Variant of Sierpiński's result on non-atomic measures
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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
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
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add a comment |
$begingroup$
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
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
$endgroup$
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.
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– Will Brian
5 hours ago
add a comment |
$begingroup$
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
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
$endgroup$
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
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
asked 5 hours ago
Aryeh KontorovichAryeh Kontorovich
2,4981527
2,4981527
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
5 hours ago
add a comment |
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
5 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
5 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
5 hours ago
add a comment |
1 Answer
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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."
answered 4 hours ago
Will BrianWill Brian
8,63923753
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add a comment |
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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."
answered 4 hours ago
Will BrianWill Brian
8,63923753
$endgroup$
add a comment |
$begingroup$
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."
answered 4 hours ago
Will BrianWill Brian
8,63923753
$endgroup$
add a comment |
$begingroup$
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."
answered 4 hours ago
Will BrianWill Brian
8,63923753
$endgroup$
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."
answered 4 hours ago
Will BrianWill Brian
8,63923753
edited 3 hours ago
answered 4 hours ago
Will BrianWill Brian
8,63923753
answered 4 hours ago
Will BrianWill Brian
8,63923753
answered 4 hours ago
Will BrianWill Brian
8,63923753
8,63923753
add a comment |
add a comment |
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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 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
5 hours ago