Measure theory: integrability and limit
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I'm an undergraduate student in math and I've been stuck on the following question for a while. Could anyone help me show that? Thanks.
Given
- a measure space $(X,mathscr{A} ,mu) $
- a Borel-measurable function $f:Xrightarrow R$
$phi:[0,+infty]
rightarrow[0,+infty] $a monotone non-decreasing function,
Show that if $intphi(|f|)dmu < + infty$, then
$ lim_{a to infty } phi(a) mu({x in X: |f(x)|>a} = 0$
measure-theory
asked 4 hours ago
jfiore
11
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I'm an undergraduate student in math and I've been stuck on the following question for a while. Could anyone help me show that? Thanks.
Given
- a measure space $(X,mathscr{A} ,mu) $
- a Borel-measurable function $f:Xrightarrow R$
$phi:[0,+infty]
rightarrow[0,+infty] $a monotone non-decreasing function,
Show that if $intphi(|f|)dmu < + infty$, then
$ lim_{a to infty } phi(a) mu({x in X: |f(x)|>a} = 0$
measure-theory
asked 4 hours ago
jfiore
11
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
add a comment |
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down vote
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up vote
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down vote
favorite
I'm an undergraduate student in math and I've been stuck on the following question for a while. Could anyone help me show that? Thanks.
Given
- a measure space $(X,mathscr{A} ,mu) $
- a Borel-measurable function $f:Xrightarrow R$
$phi:[0,+infty]
rightarrow[0,+infty] $a monotone non-decreasing function,
Show that if $intphi(|f|)dmu < + infty$, then
$ lim_{a to infty } phi(a) mu({x in X: |f(x)|>a} = 0$
measure-theory
asked 4 hours ago
jfiore
11
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm an undergraduate student in math and I've been stuck on the following question for a while. Could anyone help me show that? Thanks.
Given
- a measure space $(X,mathscr{A} ,mu) $
- a Borel-measurable function $f:Xrightarrow R$
$phi:[0,+infty]
rightarrow[0,+infty] $a monotone non-decreasing function,
Show that if $intphi(|f|)dmu < + infty$, then
$ lim_{a to infty } phi(a) mu({x in X: |f(x)|>a} = 0$
measure-theory
measure-theory
asked 4 hours ago
jfiore
11
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
jfiore
11
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
jfiore
11
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
jfiore
11
asked 4 hours ago
jfiore
11
11
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
jfiore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
add a comment |
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago
add a comment |
1 Answer
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$phi(a)mu(|{x:|f(x)| >a} leq int_{{x:|f(x)| >a}} phi(|f|)dmu to 0$ as $a to infty$. The inequality holds because $|f(x)| >a$ implies $phi(|f(x)|)>phi(a)$. The last step follows by DCT and the fact that the set ${x:|f(x)| >a}$ decreases to empty set as $ a$ increases to $infty$.
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$phi(a)mu(|{x:|f(x)| >a} leq int_{{x:|f(x)| >a}} phi(|f|)dmu to 0$ as $a to infty$. The inequality holds because $|f(x)| >a$ implies $phi(|f(x)|)>phi(a)$. The last step follows by DCT and the fact that the set ${x:|f(x)| >a}$ decreases to empty set as $ a$ increases to $infty$.
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
add a comment |
up vote
0
down vote
$phi(a)mu(|{x:|f(x)| >a} leq int_{{x:|f(x)| >a}} phi(|f|)dmu to 0$ as $a to infty$. The inequality holds because $|f(x)| >a$ implies $phi(|f(x)|)>phi(a)$. The last step follows by DCT and the fact that the set ${x:|f(x)| >a}$ decreases to empty set as $ a$ increases to $infty$.
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
add a comment |
up vote
0
down vote
up vote
0
down vote
$phi(a)mu(|{x:|f(x)| >a} leq int_{{x:|f(x)| >a}} phi(|f|)dmu to 0$ as $a to infty$. The inequality holds because $|f(x)| >a$ implies $phi(|f(x)|)>phi(a)$. The last step follows by DCT and the fact that the set ${x:|f(x)| >a}$ decreases to empty set as $ a$ increases to $infty$.
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
$phi(a)mu(|{x:|f(x)| >a} leq int_{{x:|f(x)| >a}} phi(|f|)dmu to 0$ as $a to infty$. The inequality holds because $|f(x)| >a$ implies $phi(|f(x)|)>phi(a)$. The last step follows by DCT and the fact that the set ${x:|f(x)| >a}$ decreases to empty set as $ a$ increases to $infty$.
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
answered 4 hours ago
Kavi Rama Murthy
38.5k31747
38.5k31747
add a comment |
add a comment |
jfiore is a new contributor. Be nice, and check out our Code of Conduct.
jfiore is a new contributor. Be nice, and check out our Code of Conduct.
jfiore is a new contributor. Be nice, and check out our Code of Conduct.
jfiore is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
4 hours ago