Existence of weak limit of measuresWhen weak convergence implies moment convergence?weak*-limit of bounded sequence of measuresMoments and weak convergence of probability measures IIAre vague convergence and weak convergence of measures both weak* convergence?Techniques for determining whether the weak limit of an absolutely continuous sequence of probability measures is absolutely continuous?Does weak convergence with uniformly bounded densities imply absolute continuity of the limit?Does convergence of $lim_n rightarrow infty int h dmu_n$ for all continuous, bounded $h$ imply weak convergence of $(mu_n)$?Weak convergence of linear combinations of Dirac measures to a signed measureWeak convergence of measures and boundednessweak-converging sequence of measures with ascending supports
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Existence of weak limit of measures
When weak convergence implies moment convergence?weak*-limit of bounded sequence of measuresMoments and weak convergence of probability measures IIAre vague convergence and weak convergence of measures both weak* convergence?Techniques for determining whether the weak limit of an absolutely continuous sequence of probability measures is absolutely continuous?Does weak convergence with uniformly bounded densities imply absolute continuity of the limit?Does convergence of $lim_n rightarrow infty int h dmu_n$ for all continuous, bounded $h$ imply weak convergence of $(mu_n)$?Weak convergence of linear combinations of Dirac measures to a signed measureWeak convergence of measures and boundednessweak-converging sequence of measures with ascending supports
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
$endgroup$
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
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 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.
$endgroup$
– Jason Gaitonde
7 hours ago
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
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
$endgroup$
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
analysis probability-theory measure-theory
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
edited 8 hours ago
WeakestTopology
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
asked 8 hours ago
WeakestTopologyWeakestTopology
5533 silver badges13 bronze badges
5533 silver badges13 bronze badges
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.
$endgroup$
– Jason Gaitonde
7 hours ago
add a comment |
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.
$endgroup$
– Jason Gaitonde
7 hours ago
3
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.
$endgroup$
– Jason Gaitonde
7 hours ago
$begingroup$
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.
$endgroup$
– Jason Gaitonde
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Define $l: C_c(mathbbC,mathbbR) to mathbbR$ via $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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votes
$begingroup$
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.
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 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.
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 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.
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 bronze badges
$endgroup$
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.
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 bronze badges
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 bronze badges
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 bronze badges
answered 8 hours ago
TheOscillatorTheOscillator
2,2241 gold badge8 silver badges18 bronze badges
2,2241 gold badge8 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
Define $l: C_c(mathbbC,mathbbR) to mathbbR$ via $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
Define $l: C_c(mathbbC,mathbbR) to mathbbR$ via $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
Define $l: C_c(mathbbC,mathbbR) to mathbbR$ via $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
$endgroup$
Define $l: C_c(mathbbC,mathbbR) to mathbbR$ via $l(f) = lim_n to infty int fdmu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $mu$ with $l(f) = int fdmu$.
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
edited 7 hours ago
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
answered 8 hours ago
mathworker21mathworker21
10.9k1 gold badge9 silver badges31 bronze badges
10.9k1 gold badge9 silver badges31 bronze badges
add a comment |
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$begingroup$
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.
$endgroup$
– Jason Gaitonde
7 hours ago