Sets of null (Lebesgue)-measure and sigma compacts











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Let $Esubset{mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure.



¿It is possible to find $Fsubset{mathbb R}$ such that $Esubset F$, $m(F)=0$ and $F$ is $sigma$-compact (that is, numerable union of compact sets)?



Obviously if $E$ is numerable, this is trivial.



Moreover, if $m(overline{E})=0$ (where $overline{E}$ is the clousure of $E$), it is also true. Just take $F=overline{E}$ and $overline{E}=bigcup_{ninmathbb N} overline{E}cap[-n,n]$.



I am also interested when $Esubsetmathbb T$, where $mathbb T$ is the unit circle.










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  • The real question is if there existe a non numerable nulo set whose clausure has positive mensure and such that it is not contained on any null sigma compact set.
    – Tito Eliatron
    Nov 18 at 11:02















up vote
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down vote

favorite
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Let $Esubset{mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure.



¿It is possible to find $Fsubset{mathbb R}$ such that $Esubset F$, $m(F)=0$ and $F$ is $sigma$-compact (that is, numerable union of compact sets)?



Obviously if $E$ is numerable, this is trivial.



Moreover, if $m(overline{E})=0$ (where $overline{E}$ is the clousure of $E$), it is also true. Just take $F=overline{E}$ and $overline{E}=bigcup_{ninmathbb N} overline{E}cap[-n,n]$.



I am also interested when $Esubsetmathbb T$, where $mathbb T$ is the unit circle.










share|cite|improve this question
























  • The real question is if there existe a non numerable nulo set whose clausure has positive mensure and such that it is not contained on any null sigma compact set.
    – Tito Eliatron
    Nov 18 at 11:02













up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
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Let $Esubset{mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure.



¿It is possible to find $Fsubset{mathbb R}$ such that $Esubset F$, $m(F)=0$ and $F$ is $sigma$-compact (that is, numerable union of compact sets)?



Obviously if $E$ is numerable, this is trivial.



Moreover, if $m(overline{E})=0$ (where $overline{E}$ is the clousure of $E$), it is also true. Just take $F=overline{E}$ and $overline{E}=bigcup_{ninmathbb N} overline{E}cap[-n,n]$.



I am also interested when $Esubsetmathbb T$, where $mathbb T$ is the unit circle.










share|cite|improve this question















Let $Esubset{mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure.



¿It is possible to find $Fsubset{mathbb R}$ such that $Esubset F$, $m(F)=0$ and $F$ is $sigma$-compact (that is, numerable union of compact sets)?



Obviously if $E$ is numerable, this is trivial.



Moreover, if $m(overline{E})=0$ (where $overline{E}$ is the clousure of $E$), it is also true. Just take $F=overline{E}$ and $overline{E}=bigcup_{ninmathbb N} overline{E}cap[-n,n]$.



I am also interested when $Esubsetmathbb T$, where $mathbb T$ is the unit circle.







measure-theory lebesgue-measure compactness






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share|cite|improve this question













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edited Nov 16 at 12:24

























asked Nov 16 at 10:41









Tito Eliatron

860418




860418












  • The real question is if there existe a non numerable nulo set whose clausure has positive mensure and such that it is not contained on any null sigma compact set.
    – Tito Eliatron
    Nov 18 at 11:02


















  • The real question is if there existe a non numerable nulo set whose clausure has positive mensure and such that it is not contained on any null sigma compact set.
    – Tito Eliatron
    Nov 18 at 11:02
















The real question is if there existe a non numerable nulo set whose clausure has positive mensure and such that it is not contained on any null sigma compact set.
– Tito Eliatron
Nov 18 at 11:02




The real question is if there existe a non numerable nulo set whose clausure has positive mensure and such that it is not contained on any null sigma compact set.
– Tito Eliatron
Nov 18 at 11:02















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