Fatou's lemma and convergence in measure [duplicate]
This question already has an answer here:
Exercise on convergence in measure (Folland, Real Analysis)
2 answers
Convergence in probability implies Fatou's lemma?
1 answer
How to prove that Fatou's lemma holds if almost everywhere convergence is replaced by convergence in measure??? This is an exercise problem in Battle's Elements of integration.Convergence in measure doesn't imply almost everywhere convergence. But if a sequence of function converges in measure then there exists a subsequence which converges almost everywhere. I tried to prove using this fact ,but can't. Can anyone help me...
measure-theory
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
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marked as duplicate by saz, ncmathsadist, José Carlos Santos, Song, drhab 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Exercise on convergence in measure (Folland, Real Analysis)
2 answers
Convergence in probability implies Fatou's lemma?
1 answer
How to prove that Fatou's lemma holds if almost everywhere convergence is replaced by convergence in measure??? This is an exercise problem in Battle's Elements of integration.Convergence in measure doesn't imply almost everywhere convergence. But if a sequence of function converges in measure then there exists a subsequence which converges almost everywhere. I tried to prove using this fact ,but can't. Can anyone help me...
measure-theory
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
marked as duplicate by saz, ncmathsadist, José Carlos Santos, Song, drhab 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
You want to show that if $f_n to f$ in measure, with $f_n,f$ non-negative, then $int f le liminf_n int f_n$. For the proof, first take a subsequence $(f_{n_k})_k$ with $int f_{n_k} to liminf_n int f_n$. Then since $f_{n_k} to f$ in measure, we can find a (further) subsequence $f_{n_{k_j}}$ converging to $f$ pointwise. Of course we have $int f_{n_{k_j}} to liminf_n int f_n$, and therefore, by applying Fatou to $f_{n_{k_j}}$, we get $int f le lim_j int f_{n_{k_j}} = liminf_n int f_n$.
– mathworker21
2 days ago
add a comment |
This question already has an answer here:
Exercise on convergence in measure (Folland, Real Analysis)
2 answers
Convergence in probability implies Fatou's lemma?
1 answer
How to prove that Fatou's lemma holds if almost everywhere convergence is replaced by convergence in measure??? This is an exercise problem in Battle's Elements of integration.Convergence in measure doesn't imply almost everywhere convergence. But if a sequence of function converges in measure then there exists a subsequence which converges almost everywhere. I tried to prove using this fact ,but can't. Can anyone help me...
measure-theory
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This question already has an answer here:
Exercise on convergence in measure (Folland, Real Analysis)
2 answers
Convergence in probability implies Fatou's lemma?
1 answer
How to prove that Fatou's lemma holds if almost everywhere convergence is replaced by convergence in measure??? This is an exercise problem in Battle's Elements of integration.Convergence in measure doesn't imply almost everywhere convergence. But if a sequence of function converges in measure then there exists a subsequence which converges almost everywhere. I tried to prove using this fact ,but can't. Can anyone help me...
This question already has an answer here:
Exercise on convergence in measure (Folland, Real Analysis)
2 answers
Convergence in probability implies Fatou's lemma?
1 answer
measure-theory
measure-theory
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
asked 2 days ago
Balasubramannyan SBalasubramannyan S
11
11
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Balasubramannyan S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
marked as duplicate by saz, ncmathsadist, José Carlos Santos, Song, drhab 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by saz, ncmathsadist, José Carlos Santos, Song, drhab 2 days ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
You want to show that if $f_n to f$ in measure, with $f_n,f$ non-negative, then $int f le liminf_n int f_n$. For the proof, first take a subsequence $(f_{n_k})_k$ with $int f_{n_k} to liminf_n int f_n$. Then since $f_{n_k} to f$ in measure, we can find a (further) subsequence $f_{n_{k_j}}$ converging to $f$ pointwise. Of course we have $int f_{n_{k_j}} to liminf_n int f_n$, and therefore, by applying Fatou to $f_{n_{k_j}}$, we get $int f le lim_j int f_{n_{k_j}} = liminf_n int f_n$.
– mathworker21
2 days ago
add a comment |
You want to show that if $f_n to f$ in measure, with $f_n,f$ non-negative, then $int f le liminf_n int f_n$. For the proof, first take a subsequence $(f_{n_k})_k$ with $int f_{n_k} to liminf_n int f_n$. Then since $f_{n_k} to f$ in measure, we can find a (further) subsequence $f_{n_{k_j}}$ converging to $f$ pointwise. Of course we have $int f_{n_{k_j}} to liminf_n int f_n$, and therefore, by applying Fatou to $f_{n_{k_j}}$, we get $int f le lim_j int f_{n_{k_j}} = liminf_n int f_n$.
– mathworker21
2 days ago
You want to show that if $f_n to f$ in measure, with $f_n,f$ non-negative, then $int f le liminf_n int f_n$. For the proof, first take a subsequence $(f_{n_k})_k$ with $int f_{n_k} to liminf_n int f_n$. Then since $f_{n_k} to f$ in measure, we can find a (further) subsequence $f_{n_{k_j}}$ converging to $f$ pointwise. Of course we have $int f_{n_{k_j}} to liminf_n int f_n$, and therefore, by applying Fatou to $f_{n_{k_j}}$, we get $int f le lim_j int f_{n_{k_j}} = liminf_n int f_n$.
– mathworker21
2 days ago
You want to show that if $f_n to f$ in measure, with $f_n,f$ non-negative, then $int f le liminf_n int f_n$. For the proof, first take a subsequence $(f_{n_k})_k$ with $int f_{n_k} to liminf_n int f_n$. Then since $f_{n_k} to f$ in measure, we can find a (further) subsequence $f_{n_{k_j}}$ converging to $f$ pointwise. Of course we have $int f_{n_{k_j}} to liminf_n int f_n$, and therefore, by applying Fatou to $f_{n_{k_j}}$, we get $int f le lim_j int f_{n_{k_j}} = liminf_n int f_n$.
– mathworker21
2 days ago
add a comment |
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You want to show that if $f_n to f$ in measure, with $f_n,f$ non-negative, then $int f le liminf_n int f_n$. For the proof, first take a subsequence $(f_{n_k})_k$ with $int f_{n_k} to liminf_n int f_n$. Then since $f_{n_k} to f$ in measure, we can find a (further) subsequence $f_{n_{k_j}}$ converging to $f$ pointwise. Of course we have $int f_{n_{k_j}} to liminf_n int f_n$, and therefore, by applying Fatou to $f_{n_{k_j}}$, we get $int f le lim_j int f_{n_{k_j}} = liminf_n int f_n$.
– mathworker21
2 days ago