Variant of Sierpiński's result on non-atomic measures












3












$begingroup$


Sierpiński's theorem states that nonatomic probability measures take a continuum of values. What if I assume that $mu$ is a countably additive probability measure on $(X,2^X)$ and further that $mu({x})=0$ for all $xin X$ (a weaker assumption than non-atomicity). Does it follow that $mu$ takes on every value in $[0,1]$?






set-theory measure-theory







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$endgroup$








  • 2




    $begingroup$
    If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals.
    $endgroup$
    – Will Brian
    15 hours ago










  • $begingroup$
    Nice trolling of the future, ghost of Ulam.
    $endgroup$
    – Monroe Eskew
    6 hours ago
















3












$begingroup$


Sierpiński's theorem states that nonatomic probability measures take a continuum of values. What if I assume that $mu$ is a countably additive probability measure on $(X,2^X)$ and further that $mu({x})=0$ for all $xin X$ (a weaker assumption than non-atomicity). Does it follow that $mu$ takes on every value in $[0,1]$?






set-theory measure-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals.
    $endgroup$
    – Will Brian
    15 hours ago










  • $begingroup$
    Nice trolling of the future, ghost of Ulam.
    $endgroup$
    – Monroe Eskew
    6 hours ago














3












3








3





$begingroup$


Sierpiński's theorem states that nonatomic probability measures take a continuum of values. What if I assume that $mu$ is a countably additive probability measure on $(X,2^X)$ and further that $mu({x})=0$ for all $xin X$ (a weaker assumption than non-atomicity). Does it follow that $mu$ takes on every value in $[0,1]$?






set-theory measure-theory







share|cite|improve this question









$endgroup$




Sierpiński's theorem states that nonatomic probability measures take a continuum of values. What if I assume that $mu$ is a countably additive probability measure on $(X,2^X)$ and further that $mu({x})=0$ for all $xin X$ (a weaker assumption than non-atomicity). Does it follow that $mu$ takes on every value in $[0,1]$?






set-theory measure-theory




set-theory measure-theory






share|cite|improve this question













share|cite|improve this question











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asked 15 hours ago









Aryeh KontorovichAryeh Kontorovich

2,5081527




2,5081527








  • 2




    $begingroup$
    If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals.
    $endgroup$
    – Will Brian
    15 hours ago










  • $begingroup$
    Nice trolling of the future, ghost of Ulam.
    $endgroup$
    – Monroe Eskew
    6 hours ago














  • 2




    $begingroup$
    If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals.
    $endgroup$
    – Will Brian
    15 hours ago










  • $begingroup$
    Nice trolling of the future, ghost of Ulam.
    $endgroup$
    – Monroe Eskew
    6 hours ago








2




2




$begingroup$
If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals.
$endgroup$
– Will Brian
15 hours ago




$begingroup$
If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals.
$endgroup$
– Will Brian
15 hours ago












$begingroup$
Nice trolling of the future, ghost of Ulam.
$endgroup$
– Monroe Eskew
6 hours ago




$begingroup$
Nice trolling of the future, ghost of Ulam.
$endgroup$
– Monroe Eskew
6 hours ago










1 Answer
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10












$begingroup$

If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).



Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $mu$ must be atomic, by the theorem you quoted in your post. Fix $Y subseteq X$ such that $mu(Y) = c > 0$ and if $Z subseteq Y$ then either $mu(Z) = 0$ or $mu(Z) = c$. (This is what it means for a measure to be atomic.) Letting $mathcal U = {Z subseteq Y : mu(Z) = c}$, it is not difficult to show that $mathcal U$ is a $sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $geq$ the least measurable cardinal.



Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."






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    $begingroup$

    If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).



    Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $mu$ must be atomic, by the theorem you quoted in your post. Fix $Y subseteq X$ such that $mu(Y) = c > 0$ and if $Z subseteq Y$ then either $mu(Z) = 0$ or $mu(Z) = c$. (This is what it means for a measure to be atomic.) Letting $mathcal U = {Z subseteq Y : mu(Z) = c}$, it is not difficult to show that $mathcal U$ is a $sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $geq$ the least measurable cardinal.



    Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."






    share|cite|improve this answer











    $endgroup$


















      10












      $begingroup$

      If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).



      Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $mu$ must be atomic, by the theorem you quoted in your post. Fix $Y subseteq X$ such that $mu(Y) = c > 0$ and if $Z subseteq Y$ then either $mu(Z) = 0$ or $mu(Z) = c$. (This is what it means for a measure to be atomic.) Letting $mathcal U = {Z subseteq Y : mu(Z) = c}$, it is not difficult to show that $mathcal U$ is a $sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $geq$ the least measurable cardinal.



      Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."






      share|cite|improve this answer











      $endgroup$
















        10












        10








        10





        $begingroup$

        If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).



        Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $mu$ must be atomic, by the theorem you quoted in your post. Fix $Y subseteq X$ such that $mu(Y) = c > 0$ and if $Z subseteq Y$ then either $mu(Z) = 0$ or $mu(Z) = c$. (This is what it means for a measure to be atomic.) Letting $mathcal U = {Z subseteq Y : mu(Z) = c}$, it is not difficult to show that $mathcal U$ is a $sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $geq$ the least measurable cardinal.



        Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."






        share|cite|improve this answer











        $endgroup$



        If $|X|$ is a measurable cardinal, then there is a $sigma$-complete ultrafilter $mathcal U$ on $X$. You may define a countably additive probability measure $mu$ on $(X,2^X)$ by setting $mu(Y) = 1$ if $Y in mathcal U$ and $mu(Y) = 0$ if $Y notin mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).



        Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $mu$ must be atomic, by the theorem you quoted in your post. Fix $Y subseteq X$ such that $mu(Y) = c > 0$ and if $Z subseteq Y$ then either $mu(Z) = 0$ or $mu(Z) = c$. (This is what it means for a measure to be atomic.) Letting $mathcal U = {Z subseteq Y : mu(Z) = c}$, it is not difficult to show that $mathcal U$ is a $sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $geq$ the least measurable cardinal.



        Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 13 hours ago

























        answered 14 hours ago









        Will BrianWill Brian

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        8,68423854






























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