Signed measure of uncountable set

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I have a question and hope some of you can help me :)



Consider a signed measure $nu$ on $(Omega, bfA)$ and let be $P_i in bfA$ positive sets, such that $ forall B subset P_i: nu(B) geq0 $. Then we state that $bigcup_i in I P_i$ is also a positive set.



If $I$ is countable ( $|I| leq |mathbbN|$) this seems to be true.
To show this I considered an arbitrary $M in bigcup_i in I P_i $. I know that there exists $M_i := M cap P_i subset P_i$. So we get $nu(M) = nu left( bigcup_i in IM_i right) = sum_i in I nu(M_i) geq 0$.



The last $=$ follows due to sigma-additivity of $nu$ and the $geq$ follows due to $M_i subset P_i$ and $P_i$ is positive.



Now I wonder, what happens, if $I$ is uncountable ( $|I| > |mathbbN|$). I think there must be a counterexample, because the sigma-additivity doesn't hold anymore, but I can't find one.



Does anybody have a counterexample for me, or is my assumption wrong, and one can proof, that the statement is still true for overcountable I's.



Thanks a lot!






measure-theory signed-measures







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  • Check the first answer given in math.stackexchange.com/questions/1605076/…
    – UserS
    2 hours ago















up vote
1
down vote

favorite












I have a question and hope some of you can help me :)



Consider a signed measure $nu$ on $(Omega, bfA)$ and let be $P_i in bfA$ positive sets, such that $ forall B subset P_i: nu(B) geq0 $. Then we state that $bigcup_i in I P_i$ is also a positive set.



If $I$ is countable ( $|I| leq |mathbbN|$) this seems to be true.
To show this I considered an arbitrary $M in bigcup_i in I P_i $. I know that there exists $M_i := M cap P_i subset P_i$. So we get $nu(M) = nu left( bigcup_i in IM_i right) = sum_i in I nu(M_i) geq 0$.



The last $=$ follows due to sigma-additivity of $nu$ and the $geq$ follows due to $M_i subset P_i$ and $P_i$ is positive.



Now I wonder, what happens, if $I$ is uncountable ( $|I| > |mathbbN|$). I think there must be a counterexample, because the sigma-additivity doesn't hold anymore, but I can't find one.



Does anybody have a counterexample for me, or is my assumption wrong, and one can proof, that the statement is still true for overcountable I's.



Thanks a lot!






measure-theory signed-measures







share|cite|improve this question





















  • Check the first answer given in math.stackexchange.com/questions/1605076/…
    – UserS
    2 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have a question and hope some of you can help me :)



Consider a signed measure $nu$ on $(Omega, bfA)$ and let be $P_i in bfA$ positive sets, such that $ forall B subset P_i: nu(B) geq0 $. Then we state that $bigcup_i in I P_i$ is also a positive set.



If $I$ is countable ( $|I| leq |mathbbN|$) this seems to be true.
To show this I considered an arbitrary $M in bigcup_i in I P_i $. I know that there exists $M_i := M cap P_i subset P_i$. So we get $nu(M) = nu left( bigcup_i in IM_i right) = sum_i in I nu(M_i) geq 0$.



The last $=$ follows due to sigma-additivity of $nu$ and the $geq$ follows due to $M_i subset P_i$ and $P_i$ is positive.



Now I wonder, what happens, if $I$ is uncountable ( $|I| > |mathbbN|$). I think there must be a counterexample, because the sigma-additivity doesn't hold anymore, but I can't find one.



Does anybody have a counterexample for me, or is my assumption wrong, and one can proof, that the statement is still true for overcountable I's.



Thanks a lot!






measure-theory signed-measures







share|cite|improve this question













I have a question and hope some of you can help me :)



Consider a signed measure $nu$ on $(Omega, bfA)$ and let be $P_i in bfA$ positive sets, such that $ forall B subset P_i: nu(B) geq0 $. Then we state that $bigcup_i in I P_i$ is also a positive set.



If $I$ is countable ( $|I| leq |mathbbN|$) this seems to be true.
To show this I considered an arbitrary $M in bigcup_i in I P_i $. I know that there exists $M_i := M cap P_i subset P_i$. So we get $nu(M) = nu left( bigcup_i in IM_i right) = sum_i in I nu(M_i) geq 0$.



The last $=$ follows due to sigma-additivity of $nu$ and the $geq$ follows due to $M_i subset P_i$ and $P_i$ is positive.



Now I wonder, what happens, if $I$ is uncountable ( $|I| > |mathbbN|$). I think there must be a counterexample, because the sigma-additivity doesn't hold anymore, but I can't find one.



Does anybody have a counterexample for me, or is my assumption wrong, and one can proof, that the statement is still true for overcountable I's.



Thanks a lot!






measure-theory signed-measures




measure-theory signed-measures






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asked 2 hours ago









pcalc

975




975











  • Check the first answer given in math.stackexchange.com/questions/1605076/…
    – UserS
    2 hours ago

















  • Check the first answer given in math.stackexchange.com/questions/1605076/…
    – UserS
    2 hours ago
















Check the first answer given in math.stackexchange.com/questions/1605076/…
– UserS
2 hours ago





Check the first answer given in math.stackexchange.com/questions/1605076/…
– UserS
2 hours ago











2 Answers
2






active

oldest

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up vote
3
down vote



accepted










Let $[0,1]$ be equipped with negative Lebesgue measure.



All singletons are positive sets but the union of all singletons is not.






share|cite|improve this answer




















  • Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
    – pcalc
    2 hours ago










  • You are welcome.
    – drhab
    2 hours ago

















up vote
1
down vote













Consider $(Bbb R,textLebesgue)$, $I=[0,1]$ and $P_i=i$ with the measure $nu(A)=-int_A e^-x^2sin x,dx$.






share|cite|improve this answer




















  • Thank you! It seems as I have overseen quite a lot of good examples.
    – pcalc
    2 hours ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Let $[0,1]$ be equipped with negative Lebesgue measure.



All singletons are positive sets but the union of all singletons is not.






share|cite|improve this answer




















  • Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
    – pcalc
    2 hours ago










  • You are welcome.
    – drhab
    2 hours ago














up vote
3
down vote



accepted










Let $[0,1]$ be equipped with negative Lebesgue measure.



All singletons are positive sets but the union of all singletons is not.






share|cite|improve this answer




















  • Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
    – pcalc
    2 hours ago










  • You are welcome.
    – drhab
    2 hours ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Let $[0,1]$ be equipped with negative Lebesgue measure.



All singletons are positive sets but the union of all singletons is not.






share|cite|improve this answer












Let $[0,1]$ be equipped with negative Lebesgue measure.



All singletons are positive sets but the union of all singletons is not.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









drhab

91.6k542124




91.6k542124











  • Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
    – pcalc
    2 hours ago










  • You are welcome.
    – drhab
    2 hours ago
















  • Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
    – pcalc
    2 hours ago










  • You are welcome.
    – drhab
    2 hours ago















Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
– pcalc
2 hours ago




Aaah thanks! I've already thought about this example, but rejected it, because I thought I want that it has to hold for ALL subsets and their union - what obviously is nonsense. Thanks a lot!
– pcalc
2 hours ago












You are welcome.
– drhab
2 hours ago




You are welcome.
– drhab
2 hours ago










up vote
1
down vote













Consider $(Bbb R,textLebesgue)$, $I=[0,1]$ and $P_i=i$ with the measure $nu(A)=-int_A e^-x^2sin x,dx$.






share|cite|improve this answer




















  • Thank you! It seems as I have overseen quite a lot of good examples.
    – pcalc
    2 hours ago














up vote
1
down vote













Consider $(Bbb R,textLebesgue)$, $I=[0,1]$ and $P_i=i$ with the measure $nu(A)=-int_A e^-x^2sin x,dx$.






share|cite|improve this answer




















  • Thank you! It seems as I have overseen quite a lot of good examples.
    – pcalc
    2 hours ago












up vote
1
down vote










up vote
1
down vote









Consider $(Bbb R,textLebesgue)$, $I=[0,1]$ and $P_i=i$ with the measure $nu(A)=-int_A e^-x^2sin x,dx$.






share|cite|improve this answer












Consider $(Bbb R,textLebesgue)$, $I=[0,1]$ and $P_i=i$ with the measure $nu(A)=-int_A e^-x^2sin x,dx$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Saucy O'Path

4,649424




4,649424











  • Thank you! It seems as I have overseen quite a lot of good examples.
    – pcalc
    2 hours ago
















  • Thank you! It seems as I have overseen quite a lot of good examples.
    – pcalc
    2 hours ago















Thank you! It seems as I have overseen quite a lot of good examples.
– pcalc
2 hours ago




Thank you! It seems as I have overseen quite a lot of good examples.
– pcalc
2 hours ago

















 

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