Covering null sets by a finite number of intervalsSigma-algebra from a finite class of setsIrrationals in...
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Covering null sets by a finite number of intervals
Sigma-algebra from a finite class of setsIrrationals in $[0,1]$ does not have measure zeroCharacterisation of absolutely continuous measure on the real lineInequality with outer measureBorel sets and absolutely continuous functionsfun measure theory problem mixed with elementary set theoryconstruct a set $H$ such that $Scup H$ is open and $lambda(H) < epsilon$Absolute continuity and continuitymeasure zero on a metric spaceopen set in $mathbb{R}^n$ is a countable union of non-overlapping closed intervals in $mathbb{R}^n$
$begingroup$
Let us say that a subset $A$ of $mathbb R$ has property $P$ if every $epsilon>0$ there is a finite collection of open intervals $(a_1,b_1),(a_2,b_2),cdots,(a_n,b_n)$ such that $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$. My question: if $A$ is a nowhere dense set with measure $0$ does it have property $P$?
Some basics: every compact set of measure $0$ (obviously) has property $P$.
No dense set of measure $0$ can have property $P$. More generally, if $A$ has property $P$ then $A$ is nowhere dense. Proof: if $(alpha,beta) subset overset {-} {A}$ take $epsilon <beta -alpha$. If $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$ then $(alpha,beta) subset cup_i [a_i,b_i]$ so $beta -alpha <epsilon$, a contradiction.
My question is if every nowhere dense set of measure $0$ has property $P$. My guess that the implication does not hold but I don't have a counterexample.
measure-theory
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
$endgroup$
add a comment |
$begingroup$
Let us say that a subset $A$ of $mathbb R$ has property $P$ if every $epsilon>0$ there is a finite collection of open intervals $(a_1,b_1),(a_2,b_2),cdots,(a_n,b_n)$ such that $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$. My question: if $A$ is a nowhere dense set with measure $0$ does it have property $P$?
Some basics: every compact set of measure $0$ (obviously) has property $P$.
No dense set of measure $0$ can have property $P$. More generally, if $A$ has property $P$ then $A$ is nowhere dense. Proof: if $(alpha,beta) subset overset {-} {A}$ take $epsilon <beta -alpha$. If $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$ then $(alpha,beta) subset cup_i [a_i,b_i]$ so $beta -alpha <epsilon$, a contradiction.
My question is if every nowhere dense set of measure $0$ has property $P$. My guess that the implication does not hold but I don't have a counterexample.
measure-theory
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
$endgroup$
$begingroup$
As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero.
$endgroup$
– bof
yesterday
$begingroup$
@bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove.
$endgroup$
– Kavi Rama Murthy
yesterday
add a comment |
$begingroup$
Let us say that a subset $A$ of $mathbb R$ has property $P$ if every $epsilon>0$ there is a finite collection of open intervals $(a_1,b_1),(a_2,b_2),cdots,(a_n,b_n)$ such that $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$. My question: if $A$ is a nowhere dense set with measure $0$ does it have property $P$?
Some basics: every compact set of measure $0$ (obviously) has property $P$.
No dense set of measure $0$ can have property $P$. More generally, if $A$ has property $P$ then $A$ is nowhere dense. Proof: if $(alpha,beta) subset overset {-} {A}$ take $epsilon <beta -alpha$. If $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$ then $(alpha,beta) subset cup_i [a_i,b_i]$ so $beta -alpha <epsilon$, a contradiction.
My question is if every nowhere dense set of measure $0$ has property $P$. My guess that the implication does not hold but I don't have a counterexample.
measure-theory
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
$endgroup$
Let us say that a subset $A$ of $mathbb R$ has property $P$ if every $epsilon>0$ there is a finite collection of open intervals $(a_1,b_1),(a_2,b_2),cdots,(a_n,b_n)$ such that $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$. My question: if $A$ is a nowhere dense set with measure $0$ does it have property $P$?
Some basics: every compact set of measure $0$ (obviously) has property $P$.
No dense set of measure $0$ can have property $P$. More generally, if $A$ has property $P$ then $A$ is nowhere dense. Proof: if $(alpha,beta) subset overset {-} {A}$ take $epsilon <beta -alpha$. If $A subset cup_i (a_i,b_i)$ and $sum (b_i-a_i) <epsilon$ then $(alpha,beta) subset cup_i [a_i,b_i]$ so $beta -alpha <epsilon$, a contradiction.
My question is if every nowhere dense set of measure $0$ has property $P$. My guess that the implication does not hold but I don't have a counterexample.
measure-theory
measure-theory
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
asked yesterday
Kavi Rama MurthyKavi Rama Murthy
78.4k53572
78.4k53572
$begingroup$
As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero.
$endgroup$
– bof
yesterday
$begingroup$
@bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove.
$endgroup$
– Kavi Rama Murthy
yesterday
add a comment |
$begingroup$
As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero.
$endgroup$
– bof
yesterday
$begingroup$
@bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero.
$endgroup$
– bof
yesterday
$begingroup$
As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero.
$endgroup$
– bof
yesterday
$begingroup$
@bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
@bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove.
$endgroup$
– Kavi Rama Murthy
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.
In fact, it's easy to see that a set $Asubseteqmathbb R$ has property $P$ if and only the closure of $A$ is a compact set of measure zero.
answered yesterday
bofbof
52.8k559121
$endgroup$
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
The set $mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $mathbb Z$.
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.
In fact, it's easy to see that a set $Asubseteqmathbb R$ has property $P$ if and only the closure of $A$ is a compact set of measure zero.
answered yesterday
bofbof
52.8k559121
$endgroup$
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.
In fact, it's easy to see that a set $Asubseteqmathbb R$ has property $P$ if and only the closure of $A$ is a compact set of measure zero.
answered yesterday
bofbof
52.8k559121
$endgroup$
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.
In fact, it's easy to see that a set $Asubseteqmathbb R$ has property $P$ if and only the closure of $A$ is a compact set of measure zero.
answered yesterday
bofbof
52.8k559121
$endgroup$
For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.
In fact, it's easy to see that a set $Asubseteqmathbb R$ has property $P$ if and only the closure of $A$ is a compact set of measure zero.
answered yesterday
bofbof
52.8k559121
edited yesterday
answered yesterday
bofbof
52.8k559121
answered yesterday
bofbof
52.8k559121
answered yesterday
bofbof
52.8k559121
52.8k559121
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
(+1) Nice answer! I wish I had thought of that.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
The set $mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $mathbb Z$.
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
The set $mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $mathbb Z$.
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
The set $mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $mathbb Z$.
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
The set $mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $mathbb Z$.
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
answered yesterday
José Carlos SantosJosé Carlos Santos
178k24139251
178k24139251
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
$begingroup$
No problem. That answer is better than mine.
$endgroup$
– José Carlos Santos
yesterday
add a comment |
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$begingroup$
As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero.
$endgroup$
– bof
yesterday
$begingroup$
@bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove.
$endgroup$
– Kavi Rama Murthy
yesterday