A question about a sequence of sets with positive measure
$begingroup$
For each $ninmathbb{N}$, $E_n$ is a measurable subset of $[0,1]$. Let $m$ be the Lebesuge measure, suppose
$$
m(E_n)geq delta>0, quad forall nin mathbb{N}.
$$
My question is the following: is there a subsequence ${n_k}_{kinmathbb{N}}$ such that $n_kto infty(ktoinfty)$ and
$$
m(cap_{k=0}^infty E_{n_k})>0?
$$
real-analysis measure-theory lebesgue-measure
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
$endgroup$
add a comment |
$begingroup$
For each $ninmathbb{N}$, $E_n$ is a measurable subset of $[0,1]$. Let $m$ be the Lebesuge measure, suppose
$$
m(E_n)geq delta>0, quad forall nin mathbb{N}.
$$
My question is the following: is there a subsequence ${n_k}_{kinmathbb{N}}$ such that $n_kto infty(ktoinfty)$ and
$$
m(cap_{k=0}^infty E_{n_k})>0?
$$
real-analysis measure-theory lebesgue-measure
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
$endgroup$
$begingroup$
I give a fairly thorough analysis, with many references, of results related to your question in this 13 May 2005 sci.math post archived at Math Forum.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:53
add a comment |
$begingroup$
For each $ninmathbb{N}$, $E_n$ is a measurable subset of $[0,1]$. Let $m$ be the Lebesuge measure, suppose
$$
m(E_n)geq delta>0, quad forall nin mathbb{N}.
$$
My question is the following: is there a subsequence ${n_k}_{kinmathbb{N}}$ such that $n_kto infty(ktoinfty)$ and
$$
m(cap_{k=0}^infty E_{n_k})>0?
$$
real-analysis measure-theory lebesgue-measure
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
$endgroup$
For each $ninmathbb{N}$, $E_n$ is a measurable subset of $[0,1]$. Let $m$ be the Lebesuge measure, suppose
$$
m(E_n)geq delta>0, quad forall nin mathbb{N}.
$$
My question is the following: is there a subsequence ${n_k}_{kinmathbb{N}}$ such that $n_kto infty(ktoinfty)$ and
$$
m(cap_{k=0}^infty E_{n_k})>0?
$$
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
asked Jan 8 at 8:12
Guohuan ZhaoGuohuan Zhao
283
283
$begingroup$
I give a fairly thorough analysis, with many references, of results related to your question in this 13 May 2005 sci.math post archived at Math Forum.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:53
add a comment |
$begingroup$
I give a fairly thorough analysis, with many references, of results related to your question in this 13 May 2005 sci.math post archived at Math Forum.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:53
$begingroup$
I give a fairly thorough analysis, with many references, of results related to your question in this 13 May 2005 sci.math post archived at Math Forum.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:53
$begingroup$
I give a fairly thorough analysis, with many references, of results related to your question in this 13 May 2005 sci.math post archived at Math Forum.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:53
add a comment |
1 Answer
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No. I will give a proof using Probability Theory. There exist i.i.d random variables ${X_n}$ on $[0,1]$ with Lebesgue measure such that $P{X_n=1}=P{X_n=-1}=frac 1 2$ for all $n$. Let $E_n={X_n=1}$. Then $P(E_n)=frac 1 2$ for all $n$ but $P{ X_{n_k}=1,k=1,2...}=0$ for any subsequence $(n_k)$.
Translated to a non-probabilistic setting this says the following: expand each $x in (0,1)$ to base $2$ as $x= sum frac {a_n(x)} {2^{n}}$ with $a_n(x) in {0,1}$. Then $E_n{x:a_n(x)=1}$ serves as counterexample to your statement.
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
No. I will give a proof using Probability Theory. There exist i.i.d random variables ${X_n}$ on $[0,1]$ with Lebesgue measure such that $P{X_n=1}=P{X_n=-1}=frac 1 2$ for all $n$. Let $E_n={X_n=1}$. Then $P(E_n)=frac 1 2$ for all $n$ but $P{ X_{n_k}=1,k=1,2...}=0$ for any subsequence $(n_k)$.
Translated to a non-probabilistic setting this says the following: expand each $x in (0,1)$ to base $2$ as $x= sum frac {a_n(x)} {2^{n}}$ with $a_n(x) in {0,1}$. Then $E_n{x:a_n(x)=1}$ serves as counterexample to your statement.
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
add a comment |
$begingroup$
No. I will give a proof using Probability Theory. There exist i.i.d random variables ${X_n}$ on $[0,1]$ with Lebesgue measure such that $P{X_n=1}=P{X_n=-1}=frac 1 2$ for all $n$. Let $E_n={X_n=1}$. Then $P(E_n)=frac 1 2$ for all $n$ but $P{ X_{n_k}=1,k=1,2...}=0$ for any subsequence $(n_k)$.
Translated to a non-probabilistic setting this says the following: expand each $x in (0,1)$ to base $2$ as $x= sum frac {a_n(x)} {2^{n}}$ with $a_n(x) in {0,1}$. Then $E_n{x:a_n(x)=1}$ serves as counterexample to your statement.
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
add a comment |
$begingroup$
No. I will give a proof using Probability Theory. There exist i.i.d random variables ${X_n}$ on $[0,1]$ with Lebesgue measure such that $P{X_n=1}=P{X_n=-1}=frac 1 2$ for all $n$. Let $E_n={X_n=1}$. Then $P(E_n)=frac 1 2$ for all $n$ but $P{ X_{n_k}=1,k=1,2...}=0$ for any subsequence $(n_k)$.
Translated to a non-probabilistic setting this says the following: expand each $x in (0,1)$ to base $2$ as $x= sum frac {a_n(x)} {2^{n}}$ with $a_n(x) in {0,1}$. Then $E_n{x:a_n(x)=1}$ serves as counterexample to your statement.
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
No. I will give a proof using Probability Theory. There exist i.i.d random variables ${X_n}$ on $[0,1]$ with Lebesgue measure such that $P{X_n=1}=P{X_n=-1}=frac 1 2$ for all $n$. Let $E_n={X_n=1}$. Then $P(E_n)=frac 1 2$ for all $n$ but $P{ X_{n_k}=1,k=1,2...}=0$ for any subsequence $(n_k)$.
Translated to a non-probabilistic setting this says the following: expand each $x in (0,1)$ to base $2$ as $x= sum frac {a_n(x)} {2^{n}}$ with $a_n(x) in {0,1}$. Then $E_n{x:a_n(x)=1}$ serves as counterexample to your statement.
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
edited Jan 8 at 8:37
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 8 at 8:32
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
add a comment |
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
$begingroup$
Thank you very much for your answer!
$endgroup$
– Guohuan Zhao
Jan 8 at 8:51
add a comment |
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$begingroup$
I give a fairly thorough analysis, with many references, of results related to your question in this 13 May 2005 sci.math post archived at Math Forum.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:53