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Statement of Vitali convergence theorem
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Let $f,f_n in L^p(X,mu)$, where $pin[0,infty)$ and we assume that $mu(X)<infty$. From the wikipedia page, the statement of Vitali convergence theorem is written as
$f_nto f$ in $L^p$ if and only if we have
(i) $f_n$ converges in measure to $f$.
(ii) For every $varepsilon>0$, there exists a $delta>0$ such that if a measurable set $E$ satisfies $mu(E)<delta$, then for every $n$ we have
$$
int_E |f_n|^p dmu < varepsilon^p.
$$
Remark: (i) and (ii) implies uniform integrability of $(|f_n|^p)_n$ whereas the uniform integrability of $(|f_n|^p)_n$ implies (ii).
This baffles me because isn't uniform integrability of $(|f_n|^p)_n$ simply equivalent to (ii)? Why do we need (i) at all?
real-analysis probability functional-analysis measure-theory
asked Jul 20 at 5:14
BigbearZzz
5,70311344
add a comment |Â
up vote
0
down vote
favorite
Let $f,f_n in L^p(X,mu)$, where $pin[0,infty)$ and we assume that $mu(X)<infty$. From the wikipedia page, the statement of Vitali convergence theorem is written as
$f_nto f$ in $L^p$ if and only if we have
(i) $f_n$ converges in measure to $f$.
(ii) For every $varepsilon>0$, there exists a $delta>0$ such that if a measurable set $E$ satisfies $mu(E)<delta$, then for every $n$ we have
$$
int_E |f_n|^p dmu < varepsilon^p.
$$
Remark: (i) and (ii) implies uniform integrability of $(|f_n|^p)_n$ whereas the uniform integrability of $(|f_n|^p)_n$ implies (ii).
This baffles me because isn't uniform integrability of $(|f_n|^p)_n$ simply equivalent to (ii)? Why do we need (i) at all?
real-analysis probability functional-analysis measure-theory
asked Jul 20 at 5:14
BigbearZzz
5,70311344
It seems like you're right. The remark is true though ;)
– mathworker21
Jul 20 at 5:18
That's an interesting way to look at the remark haha. Anyway, perhaps there might be a different definition of uniform integrability that I am not familiar with?
– BigbearZzz
Jul 20 at 5:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f,f_n in L^p(X,mu)$, where $pin[0,infty)$ and we assume that $mu(X)<infty$. From the wikipedia page, the statement of Vitali convergence theorem is written as
$f_nto f$ in $L^p$ if and only if we have
(i) $f_n$ converges in measure to $f$.
(ii) For every $varepsilon>0$, there exists a $delta>0$ such that if a measurable set $E$ satisfies $mu(E)<delta$, then for every $n$ we have
$$
int_E |f_n|^p dmu < varepsilon^p.
$$
Remark: (i) and (ii) implies uniform integrability of $(|f_n|^p)_n$ whereas the uniform integrability of $(|f_n|^p)_n$ implies (ii).
This baffles me because isn't uniform integrability of $(|f_n|^p)_n$ simply equivalent to (ii)? Why do we need (i) at all?
real-analysis probability functional-analysis measure-theory
asked Jul 20 at 5:14
BigbearZzz
5,70311344
Let $f,f_n in L^p(X,mu)$, where $pin[0,infty)$ and we assume that $mu(X)<infty$. From the wikipedia page, the statement of Vitali convergence theorem is written as
$f_nto f$ in $L^p$ if and only if we have
(i) $f_n$ converges in measure to $f$.
(ii) For every $varepsilon>0$, there exists a $delta>0$ such that if a measurable set $E$ satisfies $mu(E)<delta$, then for every $n$ we have
$$
int_E |f_n|^p dmu < varepsilon^p.
$$
Remark: (i) and (ii) implies uniform integrability of $(|f_n|^p)_n$ whereas the uniform integrability of $(|f_n|^p)_n$ implies (ii).
This baffles me because isn't uniform integrability of $(|f_n|^p)_n$ simply equivalent to (ii)? Why do we need (i) at all?
real-analysis probability functional-analysis measure-theory
asked Jul 20 at 5:14
BigbearZzz
5,70311344
asked Jul 20 at 5:14
BigbearZzz
5,70311344
asked Jul 20 at 5:14
BigbearZzz
5,70311344
asked Jul 20 at 5:14
BigbearZzz
5,70311344
5,70311344
It seems like you're right. The remark is true though ;)
– mathworker21
Jul 20 at 5:18
That's an interesting way to look at the remark haha. Anyway, perhaps there might be a different definition of uniform integrability that I am not familiar with?
– BigbearZzz
Jul 20 at 5:42
add a comment |Â
It seems like you're right. The remark is true though ;)
– mathworker21
Jul 20 at 5:18
That's an interesting way to look at the remark haha. Anyway, perhaps there might be a different definition of uniform integrability that I am not familiar with?
– BigbearZzz
Jul 20 at 5:42
It seems like you're right. The remark is true though ;)
– mathworker21
Jul 20 at 5:18
It seems like you're right. The remark is true though ;)
– mathworker21
Jul 20 at 5:18
That's an interesting way to look at the remark haha. Anyway, perhaps there might be a different definition of uniform integrability that I am not familiar with?
– BigbearZzz
Jul 20 at 5:42
That's an interesting way to look at the remark haha. Anyway, perhaps there might be a different definition of uniform integrability that I am not familiar with?
– BigbearZzz
Jul 20 at 5:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Uniform integrabilty is not equivalent to ii). Definition of uniform integrabilty is $int_f_n |f_n| to 0$ uniformly in $n$ as $M to infty$. This is equivalent to ii) plus the condition $sup_n int |f_n| <infty$
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Uniform integrabilty is not equivalent to ii). Definition of uniform integrabilty is $int_f_n |f_n| to 0$ uniformly in $n$ as $M to infty$. This is equivalent to ii) plus the condition $sup_n int |f_n| <infty$
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
add a comment |Â
up vote
1
down vote
accepted
Uniform integrabilty is not equivalent to ii). Definition of uniform integrabilty is $int_f_n |f_n| to 0$ uniformly in $n$ as $M to infty$. This is equivalent to ii) plus the condition $sup_n int |f_n| <infty$
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Uniform integrabilty is not equivalent to ii). Definition of uniform integrabilty is $int_f_n |f_n| to 0$ uniformly in $n$ as $M to infty$. This is equivalent to ii) plus the condition $sup_n int |f_n| <infty$
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
Uniform integrabilty is not equivalent to ii). Definition of uniform integrabilty is $int_f_n |f_n| to 0$ uniformly in $n$ as $M to infty$. This is equivalent to ii) plus the condition $sup_n int |f_n| <infty$
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
answered Jul 20 at 5:45
Kavi Rama Murthy
20.7k2830
20.7k2830
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
add a comment |Â
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
The definition I am familiar with is this "measure theoretic" definition that is found here en.wikipedia.org/wiki/… . Is the one you are using equivalent to the one that falls under the title "Probability definition" on the same page?
– BigbearZzz
Jul 20 at 5:49
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
My definition of UI is the standard definition founf in al standard texts in measure theortic probability (like Chung, Billilgsley, Meyer etc). I think you shouldn't follow the definition in the Wikipedia page. See, for example, Theorem 4.5.3 in Chung.
– Kavi Rama Murthy
Jul 20 at 5:53
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I was using the definition in Rudin's book. It could be that the one I am using is outdated, I think I'll take your words for that. Thank you.
– BigbearZzz
Jul 20 at 5:56
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
I looked at the definition in Rudin; you have quoted it correctly, but here is my take on this: UI is heavily used in Probability and rarely used outside Probability. Since all Probabilists agree on the definition in Chung's book I would say that you should ignore Rudin's definition.
– Kavi Rama Murthy
Jul 20 at 6:02
add a comment |Â
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It seems like you're right. The remark is true though ;)
– mathworker21
Jul 20 at 5:18
That's an interesting way to look at the remark haha. Anyway, perhaps there might be a different definition of uniform integrability that I am not familiar with?
– BigbearZzz
Jul 20 at 5:42