segments of arbitrary finite measure on a nonatomic measure space
$begingroup$
Let $(E,Sigma,mu)$ be a completely non-atomic and $sigma$-finite measure space. Suppose that $leq$ is a total order on $E$ and write $<$ for the corresponding strict order. We say that $E'$ is an initial segment of $E$ if $E'<Esetminus E'$, i.e. for every $xin E'$ and $yin Esetminus E'$ we have $x<y$. Suppose that every initial segment of $E$ is $mu$-measurable, and furthermore that any proper initial segment (i.e. where $E'neq E$) has finite $mu$-measure.
Conjecture 1. If $tin[0,mu(E)]$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Probably this is not true in general, so I would like to find sufficient conditions we can impose on $E$ to guarantee that Conjecture 1 holds.
I was thinking that perhaps maybe it would help if $mu$ was separable, but I can't get that to work either. Nor can I see an argument assuming that $mu(E)<infty$.
It would be good enough if I could prove this:
Conjecture 2. There exists a subset $D_0$ of $[0,mu(E)]$ of Lebesgue-measure zero such that if $tin[0,mu(E)]setminus D_0$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Again, probably some additional conditions on $E$ will be required.
Thanks guys!
EDIT:
So, I believe I have an idea, but it is not fully-formed. Suppose that $X$ is a totally-ordered set which is connected in the order topology, and that $E$ is a subspace of $X$ which inherits its order from $X$. Suppose further that $E$ is bounded in $X$, i.e. there are $u,vin X$ such that $uleq Eleq v$. Now we define $f:Xto[0,mu(E)]$ by the rule $f(y)=mu{xin E:xgeq y}$. This function is well-defined since every initial segment (and hence every final segment) of $E$ is measurable. Since $f(u)=0$ and $f(v)=mu(E)$, it suffices to show that $f$ is continuous on $X$, whence by the Intermediate Value Theorem for connected spaces we will obtain the desired conclusion.
Since $mu$ is nonatomic, it seems like the function $f$ should be continuous. Too bad "should be" doesn't count as a proof!
real-analysis measure-theory order-theory
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
$endgroup$
add a comment |
$begingroup$
Let $(E,Sigma,mu)$ be a completely non-atomic and $sigma$-finite measure space. Suppose that $leq$ is a total order on $E$ and write $<$ for the corresponding strict order. We say that $E'$ is an initial segment of $E$ if $E'<Esetminus E'$, i.e. for every $xin E'$ and $yin Esetminus E'$ we have $x<y$. Suppose that every initial segment of $E$ is $mu$-measurable, and furthermore that any proper initial segment (i.e. where $E'neq E$) has finite $mu$-measure.
Conjecture 1. If $tin[0,mu(E)]$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Probably this is not true in general, so I would like to find sufficient conditions we can impose on $E$ to guarantee that Conjecture 1 holds.
I was thinking that perhaps maybe it would help if $mu$ was separable, but I can't get that to work either. Nor can I see an argument assuming that $mu(E)<infty$.
It would be good enough if I could prove this:
Conjecture 2. There exists a subset $D_0$ of $[0,mu(E)]$ of Lebesgue-measure zero such that if $tin[0,mu(E)]setminus D_0$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Again, probably some additional conditions on $E$ will be required.
Thanks guys!
EDIT:
So, I believe I have an idea, but it is not fully-formed. Suppose that $X$ is a totally-ordered set which is connected in the order topology, and that $E$ is a subspace of $X$ which inherits its order from $X$. Suppose further that $E$ is bounded in $X$, i.e. there are $u,vin X$ such that $uleq Eleq v$. Now we define $f:Xto[0,mu(E)]$ by the rule $f(y)=mu{xin E:xgeq y}$. This function is well-defined since every initial segment (and hence every final segment) of $E$ is measurable. Since $f(u)=0$ and $f(v)=mu(E)$, it suffices to show that $f$ is continuous on $X$, whence by the Intermediate Value Theorem for connected spaces we will obtain the desired conclusion.
Since $mu$ is nonatomic, it seems like the function $f$ should be continuous. Too bad "should be" doesn't count as a proof!
real-analysis measure-theory order-theory
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
$endgroup$
add a comment |
$begingroup$
Let $(E,Sigma,mu)$ be a completely non-atomic and $sigma$-finite measure space. Suppose that $leq$ is a total order on $E$ and write $<$ for the corresponding strict order. We say that $E'$ is an initial segment of $E$ if $E'<Esetminus E'$, i.e. for every $xin E'$ and $yin Esetminus E'$ we have $x<y$. Suppose that every initial segment of $E$ is $mu$-measurable, and furthermore that any proper initial segment (i.e. where $E'neq E$) has finite $mu$-measure.
Conjecture 1. If $tin[0,mu(E)]$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Probably this is not true in general, so I would like to find sufficient conditions we can impose on $E$ to guarantee that Conjecture 1 holds.
I was thinking that perhaps maybe it would help if $mu$ was separable, but I can't get that to work either. Nor can I see an argument assuming that $mu(E)<infty$.
It would be good enough if I could prove this:
Conjecture 2. There exists a subset $D_0$ of $[0,mu(E)]$ of Lebesgue-measure zero such that if $tin[0,mu(E)]setminus D_0$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Again, probably some additional conditions on $E$ will be required.
Thanks guys!
EDIT:
So, I believe I have an idea, but it is not fully-formed. Suppose that $X$ is a totally-ordered set which is connected in the order topology, and that $E$ is a subspace of $X$ which inherits its order from $X$. Suppose further that $E$ is bounded in $X$, i.e. there are $u,vin X$ such that $uleq Eleq v$. Now we define $f:Xto[0,mu(E)]$ by the rule $f(y)=mu{xin E:xgeq y}$. This function is well-defined since every initial segment (and hence every final segment) of $E$ is measurable. Since $f(u)=0$ and $f(v)=mu(E)$, it suffices to show that $f$ is continuous on $X$, whence by the Intermediate Value Theorem for connected spaces we will obtain the desired conclusion.
Since $mu$ is nonatomic, it seems like the function $f$ should be continuous. Too bad "should be" doesn't count as a proof!
real-analysis measure-theory order-theory
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
$endgroup$
Let $(E,Sigma,mu)$ be a completely non-atomic and $sigma$-finite measure space. Suppose that $leq$ is a total order on $E$ and write $<$ for the corresponding strict order. We say that $E'$ is an initial segment of $E$ if $E'<Esetminus E'$, i.e. for every $xin E'$ and $yin Esetminus E'$ we have $x<y$. Suppose that every initial segment of $E$ is $mu$-measurable, and furthermore that any proper initial segment (i.e. where $E'neq E$) has finite $mu$-measure.
Conjecture 1. If $tin[0,mu(E)]$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Probably this is not true in general, so I would like to find sufficient conditions we can impose on $E$ to guarantee that Conjecture 1 holds.
I was thinking that perhaps maybe it would help if $mu$ was separable, but I can't get that to work either. Nor can I see an argument assuming that $mu(E)<infty$.
It would be good enough if I could prove this:
Conjecture 2. There exists a subset $D_0$ of $[0,mu(E)]$ of Lebesgue-measure zero such that if $tin[0,mu(E)]setminus D_0$ then there is an initial segment $E'$ of $E$ such that $mu(E')=t$.
Again, probably some additional conditions on $E$ will be required.
Thanks guys!
EDIT:
So, I believe I have an idea, but it is not fully-formed. Suppose that $X$ is a totally-ordered set which is connected in the order topology, and that $E$ is a subspace of $X$ which inherits its order from $X$. Suppose further that $E$ is bounded in $X$, i.e. there are $u,vin X$ such that $uleq Eleq v$. Now we define $f:Xto[0,mu(E)]$ by the rule $f(y)=mu{xin E:xgeq y}$. This function is well-defined since every initial segment (and hence every final segment) of $E$ is measurable. Since $f(u)=0$ and $f(v)=mu(E)$, it suffices to show that $f$ is continuous on $X$, whence by the Intermediate Value Theorem for connected spaces we will obtain the desired conclusion.
Since $mu$ is nonatomic, it seems like the function $f$ should be continuous. Too bad "should be" doesn't count as a proof!
real-analysis measure-theory order-theory
real-analysis measure-theory order-theory
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
edited Dec 24 '18 at 15:30
Ben W
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
asked Dec 23 '18 at 20:59
Ben WBen W
2,290615
2,290615
add a comment |
add a comment |
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