spaces be avoided by changing the definition of sigma algebra?
$begingroup$
I understand the definition of a sigma ring (the infinite unions of elements in ring is still in said ring)
And I believe I also understand the definition of countably additive. For elements in ring with the infinite union of elements also in said ring. Then the function satisfy ..ect. Then the function is countably additive.
What I’m confused about is doesn’t this require measures to only exist on sigma rings? Since rings in general don’t have closure for infinite unions. But I know this isn’t true, the ring (collection of all finite unions of disjoint intervals) is a ring and not a sigma ring but it has a lebesgue measure.
I don’t know what I’m not understanding correctly.
measure-theory lebesgue-measure
edited yesterday
Jean Marie
30.6k42154
asked yesterday
Seth MSeth M
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I understand the definition of a sigma ring (the infinite unions of elements in ring is still in said ring)
And I believe I also understand the definition of countably additive. For elements in ring with the infinite union of elements also in said ring. Then the function satisfy ..ect. Then the function is countably additive.
What I’m confused about is doesn’t this require measures to only exist on sigma rings? Since rings in general don’t have closure for infinite unions. But I know this isn’t true, the ring (collection of all finite unions of disjoint intervals) is a ring and not a sigma ring but it has a lebesgue measure.
I don’t know what I’m not understanding correctly.
measure-theory lebesgue-measure
edited yesterday
Jean Marie
30.6k42154
asked yesterday
Seth MSeth M
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
I have suppressed the tag "ring-theory" : it is not this kind of rings that are concerned by ring theory...
$endgroup$
– Jean Marie
yesterday
$begingroup$
Right. This is a set-theoretic ring not an algebraic structure.
$endgroup$
– GReyes
yesterday
$begingroup$
You can define $sigma$-additivity for a function $phi:DtoBbb R$ with arbitrary $Dsubseteq P(X)$, as whenever $bigcup_i A_iin D$ for countably many pairwise disjoint sets $A_iin D$, we have $phi(bigcup_i A_i) =sum_iphi(A_i)$.
$endgroup$
– Berci
yesterday
add a comment |
$begingroup$
I understand the definition of a sigma ring (the infinite unions of elements in ring is still in said ring)
And I believe I also understand the definition of countably additive. For elements in ring with the infinite union of elements also in said ring. Then the function satisfy ..ect. Then the function is countably additive.
What I’m confused about is doesn’t this require measures to only exist on sigma rings? Since rings in general don’t have closure for infinite unions. But I know this isn’t true, the ring (collection of all finite unions of disjoint intervals) is a ring and not a sigma ring but it has a lebesgue measure.
I don’t know what I’m not understanding correctly.
measure-theory lebesgue-measure
edited yesterday
Jean Marie
30.6k42154
asked yesterday
Seth MSeth M
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I understand the definition of a sigma ring (the infinite unions of elements in ring is still in said ring)
And I believe I also understand the definition of countably additive. For elements in ring with the infinite union of elements also in said ring. Then the function satisfy ..ect. Then the function is countably additive.
What I’m confused about is doesn’t this require measures to only exist on sigma rings? Since rings in general don’t have closure for infinite unions. But I know this isn’t true, the ring (collection of all finite unions of disjoint intervals) is a ring and not a sigma ring but it has a lebesgue measure.
I don’t know what I’m not understanding correctly.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited yesterday
Jean Marie
30.6k42154
asked yesterday
Seth MSeth M
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jean Marie
30.6k42154
asked yesterday
Seth MSeth M
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jean Marie
30.6k42154
edited yesterday
Jean Marie
30.6k42154
edited yesterday
Jean Marie
30.6k42154
30.6k42154
asked yesterday
Seth MSeth M
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Seth MSeth M
11
asked yesterday
Seth MSeth M
11
11
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Seth M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
I have suppressed the tag "ring-theory" : it is not this kind of rings that are concerned by ring theory...
$endgroup$
– Jean Marie
yesterday
$begingroup$
Right. This is a set-theoretic ring not an algebraic structure.
$endgroup$
– GReyes
yesterday
$begingroup$
You can define $sigma$-additivity for a function $phi:DtoBbb R$ with arbitrary $Dsubseteq P(X)$, as whenever $bigcup_i A_iin D$ for countably many pairwise disjoint sets $A_iin D$, we have $phi(bigcup_i A_i) =sum_iphi(A_i)$.
$endgroup$
– Berci
yesterday
add a comment |
1
$begingroup$
I have suppressed the tag "ring-theory" : it is not this kind of rings that are concerned by ring theory...
$endgroup$
– Jean Marie
yesterday
$begingroup$
Right. This is a set-theoretic ring not an algebraic structure.
$endgroup$
– GReyes
yesterday
$begingroup$
You can define $sigma$-additivity for a function $phi:DtoBbb R$ with arbitrary $Dsubseteq P(X)$, as whenever $bigcup_i A_iin D$ for countably many pairwise disjoint sets $A_iin D$, we have $phi(bigcup_i A_i) =sum_iphi(A_i)$.
$endgroup$
– Berci
yesterday
1
1
$begingroup$
I have suppressed the tag "ring-theory" : it is not this kind of rings that are concerned by ring theory...
$endgroup$
– Jean Marie
yesterday
$begingroup$
I have suppressed the tag "ring-theory" : it is not this kind of rings that are concerned by ring theory...
$endgroup$
– Jean Marie
yesterday
$begingroup$
Right. This is a set-theoretic ring not an algebraic structure.
$endgroup$
– GReyes
yesterday
$begingroup$
Right. This is a set-theoretic ring not an algebraic structure.
$endgroup$
– GReyes
yesterday
$begingroup$
You can define $sigma$-additivity for a function $phi:DtoBbb R$ with arbitrary $Dsubseteq P(X)$, as whenever $bigcup_i A_iin D$ for countably many pairwise disjoint sets $A_iin D$, we have $phi(bigcup_i A_i) =sum_iphi(A_i)$.
$endgroup$
– Berci
yesterday
$begingroup$
You can define $sigma$-additivity for a function $phi:DtoBbb R$ with arbitrary $Dsubseteq P(X)$, as whenever $bigcup_i A_iin D$ for countably many pairwise disjoint sets $A_iin D$, we have $phi(bigcup_i A_i) =sum_iphi(A_i)$.
$endgroup$
– Berci
yesterday
add a comment |
1 Answer
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$begingroup$
You are mixing two things. For a measure to be countable additive you require that IF the countable union of disjoint measurable sets happens to be measurable then the measure of the union is the sum of the measures. For a sigma ring it is always the case that countable unions belong to the ring so you can dispense with the 'if'. But a general measure can be defined and even sigma additive over a more general substrate.
answered yesterday
GReyesGReyes
1,82015
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You are mixing two things. For a measure to be countable additive you require that IF the countable union of disjoint measurable sets happens to be measurable then the measure of the union is the sum of the measures. For a sigma ring it is always the case that countable unions belong to the ring so you can dispense with the 'if'. But a general measure can be defined and even sigma additive over a more general substrate.
answered yesterday
GReyesGReyes
1,82015
$endgroup$
add a comment |
$begingroup$
You are mixing two things. For a measure to be countable additive you require that IF the countable union of disjoint measurable sets happens to be measurable then the measure of the union is the sum of the measures. For a sigma ring it is always the case that countable unions belong to the ring so you can dispense with the 'if'. But a general measure can be defined and even sigma additive over a more general substrate.
answered yesterday
GReyesGReyes
1,82015
$endgroup$
add a comment |
$begingroup$
You are mixing two things. For a measure to be countable additive you require that IF the countable union of disjoint measurable sets happens to be measurable then the measure of the union is the sum of the measures. For a sigma ring it is always the case that countable unions belong to the ring so you can dispense with the 'if'. But a general measure can be defined and even sigma additive over a more general substrate.
answered yesterday
GReyesGReyes
1,82015
$endgroup$
You are mixing two things. For a measure to be countable additive you require that IF the countable union of disjoint measurable sets happens to be measurable then the measure of the union is the sum of the measures. For a sigma ring it is always the case that countable unions belong to the ring so you can dispense with the 'if'. But a general measure can be defined and even sigma additive over a more general substrate.
answered yesterday
GReyesGReyes
1,82015
answered yesterday
GReyesGReyes
1,82015
answered yesterday
GReyesGReyes
1,82015
answered yesterday
GReyesGReyes
1,82015
1,82015
add a comment |
add a comment |
Seth M is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
I have suppressed the tag "ring-theory" : it is not this kind of rings that are concerned by ring theory...
$endgroup$
– Jean Marie
yesterday
$begingroup$
Right. This is a set-theoretic ring not an algebraic structure.
$endgroup$
– GReyes
yesterday
$begingroup$
You can define $sigma$-additivity for a function $phi:DtoBbb R$ with arbitrary $Dsubseteq P(X)$, as whenever $bigcup_i A_iin D$ for countably many pairwise disjoint sets $A_iin D$, we have $phi(bigcup_i A_i) =sum_iphi(A_i)$.
$endgroup$
– Berci
yesterday