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
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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
New contributor
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
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up vote
0
down vote
favorite
up vote
0
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
New contributor
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
New contributor
New contributor
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asked 4 hours ago
jfiore
11
11
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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
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 $infty$.
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1 Answer
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active
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1 Answer
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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 $infty$.
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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 $infty$.
add a comment |
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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 $infty$.
$phi(a)mu(|{x:|f(x)| >a} leq int_{{x:|f(x)| >a}} phi(|f|)dmu to 0$ 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
38.5k31747
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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.
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4 hours ago