Portmanteau Theorem Proof
I am reading van der Vaart Wellner's Weak Convergence and Empirical Processes in which they discuss the Portmanteau Theorem (P17-18):
Let $(Omega_alpha,A_alpha,P_alpha)$ be a net of probability spaces and $X_alpha:alphato D$ arbitrary maps (D is a metric space). The net $X_alpha$ converges weakly to a Borel measure L if
$$E^*f(X_alpha)to int fdL, text{for every bounded continuous function on D},$$
where $E^*$ is the outer integral defined as:
$$E^*f=inf{EU,fleq U, U text{measurable}}$$
They state that the following two conditions are equivalent:
$X_alpha$ converges weakly to L
$liminf E_*f(X_alpha)geq int f dL$ for every bounded, Lipschitz continuous, nonnegative f.
where $E_*$ is the inner integral defined as
$E_*f=-E^*-f$.
They say that the proof from 1 to 2 is trivial but I don't know why this is the case. I can prove the theorem using a truncation argument but the proof requires other condition in the Portmanteau theorem. I am just wondering if there is a trivial way to argue the implication.
probability-theory measure-theory
asked Dec 27 '18 at 3:15
displayname2
505
add a comment |
I am reading van der Vaart Wellner's Weak Convergence and Empirical Processes in which they discuss the Portmanteau Theorem (P17-18):
Let $(Omega_alpha,A_alpha,P_alpha)$ be a net of probability spaces and $X_alpha:alphato D$ arbitrary maps (D is a metric space). The net $X_alpha$ converges weakly to a Borel measure L if
$$E^*f(X_alpha)to int fdL, text{for every bounded continuous function on D},$$
where $E^*$ is the outer integral defined as:
$$E^*f=inf{EU,fleq U, U text{measurable}}$$
They state that the following two conditions are equivalent:
$X_alpha$ converges weakly to L
$liminf E_*f(X_alpha)geq int f dL$ for every bounded, Lipschitz continuous, nonnegative f.
where $E_*$ is the inner integral defined as
$E_*f=-E^*-f$.
They say that the proof from 1 to 2 is trivial but I don't know why this is the case. I can prove the theorem using a truncation argument but the proof requires other condition in the Portmanteau theorem. I am just wondering if there is a trivial way to argue the implication.
probability-theory measure-theory
asked Dec 27 '18 at 3:15
displayname2
505
add a comment |
I am reading van der Vaart Wellner's Weak Convergence and Empirical Processes in which they discuss the Portmanteau Theorem (P17-18):
Let $(Omega_alpha,A_alpha,P_alpha)$ be a net of probability spaces and $X_alpha:alphato D$ arbitrary maps (D is a metric space). The net $X_alpha$ converges weakly to a Borel measure L if
$$E^*f(X_alpha)to int fdL, text{for every bounded continuous function on D},$$
where $E^*$ is the outer integral defined as:
$$E^*f=inf{EU,fleq U, U text{measurable}}$$
They state that the following two conditions are equivalent:
$X_alpha$ converges weakly to L
$liminf E_*f(X_alpha)geq int f dL$ for every bounded, Lipschitz continuous, nonnegative f.
where $E_*$ is the inner integral defined as
$E_*f=-E^*-f$.
They say that the proof from 1 to 2 is trivial but I don't know why this is the case. I can prove the theorem using a truncation argument but the proof requires other condition in the Portmanteau theorem. I am just wondering if there is a trivial way to argue the implication.
probability-theory measure-theory
asked Dec 27 '18 at 3:15
displayname2
505
I am reading van der Vaart Wellner's Weak Convergence and Empirical Processes in which they discuss the Portmanteau Theorem (P17-18):
Let $(Omega_alpha,A_alpha,P_alpha)$ be a net of probability spaces and $X_alpha:alphato D$ arbitrary maps (D is a metric space). The net $X_alpha$ converges weakly to a Borel measure L if
$$E^*f(X_alpha)to int fdL, text{for every bounded continuous function on D},$$
where $E^*$ is the outer integral defined as:
$$E^*f=inf{EU,fleq U, U text{measurable}}$$
They state that the following two conditions are equivalent:
$X_alpha$ converges weakly to L
$liminf E_*f(X_alpha)geq int f dL$ for every bounded, Lipschitz continuous, nonnegative f.
where $E_*$ is the inner integral defined as
$E_*f=-E^*-f$.
They say that the proof from 1 to 2 is trivial but I don't know why this is the case. I can prove the theorem using a truncation argument but the proof requires other condition in the Portmanteau theorem. I am just wondering if there is a trivial way to argue the implication.
probability-theory measure-theory
probability-theory measure-theory
asked Dec 27 '18 at 3:15
displayname2
505
asked Dec 27 '18 at 3:15
displayname2
505
asked Dec 27 '18 at 3:15
displayname2
505
asked Dec 27 '18 at 3:15
displayname2
505
asked Dec 27 '18 at 3:15
displayname2
505
505
add a comment |
add a comment |
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