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Can a set of positive measure and its complement both have empty interior?
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This might be silly, but I am not sure:
Does there exist a Lebesgue measurable subset $E subseteq (0,1)$ such that
$E$ and $(0,1) setminus E$ both have positive Lebesgue measure.
$E$ and $(0,1) setminus E$ both have empty interiors.
If we relax condition $1$, then $E=Qcap (0,1)$ works. If we relax condition $2$, then the fat Cantor set does the job. (Its complement have non-empty interior though).
general-topology measure-theory examples-counterexamples cantor-set
asked 2 hours ago
Asaf Shachar
4,7533939
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up vote
3
down vote
favorite
This might be silly, but I am not sure:
Does there exist a Lebesgue measurable subset $E subseteq (0,1)$ such that
$E$ and $(0,1) setminus E$ both have positive Lebesgue measure.
$E$ and $(0,1) setminus E$ both have empty interiors.
If we relax condition $1$, then $E=Qcap (0,1)$ works. If we relax condition $2$, then the fat Cantor set does the job. (Its complement have non-empty interior though).
general-topology measure-theory examples-counterexamples cantor-set
asked 2 hours ago
Asaf Shachar
4,7533939
In my answer to this question I constructed an $F_sigma$ set $Msubseteqmathbb R$ such that $0lt m(Mcap I)lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $mathbb Rsetminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=Mcap(0,1)$ and $(0,1)setminus E$.
– bof
14 secs ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This might be silly, but I am not sure:
Does there exist a Lebesgue measurable subset $E subseteq (0,1)$ such that
$E$ and $(0,1) setminus E$ both have positive Lebesgue measure.
$E$ and $(0,1) setminus E$ both have empty interiors.
If we relax condition $1$, then $E=Qcap (0,1)$ works. If we relax condition $2$, then the fat Cantor set does the job. (Its complement have non-empty interior though).
general-topology measure-theory examples-counterexamples cantor-set
asked 2 hours ago
Asaf Shachar
4,7533939
This might be silly, but I am not sure:
Does there exist a Lebesgue measurable subset $E subseteq (0,1)$ such that
$E$ and $(0,1) setminus E$ both have positive Lebesgue measure.
$E$ and $(0,1) setminus E$ both have empty interiors.
If we relax condition $1$, then $E=Qcap (0,1)$ works. If we relax condition $2$, then the fat Cantor set does the job. (Its complement have non-empty interior though).
general-topology measure-theory examples-counterexamples cantor-set
general-topology measure-theory examples-counterexamples cantor-set
asked 2 hours ago
Asaf Shachar
4,7533939
asked 2 hours ago
Asaf Shachar
4,7533939
edited 1 hour ago
asked 2 hours ago
Asaf Shachar
4,7533939
asked 2 hours ago
Asaf Shachar
4,7533939
asked 2 hours ago
Asaf Shachar
4,7533939
4,7533939
In my answer to this question I constructed an $F_sigma$ set $Msubseteqmathbb R$ such that $0lt m(Mcap I)lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $mathbb Rsetminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=Mcap(0,1)$ and $(0,1)setminus E$.
– bof
14 secs ago
add a comment |Â
In my answer to this question I constructed an $F_sigma$ set $Msubseteqmathbb R$ such that $0lt m(Mcap I)lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $mathbb Rsetminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=Mcap(0,1)$ and $(0,1)setminus E$.
– bof
14 secs ago
In my answer to this question I constructed an $F_sigma$ set $Msubseteqmathbb R$ such that $0lt m(Mcap I)lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $mathbb Rsetminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=Mcap(0,1)$ and $(0,1)setminus E$.
– bof
14 secs ago
In my answer to this question I constructed an $F_sigma$ set $Msubseteqmathbb R$ such that $0lt m(Mcap I)lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $mathbb Rsetminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=Mcap(0,1)$ and $(0,1)setminus E$.
– bof
14 secs ago
add a comment |Â
2 Answers
2
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oldest
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up vote
3
down vote
Let $E$ be the union of $(0,1) setminus mathbb Q)cap (0,frac 1 2]$ and $mathbb Qcap (frac 1 2,1)$. Then $E$ and $(0,1)setminus E$ both have positive measure and they have no interior.
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
1
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
add a comment |Â
up vote
2
down vote
Let $E$ be a fat Cantor set $C$ together with the $C^complementcapmathbb Q$.
answered 1 hour ago
José Carlos Santos
134k17108197
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $E$ be the union of $(0,1) setminus mathbb Q)cap (0,frac 1 2]$ and $mathbb Qcap (frac 1 2,1)$. Then $E$ and $(0,1)setminus E$ both have positive measure and they have no interior.
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
1
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
add a comment |Â
up vote
3
down vote
Let $E$ be the union of $(0,1) setminus mathbb Q)cap (0,frac 1 2]$ and $mathbb Qcap (frac 1 2,1)$. Then $E$ and $(0,1)setminus E$ both have positive measure and they have no interior.
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
1
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $E$ be the union of $(0,1) setminus mathbb Q)cap (0,frac 1 2]$ and $mathbb Qcap (frac 1 2,1)$. Then $E$ and $(0,1)setminus E$ both have positive measure and they have no interior.
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
Let $E$ be the union of $(0,1) setminus mathbb Q)cap (0,frac 1 2]$ and $mathbb Qcap (frac 1 2,1)$. Then $E$ and $(0,1)setminus E$ both have positive measure and they have no interior.
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
edited 41 mins ago
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
answered 1 hour ago
Kavi Rama Murthy
35.2k31746
35.2k31746
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
1
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
add a comment |Â
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
1
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
I am not sure I understand. I meant to take the complement inside $(0,1)$...
– Asaf Shachar
1 hour ago
1
1
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
No big deal. I had given an example in $mathbb R$ and now I have changed it to $(0,1)$.
– Kavi Rama Murthy
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
Neither of those two sets satisfies condition 1. Their union is supposed to be $(0,1)$. Have I misunderstood? What set do you think $E$ is?
– MJD
1 hour ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@MJD You are right. Only a slight modification was required. Please see my answer now.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
@AsafShachar Please see my revised answer.
– Kavi Rama Murthy
40 mins ago
add a comment |Â
up vote
2
down vote
Let $E$ be a fat Cantor set $C$ together with the $C^complementcapmathbb Q$.
answered 1 hour ago
José Carlos Santos
134k17108197
add a comment |Â
up vote
2
down vote
Let $E$ be a fat Cantor set $C$ together with the $C^complementcapmathbb Q$.
answered 1 hour ago
José Carlos Santos
134k17108197
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $E$ be a fat Cantor set $C$ together with the $C^complementcapmathbb Q$.
answered 1 hour ago
José Carlos Santos
134k17108197
Let $E$ be a fat Cantor set $C$ together with the $C^complementcapmathbb Q$.
answered 1 hour ago
José Carlos Santos
134k17108197
answered 1 hour ago
José Carlos Santos
134k17108197
answered 1 hour ago
José Carlos Santos
134k17108197
answered 1 hour ago
José Carlos Santos
134k17108197
134k17108197
add a comment |Â
add a comment |Â
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In my answer to this question I constructed an $F_sigma$ set $Msubseteqmathbb R$ such that $0lt m(Mcap I)lt m(I)$ for every interval $I$, where $m$ denotes Lebesgue measure. Obviously the sets $M$ and $mathbb Rsetminus M$ have positive Lebesgue measure and have empty interiors, and the same goes for the sets $E=Mcap(0,1)$ and $(0,1)setminus E$.
– bof
14 secs ago