Does every set of positive measure contain an uncountable null set?












1














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







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    1














    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







    share|cite|improve this question

























      1












      1








      1







      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







      share|cite|improve this question













      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






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      asked 1 hour ago









      Guacho Perez

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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.)






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            1 Answer
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            1 Answer
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            active

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            5














            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.)






            share|cite|improve this answer


























              5














              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.)






              share|cite|improve this answer
























                5












                5








                5






                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.)






                share|cite|improve this answer












                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.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Noah Schweber

                120k10146278




                120k10146278






























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