Convergence and Measure
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Prove that if $(f_n)$ is a sequence of nonnegative, measurable functions on $[a,b]$ such that $lim_{ntoinfty}int_a^b f_n(x)dx=0$, then $(f_n)$ converges to $0$ in measure. Show by example that we cannot replace the conclusion with the assertion that $(f_n)$ converges to $0$ almost everywhere.
I don't really know how to go about proving this. I know that convergence almost everywhere implies convergence in measure and that if a sequence converges in measure then there exists a subsequence that converges almost everywhere but I haven't done a lot with convergence in measure.
real-analysis measure-theory convergence lebesgue-measure
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
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add a comment |
$begingroup$
Prove that if $(f_n)$ is a sequence of nonnegative, measurable functions on $[a,b]$ such that $lim_{ntoinfty}int_a^b f_n(x)dx=0$, then $(f_n)$ converges to $0$ in measure. Show by example that we cannot replace the conclusion with the assertion that $(f_n)$ converges to $0$ almost everywhere.
I don't really know how to go about proving this. I know that convergence almost everywhere implies convergence in measure and that if a sequence converges in measure then there exists a subsequence that converges almost everywhere but I haven't done a lot with convergence in measure.
real-analysis measure-theory convergence lebesgue-measure
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
$endgroup$
$begingroup$
For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$.
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– p4sch
Dec 4 '18 at 22:03
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So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem?
$endgroup$
– ICanMakeYouHateME
Dec 5 '18 at 3:28
add a comment |
$begingroup$
Prove that if $(f_n)$ is a sequence of nonnegative, measurable functions on $[a,b]$ such that $lim_{ntoinfty}int_a^b f_n(x)dx=0$, then $(f_n)$ converges to $0$ in measure. Show by example that we cannot replace the conclusion with the assertion that $(f_n)$ converges to $0$ almost everywhere.
I don't really know how to go about proving this. I know that convergence almost everywhere implies convergence in measure and that if a sequence converges in measure then there exists a subsequence that converges almost everywhere but I haven't done a lot with convergence in measure.
real-analysis measure-theory convergence lebesgue-measure
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
$endgroup$
Prove that if $(f_n)$ is a sequence of nonnegative, measurable functions on $[a,b]$ such that $lim_{ntoinfty}int_a^b f_n(x)dx=0$, then $(f_n)$ converges to $0$ in measure. Show by example that we cannot replace the conclusion with the assertion that $(f_n)$ converges to $0$ almost everywhere.
I don't really know how to go about proving this. I know that convergence almost everywhere implies convergence in measure and that if a sequence converges in measure then there exists a subsequence that converges almost everywhere but I haven't done a lot with convergence in measure.
real-analysis measure-theory convergence lebesgue-measure
real-analysis measure-theory convergence lebesgue-measure
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
edited Dec 4 '18 at 22:01
Foobaz John
21.6k41351
21.6k41351
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
asked Dec 4 '18 at 21:56
ICanMakeYouHateMEICanMakeYouHateME
154
154
$begingroup$
For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$.
$endgroup$
– p4sch
Dec 4 '18 at 22:03
$begingroup$
So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem?
$endgroup$
– ICanMakeYouHateME
Dec 5 '18 at 3:28
add a comment |
$begingroup$
For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$.
$endgroup$
– p4sch
Dec 4 '18 at 22:03
$begingroup$
So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem?
$endgroup$
– ICanMakeYouHateME
Dec 5 '18 at 3:28
$begingroup$
For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$.
$endgroup$
– p4sch
Dec 4 '18 at 22:03
$begingroup$
For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$.
$endgroup$
– p4sch
Dec 4 '18 at 22:03
$begingroup$
So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem?
$endgroup$
– ICanMakeYouHateME
Dec 5 '18 at 3:28
$begingroup$
So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem?
$endgroup$
– ICanMakeYouHateME
Dec 5 '18 at 3:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use Markov's inequality namely,
$$
mu(|f_n|>varepsilon)leq frac{1}{varepsilon}int_a^b f_n(x), dx.
$$
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
$endgroup$
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
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And you have outed yourself as a statistician :). +1
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– qbert
Dec 4 '18 at 22:04
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@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use Markov's inequality namely,
$$
mu(|f_n|>varepsilon)leq frac{1}{varepsilon}int_a^b f_n(x), dx.
$$
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
$endgroup$
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
$begingroup$
And you have outed yourself as a statistician :). +1
$endgroup$
– qbert
Dec 4 '18 at 22:04
$begingroup$
@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
add a comment |
$begingroup$
Use Markov's inequality namely,
$$
mu(|f_n|>varepsilon)leq frac{1}{varepsilon}int_a^b f_n(x), dx.
$$
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
$endgroup$
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
$begingroup$
And you have outed yourself as a statistician :). +1
$endgroup$
– qbert
Dec 4 '18 at 22:04
$begingroup$
@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
add a comment |
$begingroup$
Use Markov's inequality namely,
$$
mu(|f_n|>varepsilon)leq frac{1}{varepsilon}int_a^b f_n(x), dx.
$$
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
$endgroup$
Use Markov's inequality namely,
$$
mu(|f_n|>varepsilon)leq frac{1}{varepsilon}int_a^b f_n(x), dx.
$$
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
answered Dec 4 '18 at 22:01
Foobaz JohnFoobaz John
21.6k41351
21.6k41351
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
$begingroup$
And you have outed yourself as a statistician :). +1
$endgroup$
– qbert
Dec 4 '18 at 22:04
$begingroup$
@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
add a comment |
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
$begingroup$
And you have outed yourself as a statistician :). +1
$endgroup$
– qbert
Dec 4 '18 at 22:04
$begingroup$
@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
$begingroup$
So I don't have Markov's inequality at my disposal. Some theorems I have are The Dominated Convergence Theorem, Egorov's Theorem, Almost everywhere implies in measure, and Riesz's theorem which I stated above.
$endgroup$
– ICanMakeYouHateME
Dec 4 '18 at 22:04
$begingroup$
And you have outed yourself as a statistician :). +1
$endgroup$
– qbert
Dec 4 '18 at 22:04
$begingroup$
And you have outed yourself as a statistician :). +1
$endgroup$
– qbert
Dec 4 '18 at 22:04
$begingroup$
@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
$begingroup$
@ICanMakeYouHateME It is not too hard to prove the inequality from scratch in your particular case.
$endgroup$
– Foobaz John
Dec 4 '18 at 22:13
add a comment |
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$begingroup$
For an counterexample take an example with a 'running maximum', that is $f_{n,k}(x) = 1_{[k/,(k+1)/n]}(x)$.
$endgroup$
– p4sch
Dec 4 '18 at 22:03
$begingroup$
So I understand the example that shows this but without using Markovs inequality where should I start in proving the theorem?
$endgroup$
– ICanMakeYouHateME
Dec 5 '18 at 3:28