Does every set of positive measure contain an uncountable null set?
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
asked 1 hour ago
Guacho Perez
3,80211131
add a comment |
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
asked 1 hour ago
Guacho Perez
3,80211131
add a comment |
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
asked 1 hour ago
Guacho Perez
3,80211131
If $E$ is Lebesgue measurable and $m(E)>0$, does there exist an uncountable $Csubset E$ with $m(C)=0$?
This seems intuitively clear but I cannot prove it. Since $E$ has positive measure it contains a nonmeasurable set $V$, and every measurable subset of $V$ is null, but I could not show $V$ contains uncountable measurable subsets. I also tried using the fact that for $alphain(0,1)$ there is an interval $I$ s.t $m(Ecap I)ge alpha m(I)$, attempting to construct a Cantor set inside $I$ with an uncountable intersection with $E$ but I was unsuccessful. Any hints? Is this even true?
real-analysis measure-theory
real-analysis measure-theory
asked 1 hour ago
Guacho Perez
3,80211131
asked 1 hour ago
Guacho Perez
3,80211131
asked 1 hour ago
Guacho Perez
3,80211131
asked 1 hour ago
Guacho Perez
3,80211131
asked 1 hour ago
Guacho Perez
3,80211131
3,80211131
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
answered 1 hour ago
Noah Schweber
120k10146278
add a comment |
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1 Answer
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1 Answer
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Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
answered 1 hour ago
Noah Schweber
120k10146278
add a comment |
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
answered 1 hour ago
Noah Schweber
120k10146278
add a comment |
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
answered 1 hour ago
Noah Schweber
120k10146278
Yes.
By regularity, if $m(E)>0$ then there is some closed $Csubseteq E$ with $m(C)>0$. So WLOG assume $E$ is closed (and has measure $1$).
A variant of the Cantor set construction now lets us "thin" $E$ out to a closed uncountable $Dsubseteq E$ with measure zero. E.g. at our first step we fix $a<b$ such that $m(E_{<a})=m(E_{>b})={1over 3}$, and cut down to $Esetminus (a,b)$.
(How does the closedness of $E$ matter? Well, we need to argue that the $D$ we construct is in fact as desired. Trivially this $D$ is null, since we've shrunk its measure appropriately at each stage; knowing that $D$ is closed tells us that everything that should be in $D$, is.)
answered 1 hour ago
Noah Schweber
120k10146278
answered 1 hour ago
Noah Schweber
120k10146278
answered 1 hour ago
Noah Schweber
120k10146278
answered 1 hour ago
Noah Schweber
120k10146278
120k10146278
add a comment |
add a comment |
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