Signed measure of uncountable set

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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

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
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










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2963260%2fsigned-measure-of-uncountable-set%23new-answer', 'question_page');

);

Post as a guest

















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

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2963260%2fsigned-measure-of-uncountable-set%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
Read more

Installing NextGIS Connect into QGIS 3?

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
Read more

One-line joke

A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
Read more