Convergence in measure implies uniform convergence on a set of finite measure
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Question is :
Does there exist a sequence $left(f_kright)$ of Lebesgue measurable functions such that $f_k$ converges to $0$ in measure in $mathbb{R}$ but no subsequence converges uniformly on any subset of positive measure?
See that $left(f_kright)$ converges to $0$ in measure implies there exists a subsequence $left(f_{n_k}right)$ that converges point wise almost every where to $0$.
Then by Egoroff's theorem we can say that on a set of finite measure, $left(f_{n_k}right)$ converges uniformly almost everywhere to $0$.
I am not sure if we can say it converges uniformly and not just almost everywhere.
measure-theory convergence lebesgue-measure uniform-convergence
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
Question is :
Does there exist a sequence $left(f_kright)$ of Lebesgue measurable functions such that $f_k$ converges to $0$ in measure in $mathbb{R}$ but no subsequence converges uniformly on any subset of positive measure?
See that $left(f_kright)$ converges to $0$ in measure implies there exists a subsequence $left(f_{n_k}right)$ that converges point wise almost every where to $0$.
Then by Egoroff's theorem we can say that on a set of finite measure, $left(f_{n_k}right)$ converges uniformly almost everywhere to $0$.
I am not sure if we can say it converges uniformly and not just almost everywhere.
measure-theory convergence lebesgue-measure uniform-convergence
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
$endgroup$
$begingroup$
1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere.
$endgroup$
– Ramiro
May 7 '16 at 6:08
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2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$.
$endgroup$
– Ramiro
May 7 '16 at 6:25
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@Ramiro : I understand 1st point but not second point
$endgroup$
– user311526
May 7 '16 at 7:34
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $varepsilon>0$, there is a set measurable set $A$ such that $m(A)<varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D).
$endgroup$
– Ramiro
May 8 '16 at 22:54
add a comment |
$begingroup$
Question is :
Does there exist a sequence $left(f_kright)$ of Lebesgue measurable functions such that $f_k$ converges to $0$ in measure in $mathbb{R}$ but no subsequence converges uniformly on any subset of positive measure?
See that $left(f_kright)$ converges to $0$ in measure implies there exists a subsequence $left(f_{n_k}right)$ that converges point wise almost every where to $0$.
Then by Egoroff's theorem we can say that on a set of finite measure, $left(f_{n_k}right)$ converges uniformly almost everywhere to $0$.
I am not sure if we can say it converges uniformly and not just almost everywhere.
measure-theory convergence lebesgue-measure uniform-convergence
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
$endgroup$
Question is :
Does there exist a sequence $left(f_kright)$ of Lebesgue measurable functions such that $f_k$ converges to $0$ in measure in $mathbb{R}$ but no subsequence converges uniformly on any subset of positive measure?
See that $left(f_kright)$ converges to $0$ in measure implies there exists a subsequence $left(f_{n_k}right)$ that converges point wise almost every where to $0$.
Then by Egoroff's theorem we can say that on a set of finite measure, $left(f_{n_k}right)$ converges uniformly almost everywhere to $0$.
I am not sure if we can say it converges uniformly and not just almost everywhere.
measure-theory convergence lebesgue-measure uniform-convergence
measure-theory convergence lebesgue-measure uniform-convergence
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
edited Jan 16 at 17:29
Davide Giraudo
128k17156268
128k17156268
asked May 7 '16 at 5:43
user311526
$begingroup$
1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere.
$endgroup$
– Ramiro
May 7 '16 at 6:08
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$.
$endgroup$
– Ramiro
May 7 '16 at 6:25
$begingroup$
@Ramiro : I understand 1st point but not second point
$endgroup$
– user311526
May 7 '16 at 7:34
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $varepsilon>0$, there is a set measurable set $A$ such that $m(A)<varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D).
$endgroup$
– Ramiro
May 8 '16 at 22:54
add a comment |
$begingroup$
1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere.
$endgroup$
– Ramiro
May 7 '16 at 6:08
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$.
$endgroup$
– Ramiro
May 7 '16 at 6:25
$begingroup$
@Ramiro : I understand 1st point but not second point
$endgroup$
– user311526
May 7 '16 at 7:34
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $varepsilon>0$, there is a set measurable set $A$ such that $m(A)<varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D).
$endgroup$
– Ramiro
May 8 '16 at 22:54
$begingroup$
1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere.
$endgroup$
– Ramiro
May 7 '16 at 6:08
$begingroup$
1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere.
$endgroup$
– Ramiro
May 7 '16 at 6:08
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$.
$endgroup$
– Ramiro
May 7 '16 at 6:25
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$.
$endgroup$
– Ramiro
May 7 '16 at 6:25
$begingroup$
@Ramiro : I understand 1st point but not second point
$endgroup$
– user311526
May 7 '16 at 7:34
$begingroup$
@Ramiro : I understand 1st point but not second point
$endgroup$
– user311526
May 7 '16 at 7:34
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $varepsilon>0$, there is a set measurable set $A$ such that $m(A)<varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D).
$endgroup$
– Ramiro
May 8 '16 at 22:54
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $varepsilon>0$, there is a set measurable set $A$ such that $m(A)<varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D).
$endgroup$
– Ramiro
May 8 '16 at 22:54
add a comment |
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$begingroup$
1. By Egoroff's theorem we can say that on a set of finite measure, if $(f_{n_k})$ that converges point wise almost every where to $0$, then $(f_{n_k})$ converges almost uniformly to $0$. Please, note that almost uniform convergence is not the same as uniform convergence almost everywhere.
$endgroup$
– Ramiro
May 7 '16 at 6:08
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly. As a consequence, there is a set of of finite measure where $(f_{n_k})$ converges uniformly to $0$.
$endgroup$
– Ramiro
May 7 '16 at 6:25
$begingroup$
@Ramiro : I understand 1st point but not second point
$endgroup$
– user311526
May 7 '16 at 7:34
$begingroup$
2. It is a known result that if a sequence $(f_k)$ of lebesgue measurable functions converges to $0$ in measure then there is subsequence $(f_{n_k})$ which converges almost uniformly to $0$. It means, for all $varepsilon>0$, there is a set measurable set $A$ such that $m(A)<varepsilon$ and $(f_{n_k})$ converges o $0$ uniformly on $mathbb{R}-A$. So the answer to your question is NO, there is not a sequence as you described in your question. (See, for instance, Bartle, Elements of Integration, theorem 7.11 or Halmos, Measure Theory, section 22, theorem D).
$endgroup$
– Ramiro
May 8 '16 at 22:54