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Does convergence in measure imply convergence in mean?

How do I look up applications of this type of convergence?Convergence in metric and in Borel measureConvergence in measure of productsInstance where local convergence in measure implies global convergence in measureIf $f_n to f$ and $g_n to g$ in measure and $mu$ is finite, then $f_n g_n to fg$ in measureConvergence in measure iff convergence pointwise a.e.convergence in measure implies convergence in mean?Does convergence in the mean imply convergence in measure?Possible error in Donald Cohns measure theory on definition of $mathscrL^infty$Checking work on three successive questions in Cohn, Measure Theory

0

$begingroup$

Here is the proposition. I doubt if the author's proof is correct.

$mathscrL^1(X,mathscrA,mu,mathbbR)$ is the set of all real-valued ([rather than $[-infty,+infty]$] -valued) integrable functions on $X$.

==============

==============

The first part of the proof is easy. But I doubt whether the second part is correct. That is to say, when these assumptions are satisfied, does convergence in measure imply convergence in mean?

The red line part is confusing. In Royden's Real Analysis, the red line part is true when the assumption $mu(X)<+infty$ is added. See Corollary 19 below. However, Cohn doesn't give this assumption.

==============

==============

For the second part of proof, is Cohn's approach correct? Do we need the assumption that the measure is finite?

real-analysis probability functional-analysis measure-theory stochastic-processes

share|cite|improve this question

edited Mar 11 at 8:44

DavidRock

asked Mar 11 at 2:20

DavidRockDavidRock

337

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused why Cohn argues that If $f_n$ converges to $f$ in measure, then every subsequence of $f_n$ has a subsequence converges to $f$ almost everywhere. Why does Cohn use the word EVERY?
  $endgroup$
  – DavidRock
  Mar 11 at 7:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you start with the whole sequence $f_n$ you will only get a subsequence which converges in the mean. We have to show that $f_n$ itself converges in the mean, not some subsequence. So the idea is to you the following: a sequence of real numbers converges to $0$ iff EVERY subsequence of it has a further sub sequence which converges to $0$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 8:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Kavi Rama Murthy! That's the point!! You are a great teacher!!
  $endgroup$
  – DavidRock
  Mar 11 at 8:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would surprise me if Cohn's book co tained a mistake (at least the second edition). He is a great author imo.
  $endgroup$
  – Math_QED
  Mar 11 at 14:02
  

  
  
  
  

add a comment | 

0

$begingroup$

Here is the proposition. I doubt if the author's proof is correct.

$mathscrL^1(X,mathscrA,mu,mathbbR)$ is the set of all real-valued ([rather than $[-infty,+infty]$] -valued) integrable functions on $X$.

==============

==============

The first part of the proof is easy. But I doubt whether the second part is correct. That is to say, when these assumptions are satisfied, does convergence in measure imply convergence in mean?

The red line part is confusing. In Royden's Real Analysis, the red line part is true when the assumption $mu(X)<+infty$ is added. See Corollary 19 below. However, Cohn doesn't give this assumption.

==============

==============

For the second part of proof, is Cohn's approach correct? Do we need the assumption that the measure is finite?

real-analysis probability functional-analysis measure-theory stochastic-processes

share|cite|improve this question

edited Mar 11 at 8:44

DavidRock

asked Mar 11 at 2:20

DavidRockDavidRock

337

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused why Cohn argues that If $f_n$ converges to $f$ in measure, then every subsequence of $f_n$ has a subsequence converges to $f$ almost everywhere. Why does Cohn use the word EVERY?
  $endgroup$
  – DavidRock
  Mar 11 at 7:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you start with the whole sequence $f_n$ you will only get a subsequence which converges in the mean. We have to show that $f_n$ itself converges in the mean, not some subsequence. So the idea is to you the following: a sequence of real numbers converges to $0$ iff EVERY subsequence of it has a further sub sequence which converges to $0$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 8:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, @Kavi Rama Murthy! That's the point!! You are a great teacher!!
  $endgroup$
  – DavidRock
  Mar 11 at 8:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would surprise me if Cohn's book co tained a mistake (at least the second edition). He is a great author imo.
  $endgroup$
  – Math_QED
  Mar 11 at 14:02
  

  
  
  
  

add a comment | 

$begingroup$

Here is the proposition. I doubt if the author's proof is correct.

$mathscrL^1(X,mathscrA,mu,mathbbR)$ is the set of all real-valued ([rather than $[-infty,+infty]$] -valued) integrable functions on $X$.

==============

==============

The first part of the proof is easy. But I doubt whether the second part is correct. That is to say, when these assumptions are satisfied, does convergence in measure imply convergence in mean?

The red line part is confusing. In Royden's Real Analysis, the red line part is true when the assumption $mu(X)<+infty$ is added. See Corollary 19 below. However, Cohn doesn't give this assumption.

==============

==============

For the second part of proof, is Cohn's approach correct? Do we need the assumption that the measure is finite?

real-analysis probability functional-analysis measure-theory stochastic-processes

share|cite|improve this question

edited Mar 11 at 8:44

DavidRock

asked Mar 11 at 2:20

DavidRockDavidRock

337

$endgroup$

share|cite|improve this question

edited Mar 11 at 8:44

DavidRock

asked Mar 11 at 2:20

DavidRockDavidRock

337

share|cite|improve this question

edited Mar 11 at 8:44

DavidRock

asked Mar 11 at 2:20

DavidRockDavidRock

337

asked Mar 11 at 2:20

DavidRockDavidRock

337

asked Mar 11 at 2:20

DavidRockDavidRock

337

2 Answers
2

active

oldest

votes

1

$begingroup$

In any measure space if a sequence $f_n$ converges in measure then there is a subsequence which converges almost everywhere. The given measure need not be finite. Proof: choose $k_n$ such that $k_n <k_n+1$ and $mu f_k_n(x)-f(x) <frac 1 2^n$. Then $sum_n muf_k_n(x)-f(x) <infty$. This implies that $mu (lim sup f_k_n(x)-f(x)=0$. it follows that $f_k_n to f$ almost everywhere.

share|cite|improve this answer

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, Kavi Rama Murthy!! Your approach is so concise. Theorem: If $f_n$ converges to $f$, then there is a subsequence of $f_n$ that converges to $f$ almost everywhere. This theorem is easy to prove. However, I'm confused why Cohn argues that If $f_n$ converges to $f$ in measure, then every subsequence of $f_n$ has a subsequence converges to $f$ almost everywhere. Why does Cohn use the word EVERY?
  $endgroup$
  – DavidRock
  Mar 11 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A sequence of real numbers converges to 00 iff EVERY subsequence of it has a further sub sequence which converges to 0. Thank you, Kavi Rama Murthy!! That's the point!! You are a great teacher!!
  $endgroup$
  – DavidRock
  Mar 11 at 8:43
  

  
  
  
  

add a comment | 

0

$begingroup$

The part underlined in red does not require the measure space to be finite. It is mostly a consequence of the Borel-Cantelli lemma, which works in any measure space. It's a bit vague, but I always remember it as "if a sequence of functions converges in measure fast enough, then it converges almost everywhere" and taking subsequences lets you increase the rate of convergence.

share|cite|improve this answer

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But Borel-Cantelli lemma requires that $sum_n=1^infty mu(E_n)<infty$.
  $endgroup$
  – DavidRock
  Mar 11 at 5:52
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

In any measure space if a sequence $f_n$ converges in measure then there is a subsequence which converges almost everywhere. The given measure need not be finite. Proof: choose $k_n$ such that $k_n <k_n+1$ and $mu f_k_n(x)-f(x) <frac 1 2^n$. Then $sum_n muf_k_n(x)-f(x) <infty$. This implies that $mu (lim sup f_k_n(x)-f(x)=0$. it follows that $f_k_n to f$ almost everywhere.

share|cite|improve this answer

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, Kavi Rama Murthy!! Your approach is so concise. Theorem: If $f_n$ converges to $f$, then there is a subsequence of $f_n$ that converges to $f$ almost everywhere. This theorem is easy to prove. However, I'm confused why Cohn argues that If $f_n$ converges to $f$ in measure, then every subsequence of $f_n$ has a subsequence converges to $f$ almost everywhere. Why does Cohn use the word EVERY?
  $endgroup$
  – DavidRock
  Mar 11 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A sequence of real numbers converges to 00 iff EVERY subsequence of it has a further sub sequence which converges to 0. Thank you, Kavi Rama Murthy!! That's the point!! You are a great teacher!!
  $endgroup$
  – DavidRock
  Mar 11 at 8:43
  

  
  
  
  

add a comment | 

1

$begingroup$

In any measure space if a sequence $f_n$ converges in measure then there is a subsequence which converges almost everywhere. The given measure need not be finite. Proof: choose $k_n$ such that $k_n <k_n+1$ and $mu f_k_n(x)-f(x) <frac 1 2^n$. Then $sum_n muf_k_n(x)-f(x) <infty$. This implies that $mu (lim sup f_k_n(x)-f(x)=0$. it follows that $f_k_n to f$ almost everywhere.

share|cite|improve this answer

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, Kavi Rama Murthy!! Your approach is so concise. Theorem: If $f_n$ converges to $f$, then there is a subsequence of $f_n$ that converges to $f$ almost everywhere. This theorem is easy to prove. However, I'm confused why Cohn argues that If $f_n$ converges to $f$ in measure, then every subsequence of $f_n$ has a subsequence converges to $f$ almost everywhere. Why does Cohn use the word EVERY?
  $endgroup$
  – DavidRock
  Mar 11 at 7:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A sequence of real numbers converges to 00 iff EVERY subsequence of it has a further sub sequence which converges to 0. Thank you, Kavi Rama Murthy!! That's the point!! You are a great teacher!!
  $endgroup$
  – DavidRock
  Mar 11 at 8:43
  

  
  
  
  

add a comment | 

$begingroup$

In any measure space if a sequence $f_n$ converges in measure then there is a subsequence which converges almost everywhere. The given measure need not be finite. Proof: choose $k_n$ such that $k_n <k_n+1$ and $mu f_k_n(x)-f(x) <frac 1 2^n$. Then $sum_n muf_k_n(x)-f(x) <infty$. This implies that $mu (lim sup f_k_n(x)-f(x)=0$. it follows that $f_k_n to f$ almost everywhere.

share|cite|improve this answer

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

$endgroup$

share|cite|improve this answer

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

answered Mar 11 at 7:29

Kavi Rama MurthyKavi Rama Murthy

67.2k53067

0

$begingroup$

The part underlined in red does not require the measure space to be finite. It is mostly a consequence of the Borel-Cantelli lemma, which works in any measure space. It's a bit vague, but I always remember it as "if a sequence of functions converges in measure fast enough, then it converges almost everywhere" and taking subsequences lets you increase the rate of convergence.

share|cite|improve this answer

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But Borel-Cantelli lemma requires that $sum_n=1^infty mu(E_n)<infty$.
  $endgroup$
  – DavidRock
  Mar 11 at 5:52
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The part underlined in red does not require the measure space to be finite. It is mostly a consequence of the Borel-Cantelli lemma, which works in any measure space. It's a bit vague, but I always remember it as "if a sequence of functions converges in measure fast enough, then it converges almost everywhere" and taking subsequences lets you increase the rate of convergence.

share|cite|improve this answer

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But Borel-Cantelli lemma requires that $sum_n=1^infty mu(E_n)<infty$.
  $endgroup$
  – DavidRock
  Mar 11 at 5:52
  
  

  
  
  
  

add a comment | 

$begingroup$

The part underlined in red does not require the measure space to be finite. It is mostly a consequence of the Borel-Cantelli lemma, which works in any measure space. It's a bit vague, but I always remember it as "if a sequence of functions converges in measure fast enough, then it converges almost everywhere" and taking subsequences lets you increase the rate of convergence.

share|cite|improve this answer

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315

$endgroup$

share|cite|improve this answer

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315

answered Mar 11 at 4:43

confused_walletconfused_wallet

596315