Is there a finitely additive measure on R which is not sigma-additive?

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Consider the usual measurable space of real number $( mathbbR, mathcalB(mathbbR))$.
My question is:



Is there an application $mu$ on $mathcalB( mathbbR) rightarrow [0,+infty]$ such that :



i) $mu$ is finitely additive



ii) $mu( mathbbR) < infty$



iii) $exists (A_n) subset mathcalB( mathbbR) $ that:



a) $ A_n subset A_m forall n<m $



b)
$ lim_nrightarrow infty mu(A_n) < mu( cup_n A_n)$






measure-theory







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  • Maybe this might be useful for you: Example for fintely additive but not countably additive probability measure (and other posts linked there).
    – Martin Sleziak
    4 hours ago














up vote
2
down vote

favorite












Consider the usual measurable space of real number $( mathbbR, mathcalB(mathbbR))$.
My question is:



Is there an application $mu$ on $mathcalB( mathbbR) rightarrow [0,+infty]$ such that :



i) $mu$ is finitely additive



ii) $mu( mathbbR) < infty$



iii) $exists (A_n) subset mathcalB( mathbbR) $ that:



a) $ A_n subset A_m forall n<m $



b)
$ lim_nrightarrow infty mu(A_n) < mu( cup_n A_n)$






measure-theory







share|cite|improve this question























  • Maybe this might be useful for you: Example for fintely additive but not countably additive probability measure (and other posts linked there).
    – Martin Sleziak
    4 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Consider the usual measurable space of real number $( mathbbR, mathcalB(mathbbR))$.
My question is:



Is there an application $mu$ on $mathcalB( mathbbR) rightarrow [0,+infty]$ such that :



i) $mu$ is finitely additive



ii) $mu( mathbbR) < infty$



iii) $exists (A_n) subset mathcalB( mathbbR) $ that:



a) $ A_n subset A_m forall n<m $



b)
$ lim_nrightarrow infty mu(A_n) < mu( cup_n A_n)$






measure-theory







share|cite|improve this question















Consider the usual measurable space of real number $( mathbbR, mathcalB(mathbbR))$.
My question is:



Is there an application $mu$ on $mathcalB( mathbbR) rightarrow [0,+infty]$ such that :



i) $mu$ is finitely additive



ii) $mu( mathbbR) < infty$



iii) $exists (A_n) subset mathcalB( mathbbR) $ that:



a) $ A_n subset A_m forall n<m $



b)
$ lim_nrightarrow infty mu(A_n) < mu( cup_n A_n)$






measure-theory




measure-theory






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edited 4 hours ago

























asked 4 hours ago









Taro NGUYEN

715




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  • Maybe this might be useful for you: Example for fintely additive but not countably additive probability measure (and other posts linked there).
    – Martin Sleziak
    4 hours ago
















  • Maybe this might be useful for you: Example for fintely additive but not countably additive probability measure (and other posts linked there).
    – Martin Sleziak
    4 hours ago















Maybe this might be useful for you: Example for fintely additive but not countably additive probability measure (and other posts linked there).
– Martin Sleziak
4 hours ago




Maybe this might be useful for you: Example for fintely additive but not countably additive probability measure (and other posts linked there).
– Martin Sleziak
4 hours ago










1 Answer
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4
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Yes. Take a non-principal ultrafiler $omega$ on $mathbbN$, and define $mu$ as follows:



$mu(A)=0$ if $AcapmathbbN$ does not belong to $omega$,



$mu(A)=1$ if $Acap mathbbN$ belongs to $omega$.



This measure if finitely additive, and if $A_n=0,1,ldots,n$, then $mu(A_n)=0$ for every $n$, while $mu(bigcup A_n)=mu(mathbbN)=1$.






share|cite|improve this answer




















  • On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
    – Pietro Majer
    3 hours ago











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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up vote
4
down vote













Yes. Take a non-principal ultrafiler $omega$ on $mathbbN$, and define $mu$ as follows:



$mu(A)=0$ if $AcapmathbbN$ does not belong to $omega$,



$mu(A)=1$ if $Acap mathbbN$ belongs to $omega$.



This measure if finitely additive, and if $A_n=0,1,ldots,n$, then $mu(A_n)=0$ for every $n$, while $mu(bigcup A_n)=mu(mathbbN)=1$.






share|cite|improve this answer




















  • On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
    – Pietro Majer
    3 hours ago















up vote
4
down vote













Yes. Take a non-principal ultrafiler $omega$ on $mathbbN$, and define $mu$ as follows:



$mu(A)=0$ if $AcapmathbbN$ does not belong to $omega$,



$mu(A)=1$ if $Acap mathbbN$ belongs to $omega$.



This measure if finitely additive, and if $A_n=0,1,ldots,n$, then $mu(A_n)=0$ for every $n$, while $mu(bigcup A_n)=mu(mathbbN)=1$.






share|cite|improve this answer




















  • On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
    – Pietro Majer
    3 hours ago













up vote
4
down vote










up vote
4
down vote









Yes. Take a non-principal ultrafiler $omega$ on $mathbbN$, and define $mu$ as follows:



$mu(A)=0$ if $AcapmathbbN$ does not belong to $omega$,



$mu(A)=1$ if $Acap mathbbN$ belongs to $omega$.



This measure if finitely additive, and if $A_n=0,1,ldots,n$, then $mu(A_n)=0$ for every $n$, while $mu(bigcup A_n)=mu(mathbbN)=1$.






share|cite|improve this answer












Yes. Take a non-principal ultrafiler $omega$ on $mathbbN$, and define $mu$ as follows:



$mu(A)=0$ if $AcapmathbbN$ does not belong to $omega$,



$mu(A)=1$ if $Acap mathbbN$ belongs to $omega$.



This measure if finitely additive, and if $A_n=0,1,ldots,n$, then $mu(A_n)=0$ for every $n$, while $mu(bigcup A_n)=mu(mathbbN)=1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Roberto Frigerio

2,4491521




2,4491521











  • On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
    – Pietro Majer
    3 hours ago

















  • On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
    – Pietro Majer
    3 hours ago
















On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
– Pietro Majer
3 hours ago





On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $ell_1$, meaning there are bounded additive, non $sigma$-additive measures on $mathbbN$ (signed, but then also positive)
– Pietro Majer
3 hours ago


















 

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