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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

Let $mu$ be a finite measure on $mathbbR$ satisfying,
$$int_mathbbRf(x)dmu(x)=0,~forall fin C_c(mathbbR)$$
Then is it true that $mu =0$?

We know that the result is true for $L^1(mathbbR)$, which is a subspace of the above.

Edit after the comments of Kavi Rama Murthy Sir:

Is the result also true for complex measure $mu$ on $mathbbR$?

functional-analysis measure-theory

share|cite|improve this question

edited Jul 8 at 7:41

Prof.Hijibiji

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Let $mu$ be a finite measure on $mathbbR$ satisfying,
$$int_mathbbRf(x)dmu(x)=0,~forall fin C_c(mathbbR)$$
Then is it true that $mu =0$?

We know that the result is true for $L^1(mathbbR)$, which is a subspace of the above.

Edit after the comments of Kavi Rama Murthy Sir:

Is the result also true for complex measure $mu$ on $mathbbR$?

functional-analysis measure-theory

share|cite|improve this question

edited Jul 8 at 7:41

Prof.Hijibiji

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $mu$ be a finite measure on $mathbbR$ satisfying,
$$int_mathbbRf(x)dmu(x)=0,~forall fin C_c(mathbbR)$$
Then is it true that $mu =0$?

We know that the result is true for $L^1(mathbbR)$, which is a subspace of the above.

Edit after the comments of Kavi Rama Murthy Sir:

Is the result also true for complex measure $mu$ on $mathbbR$?

functional-analysis measure-theory

share|cite|improve this question

edited Jul 8 at 7:41

Prof.Hijibiji

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 8 at 7:41

Prof.Hijibiji

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

share|cite|improve this question

edited Jul 8 at 7:41

Prof.Hijibiji

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

asked Jul 8 at 3:55

Prof.HijibijiProf.Hijibiji

801010 bronze badges

2 Answers
2

active

oldest

votes

1

$begingroup$

If we prove this for real measures the result for complex measures follows by just taking real and imaginary parts. Any real measure is the difference of two positive finite measures. So what we have to prove is the following:
if $mu$ and $nu$ are two finite positive measures such that $int f dmu=int f dnu$ for all $f in C_c(mathbb R)$ then $mu =nu$.

For this note that given $epsilon >0$ we can find $N$ such that $mu ([-N,N])^c) <epsilon$ and $nu ([-N,N])^c) <epsilon$. There exists $f in C_c(mathbb R)$ such that $0 leq f(x) leq 1$ for all $x$ and $f(x)=1$ for $|x| leq N$. If $g$ is any bounded continuous function then $fg in C_c(mathbb R)$ so $int fg dmu=int fg dnu$. I now leave it to you put these together to conclude that $int g dmu=int g dnu$. The fact that this hods for all bounded continuous$f$ implies that $mu =nu$. [This is standard and it is part of Portmanteus's Theorem characterizing weak convergence of probability measures].

share|cite|improve this answer

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sir i am thinking that, there exists h, a complex borel function such that |h|=1 and $int fdmu= int fhd|mu|$. If h is real then $hd|mu|=0$ implies $|mu|=0$ as |h|=1, implies $mu=0$. If h is complex then will break h into real and imaginary and apply above separately. Are there any flaw?
  $endgroup$
  – Prof.Hijibiji
  Jul 9 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The argument for going from the real to the complex case is simpler. We can write $mu =mu_1+imu_2$ where $mu_1$ and $mu_2$ are real measures. For any real valued function $f in C_c(mathbb R)$ we get $int f dmu_1=0$ and $int f dmu_2=0$ , so $mu_1=mu_2=0$ from the real case.
  $endgroup$
  – Kavi Rama Murthy
  Jul 9 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sir, my above comment was about the proof for complex measure using the result for finite positive measure. Can you please check whether my argument was right or not. Thanking you.
  $endgroup$
  – Prof.Hijibiji
  Jul 9 at 12:26
  

  
  
  
  

add a comment | 

4

$begingroup$

I guess that for all $n=1,2,3,dots$ you could consider a non-negative function $f in mathrmC_mathrmc(mathbbR)$ such that $f=1$ on $[-n,n]$. Therefore for all $n=1,2,3,dots$

beginequation
mu([-n,n]) leq int_mathbbR f d mu = 0.
endequation

By countable subadditivity $mu(mathbbR)=0$ and we are done.
I hope this actually works, it seems too easy but I don’t see any errors for now.

share|cite|improve this answer

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is more interesting for complex measures. For positive measures this answer is fine.
  $endgroup$
  – Kavi Rama Murthy
  Jul 8 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kavu Rama Murthy Sir, can you shed some light for complex measure?
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then I guess the total variation $|mu|$ of the complex measure $mu$ will do the job. @KaviRamaMurthy sir
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Prof.Hijibiji It is more complicated than that. It is not obvious that $int f d|mu|=0$ for $f in C_c(mathbb R)$.
  $endgroup$
  – Kavi Rama Murthy
  Jul 8 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oopps, I missed that. So can you please write an answer or give me a hint for the complex measure! Thank you.
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:35
  

  
  
  
  

 | 
show 1 more comment

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1

$begingroup$

If we prove this for real measures the result for complex measures follows by just taking real and imaginary parts. Any real measure is the difference of two positive finite measures. So what we have to prove is the following:
if $mu$ and $nu$ are two finite positive measures such that $int f dmu=int f dnu$ for all $f in C_c(mathbb R)$ then $mu =nu$.

For this note that given $epsilon >0$ we can find $N$ such that $mu ([-N,N])^c) <epsilon$ and $nu ([-N,N])^c) <epsilon$. There exists $f in C_c(mathbb R)$ such that $0 leq f(x) leq 1$ for all $x$ and $f(x)=1$ for $|x| leq N$. If $g$ is any bounded continuous function then $fg in C_c(mathbb R)$ so $int fg dmu=int fg dnu$. I now leave it to you put these together to conclude that $int g dmu=int g dnu$. The fact that this hods for all bounded continuous$f$ implies that $mu =nu$. [This is standard and it is part of Portmanteus's Theorem characterizing weak convergence of probability measures].

share|cite|improve this answer

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sir i am thinking that, there exists h, a complex borel function such that |h|=1 and $int fdmu= int fhd|mu|$. If h is real then $hd|mu|=0$ implies $|mu|=0$ as |h|=1, implies $mu=0$. If h is complex then will break h into real and imaginary and apply above separately. Are there any flaw?
  $endgroup$
  – Prof.Hijibiji
  Jul 9 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The argument for going from the real to the complex case is simpler. We can write $mu =mu_1+imu_2$ where $mu_1$ and $mu_2$ are real measures. For any real valued function $f in C_c(mathbb R)$ we get $int f dmu_1=0$ and $int f dmu_2=0$ , so $mu_1=mu_2=0$ from the real case.
  $endgroup$
  – Kavi Rama Murthy
  Jul 9 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sir, my above comment was about the proof for complex measure using the result for finite positive measure. Can you please check whether my argument was right or not. Thanking you.
  $endgroup$
  – Prof.Hijibiji
  Jul 9 at 12:26
  

  
  
  
  

add a comment | 

1

$begingroup$

If we prove this for real measures the result for complex measures follows by just taking real and imaginary parts. Any real measure is the difference of two positive finite measures. So what we have to prove is the following:
if $mu$ and $nu$ are two finite positive measures such that $int f dmu=int f dnu$ for all $f in C_c(mathbb R)$ then $mu =nu$.

For this note that given $epsilon >0$ we can find $N$ such that $mu ([-N,N])^c) <epsilon$ and $nu ([-N,N])^c) <epsilon$. There exists $f in C_c(mathbb R)$ such that $0 leq f(x) leq 1$ for all $x$ and $f(x)=1$ for $|x| leq N$. If $g$ is any bounded continuous function then $fg in C_c(mathbb R)$ so $int fg dmu=int fg dnu$. I now leave it to you put these together to conclude that $int g dmu=int g dnu$. The fact that this hods for all bounded continuous$f$ implies that $mu =nu$. [This is standard and it is part of Portmanteus's Theorem characterizing weak convergence of probability measures].

share|cite|improve this answer

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sir i am thinking that, there exists h, a complex borel function such that |h|=1 and $int fdmu= int fhd|mu|$. If h is real then $hd|mu|=0$ implies $|mu|=0$ as |h|=1, implies $mu=0$. If h is complex then will break h into real and imaginary and apply above separately. Are there any flaw?
  $endgroup$
  – Prof.Hijibiji
  Jul 9 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The argument for going from the real to the complex case is simpler. We can write $mu =mu_1+imu_2$ where $mu_1$ and $mu_2$ are real measures. For any real valued function $f in C_c(mathbb R)$ we get $int f dmu_1=0$ and $int f dmu_2=0$ , so $mu_1=mu_2=0$ from the real case.
  $endgroup$
  – Kavi Rama Murthy
  Jul 9 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sir, my above comment was about the proof for complex measure using the result for finite positive measure. Can you please check whether my argument was right or not. Thanking you.
  $endgroup$
  – Prof.Hijibiji
  Jul 9 at 12:26
  

  
  
  
  

add a comment | 

$begingroup$

If we prove this for real measures the result for complex measures follows by just taking real and imaginary parts. Any real measure is the difference of two positive finite measures. So what we have to prove is the following:
if $mu$ and $nu$ are two finite positive measures such that $int f dmu=int f dnu$ for all $f in C_c(mathbb R)$ then $mu =nu$.

For this note that given $epsilon >0$ we can find $N$ such that $mu ([-N,N])^c) <epsilon$ and $nu ([-N,N])^c) <epsilon$. There exists $f in C_c(mathbb R)$ such that $0 leq f(x) leq 1$ for all $x$ and $f(x)=1$ for $|x| leq N$. If $g$ is any bounded continuous function then $fg in C_c(mathbb R)$ so $int fg dmu=int fg dnu$. I now leave it to you put these together to conclude that $int g dmu=int g dnu$. The fact that this hods for all bounded continuous$f$ implies that $mu =nu$. [This is standard and it is part of Portmanteus's Theorem characterizing weak convergence of probability measures].

share|cite|improve this answer

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

answered Jul 9 at 7:27

Kavi Rama MurthyKavi Rama Murthy

99.2k55 gold badges4242 silver badges8080 bronze badges

4

$begingroup$

I guess that for all $n=1,2,3,dots$ you could consider a non-negative function $f in mathrmC_mathrmc(mathbbR)$ such that $f=1$ on $[-n,n]$. Therefore for all $n=1,2,3,dots$

beginequation
mu([-n,n]) leq int_mathbbR f d mu = 0.
endequation

By countable subadditivity $mu(mathbbR)=0$ and we are done.
I hope this actually works, it seems too easy but I don’t see any errors for now.

share|cite|improve this answer

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is more interesting for complex measures. For positive measures this answer is fine.
  $endgroup$
  – Kavi Rama Murthy
  Jul 8 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kavu Rama Murthy Sir, can you shed some light for complex measure?
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then I guess the total variation $|mu|$ of the complex measure $mu$ will do the job. @KaviRamaMurthy sir
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Prof.Hijibiji It is more complicated than that. It is not obvious that $int f d|mu|=0$ for $f in C_c(mathbb R)$.
  $endgroup$
  – Kavi Rama Murthy
  Jul 8 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oopps, I missed that. So can you please write an answer or give me a hint for the complex measure! Thank you.
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:35
  

  
  
  
  

 | 
show 1 more comment

4

$begingroup$

I guess that for all $n=1,2,3,dots$ you could consider a non-negative function $f in mathrmC_mathrmc(mathbbR)$ such that $f=1$ on $[-n,n]$. Therefore for all $n=1,2,3,dots$

beginequation
mu([-n,n]) leq int_mathbbR f d mu = 0.
endequation

By countable subadditivity $mu(mathbbR)=0$ and we are done.
I hope this actually works, it seems too easy but I don’t see any errors for now.

share|cite|improve this answer

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question is more interesting for complex measures. For positive measures this answer is fine.
  $endgroup$
  – Kavi Rama Murthy
  Jul 8 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kavu Rama Murthy Sir, can you shed some light for complex measure?
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then I guess the total variation $|mu|$ of the complex measure $mu$ will do the job. @KaviRamaMurthy sir
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Prof.Hijibiji It is more complicated than that. It is not obvious that $int f d|mu|=0$ for $f in C_c(mathbb R)$.
  $endgroup$
  – Kavi Rama Murthy
  Jul 8 at 7:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oopps, I missed that. So can you please write an answer or give me a hint for the complex measure! Thank you.
  $endgroup$
  – Prof.Hijibiji
  Jul 8 at 7:35
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I guess that for all $n=1,2,3,dots$ you could consider a non-negative function $f in mathrmC_mathrmc(mathbbR)$ such that $f=1$ on $[-n,n]$. Therefore for all $n=1,2,3,dots$

beginequation
mu([-n,n]) leq int_mathbbR f d mu = 0.
endequation

By countable subadditivity $mu(mathbbR)=0$ and we are done.
I hope this actually works, it seems too easy but I don’t see any errors for now.

share|cite|improve this answer

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges

answered Jul 8 at 4:35

FedericoFederico

12877 bronze badges