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Existence of weak limit of measures

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4

1

$begingroup$

Suppose $mu_n$ is a sequence of Borel probability measures on $mathbbC$ such that

$$
lim_n to infty int fdmu_n
$$

exists for all $f in C_b(mathbbC, mathbbR)$. Is it true that there exists a measure $mu$ such that

$$
int f dmu = lim_n to infty int fdmu_n?
$$

In other words, is there a measure $mu$ such that $mu_n to mu$ weakly?

In my problem, we can assume the support of the $mu_n$ is contained in the unit ball, if it helps.

analysis probability-theory measure-theory

share|cite|improve this question

edited Aug 11 at 22:33

WeakestTopology

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The answers below are right, but another thing you might care about is whether the resulting $mu$ is actually a probability measure. In general, it doesn't have to be (imagine $mu_n$ spreading out to infinity, like a uniform distribution on the ball of radius $n$), but in your case it actually is, by just taking $f$ to be any continuous function with compact support that is $1$ on the unit ball.
  $endgroup$
  – J.G
  Aug 11 at 22:49
  

  
  
  
  

add a comment | 

4

1

$begingroup$

Suppose $mu_n$ is a sequence of Borel probability measures on $mathbbC$ such that

$$
lim_n to infty int fdmu_n
$$

exists for all $f in C_b(mathbbC, mathbbR)$. Is it true that there exists a measure $mu$ such that

$$
int f dmu = lim_n to infty int fdmu_n?
$$

In other words, is there a measure $mu$ such that $mu_n to mu$ weakly?

In my problem, we can assume the support of the $mu_n$ is contained in the unit ball, if it helps.

analysis probability-theory measure-theory

share|cite|improve this question

edited Aug 11 at 22:33

WeakestTopology

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The answers below are right, but another thing you might care about is whether the resulting $mu$ is actually a probability measure. In general, it doesn't have to be (imagine $mu_n$ spreading out to infinity, like a uniform distribution on the ball of radius $n$), but in your case it actually is, by just taking $f$ to be any continuous function with compact support that is $1$ on the unit ball.
  $endgroup$
  – J.G
  Aug 11 at 22:49
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $mu_n$ is a sequence of Borel probability measures on $mathbbC$ such that

$$
lim_n to infty int fdmu_n
$$

exists for all $f in C_b(mathbbC, mathbbR)$. Is it true that there exists a measure $mu$ such that

$$
int f dmu = lim_n to infty int fdmu_n?
$$

In other words, is there a measure $mu$ such that $mu_n to mu$ weakly?

In my problem, we can assume the support of the $mu_n$ is contained in the unit ball, if it helps.

analysis probability-theory measure-theory

share|cite|improve this question

edited Aug 11 at 22:33

WeakestTopology

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

$endgroup$

share|cite|improve this question

edited Aug 11 at 22:33

WeakestTopology

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

share|cite|improve this question

edited Aug 11 at 22:33

WeakestTopology

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

asked Aug 11 at 22:23

WeakestTopologyWeakestTopology

55333 silver badges1313 bronze badges

2 Answers
2

active

oldest

votes

2

$begingroup$

Yes! Consider the restriction to the space of continuous functions with compact support, denoted by $C_c(mathbbC, mathbbR)$. Now define $$L(f):= lim_nrightarrow infty int f , dmu_nqquad , , fin C_c(mathbbC, mathbbR).$$

It is straightforward to check that $L$ defines a bounded linear functional on $C_c(mathbbC,mathbbR)$. Indeed, since the $leftmu_nright$'s are Borel probability measures on $mathbbC$, we have

$$lvert L(f) rvert leq lvert lvert frvert rvert_infty$$ Hence by the Riesz–Markov–Kakutani representation theorem, there exists a unique Borel measure $mu$,with total variation-norm less than 1, such that $$L(f) = int f , dmu qquad forall fin C_c(mathbbC,mathbbR).$$
In fact, the integral is well-defined for all $C_b(mathbbC, mathbbR)$ as well.

share|cite|improve this answer

answered Aug 11 at 22:40

TheOscillatorTheOscillator

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$endgroup$

add a comment | 

3

$begingroup$

Define $l: C_c(mathbbC,mathbbR) to mathbbR$ by $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.

share|cite|improve this answer

edited Aug 12 at 13:20

answered Aug 11 at 22:37

mathworker21mathworker21

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2

$begingroup$

Yes! Consider the restriction to the space of continuous functions with compact support, denoted by $C_c(mathbbC, mathbbR)$. Now define $$L(f):= lim_nrightarrow infty int f , dmu_nqquad , , fin C_c(mathbbC, mathbbR).$$

It is straightforward to check that $L$ defines a bounded linear functional on $C_c(mathbbC,mathbbR)$. Indeed, since the $leftmu_nright$'s are Borel probability measures on $mathbbC$, we have

$$lvert L(f) rvert leq lvert lvert frvert rvert_infty$$ Hence by the Riesz–Markov–Kakutani representation theorem, there exists a unique Borel measure $mu$,with total variation-norm less than 1, such that $$L(f) = int f , dmu qquad forall fin C_c(mathbbC,mathbbR).$$
In fact, the integral is well-defined for all $C_b(mathbbC, mathbbR)$ as well.

share|cite|improve this answer

answered Aug 11 at 22:40

TheOscillatorTheOscillator

2,23411 gold badge88 silver badges1818 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Yes! Consider the restriction to the space of continuous functions with compact support, denoted by $C_c(mathbbC, mathbbR)$. Now define $$L(f):= lim_nrightarrow infty int f , dmu_nqquad , , fin C_c(mathbbC, mathbbR).$$

It is straightforward to check that $L$ defines a bounded linear functional on $C_c(mathbbC,mathbbR)$. Indeed, since the $leftmu_nright$'s are Borel probability measures on $mathbbC$, we have

$$lvert L(f) rvert leq lvert lvert frvert rvert_infty$$ Hence by the Riesz–Markov–Kakutani representation theorem, there exists a unique Borel measure $mu$,with total variation-norm less than 1, such that $$L(f) = int f , dmu qquad forall fin C_c(mathbbC,mathbbR).$$
In fact, the integral is well-defined for all $C_b(mathbbC, mathbbR)$ as well.

share|cite|improve this answer

answered Aug 11 at 22:40

TheOscillatorTheOscillator

2,23411 gold badge88 silver badges1818 bronze badges

$endgroup$

add a comment | 

$begingroup$

Yes! Consider the restriction to the space of continuous functions with compact support, denoted by $C_c(mathbbC, mathbbR)$. Now define $$L(f):= lim_nrightarrow infty int f , dmu_nqquad , , fin C_c(mathbbC, mathbbR).$$

It is straightforward to check that $L$ defines a bounded linear functional on $C_c(mathbbC,mathbbR)$. Indeed, since the $leftmu_nright$'s are Borel probability measures on $mathbbC$, we have

$$lvert L(f) rvert leq lvert lvert frvert rvert_infty$$ Hence by the Riesz–Markov–Kakutani representation theorem, there exists a unique Borel measure $mu$,with total variation-norm less than 1, such that $$L(f) = int f , dmu qquad forall fin C_c(mathbbC,mathbbR).$$
In fact, the integral is well-defined for all $C_b(mathbbC, mathbbR)$ as well.

share|cite|improve this answer

answered Aug 11 at 22:40

TheOscillatorTheOscillator

2,23411 gold badge88 silver badges1818 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 11 at 22:40

TheOscillatorTheOscillator

2,23411 gold badge88 silver badges1818 bronze badges

answered Aug 11 at 22:40

TheOscillatorTheOscillator

2,23411 gold badge88 silver badges1818 bronze badges

answered Aug 11 at 22:40

TheOscillatorTheOscillator

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3

$begingroup$

Define $l: C_c(mathbbC,mathbbR) to mathbbR$ by $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.

share|cite|improve this answer

edited Aug 12 at 13:20

answered Aug 11 at 22:37

mathworker21mathworker21

11.1k11 gold badge1010 silver badges3131 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Define $l: C_c(mathbbC,mathbbR) to mathbbR$ by $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.

share|cite|improve this answer

edited Aug 12 at 13:20

answered Aug 11 at 22:37

mathworker21mathworker21

11.1k11 gold badge1010 silver badges3131 bronze badges

$endgroup$

add a comment | 

$begingroup$

Define $l: C_c(mathbbC,mathbbR) to mathbbR$ by $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.

share|cite|improve this answer

edited Aug 12 at 13:20

answered Aug 11 at 22:37

mathworker21mathworker21

11.1k11 gold badge1010 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

edited Aug 12 at 13:20

answered Aug 11 at 22:37

mathworker21mathworker21

11.1k11 gold badge1010 silver badges3131 bronze badges

answered Aug 11 at 22:37

mathworker21mathworker21

11.1k11 gold badge1010 silver badges3131 bronze badges

answered Aug 11 at 22:37

mathworker21mathworker21

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