Measure Theory - $mu$-integrable [closed]
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Let $(X,mathcal A,mu)$ be a measure space and let $f: X rightarrow Bbb R$ be a $mu$-integrable Mapping. Let $A = X-f^{-1} (0)$. Show that there is a sequence ${A_n}_{nin Bbb N}$ such that $A = bigcup_{nin Bbb N}A_n$
And $mu(A_n) < infty$ holds for all $nin Bbb N$.
measure-theory
edited Dec 3 at 13:31
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asked Dec 3 at 12:22
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closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus Dec 4 at 1:09
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Let $(X,mathcal A,mu)$ be a measure space and let $f: X rightarrow Bbb R$ be a $mu$-integrable Mapping. Let $A = X-f^{-1} (0)$. Show that there is a sequence ${A_n}_{nin Bbb N}$ such that $A = bigcup_{nin Bbb N}A_n$
And $mu(A_n) < infty$ holds for all $nin Bbb N$.
measure-theory
edited Dec 3 at 13:31
UserS
1,506112
asked Dec 3 at 12:22
Luuuuu
1
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus Dec 4 at 1:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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0
down vote
favorite
up vote
0
down vote
favorite
Let $(X,mathcal A,mu)$ be a measure space and let $f: X rightarrow Bbb R$ be a $mu$-integrable Mapping. Let $A = X-f^{-1} (0)$. Show that there is a sequence ${A_n}_{nin Bbb N}$ such that $A = bigcup_{nin Bbb N}A_n$
And $mu(A_n) < infty$ holds for all $nin Bbb N$.
measure-theory
edited Dec 3 at 13:31
UserS
1,506112
asked Dec 3 at 12:22
Luuuuu
1
Let $(X,mathcal A,mu)$ be a measure space and let $f: X rightarrow Bbb R$ be a $mu$-integrable Mapping. Let $A = X-f^{-1} (0)$. Show that there is a sequence ${A_n}_{nin Bbb N}$ such that $A = bigcup_{nin Bbb N}A_n$
And $mu(A_n) < infty$ holds for all $nin Bbb N$.
measure-theory
measure-theory
edited Dec 3 at 13:31
UserS
1,506112
asked Dec 3 at 12:22
Luuuuu
1
edited Dec 3 at 13:31
UserS
1,506112
asked Dec 3 at 12:22
Luuuuu
1
edited Dec 3 at 13:31
UserS
1,506112
edited Dec 3 at 13:31
UserS
1,506112
edited Dec 3 at 13:31
UserS
1,506112
1,506112
asked Dec 3 at 12:22
Luuuuu
1
asked Dec 3 at 12:22
Luuuuu
1
asked Dec 3 at 12:22
Luuuuu
1
1
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus Dec 4 at 1:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus Dec 4 at 1:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, user302797, NCh, Leucippus
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Write $A_n={xin X: |f(x)|≥frac{1}{n}}$,then $A=bigcup_n A_n$.
Now $$0≤frac{1}{n}chi_{A_n}≤|f|chi_{A_n}≤|f|,$$ so that $$int_X frac{1}{n}chi_{A_n}dmu≤int_X|f|chi_{A_n}dmu≤int_X|f|dmu.$$
Since $int_X frac{1}{n}chi_{A_n}dmu=frac{1}{n}mu(A_n)$ we have $mu(A_n)≤nint_X|f|dmu<infty$.
answered Dec 3 at 12:55
UserS
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Ahhh, thank you! That makes sense!
But what is about the following? How can I use the upper statement?
Let s ≥ 0 be a real number and f : Rn → [0,∞) a Borel- measurable mapping. Suppose
0 < ∫fdH^s above R^n <∞
Proof that s is the Hausdorff-Dimension of Rn f−1(0).
answered Dec 3 at 16:12
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2 Answers
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2 Answers
2
active
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active
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active
oldest
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up vote
0
down vote
Write $A_n={xin X: |f(x)|≥frac{1}{n}}$,then $A=bigcup_n A_n$.
Now $$0≤frac{1}{n}chi_{A_n}≤|f|chi_{A_n}≤|f|,$$ so that $$int_X frac{1}{n}chi_{A_n}dmu≤int_X|f|chi_{A_n}dmu≤int_X|f|dmu.$$
Since $int_X frac{1}{n}chi_{A_n}dmu=frac{1}{n}mu(A_n)$ we have $mu(A_n)≤nint_X|f|dmu<infty$.
answered Dec 3 at 12:55
UserS
1,506112
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up vote
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Write $A_n={xin X: |f(x)|≥frac{1}{n}}$,then $A=bigcup_n A_n$.
Now $$0≤frac{1}{n}chi_{A_n}≤|f|chi_{A_n}≤|f|,$$ so that $$int_X frac{1}{n}chi_{A_n}dmu≤int_X|f|chi_{A_n}dmu≤int_X|f|dmu.$$
Since $int_X frac{1}{n}chi_{A_n}dmu=frac{1}{n}mu(A_n)$ we have $mu(A_n)≤nint_X|f|dmu<infty$.
answered Dec 3 at 12:55
UserS
1,506112
add a comment |
up vote
0
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up vote
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Write $A_n={xin X: |f(x)|≥frac{1}{n}}$,then $A=bigcup_n A_n$.
Now $$0≤frac{1}{n}chi_{A_n}≤|f|chi_{A_n}≤|f|,$$ so that $$int_X frac{1}{n}chi_{A_n}dmu≤int_X|f|chi_{A_n}dmu≤int_X|f|dmu.$$
Since $int_X frac{1}{n}chi_{A_n}dmu=frac{1}{n}mu(A_n)$ we have $mu(A_n)≤nint_X|f|dmu<infty$.
answered Dec 3 at 12:55
UserS
1,506112
Write $A_n={xin X: |f(x)|≥frac{1}{n}}$,then $A=bigcup_n A_n$.
Now $$0≤frac{1}{n}chi_{A_n}≤|f|chi_{A_n}≤|f|,$$ so that $$int_X frac{1}{n}chi_{A_n}dmu≤int_X|f|chi_{A_n}dmu≤int_X|f|dmu.$$
Since $int_X frac{1}{n}chi_{A_n}dmu=frac{1}{n}mu(A_n)$ we have $mu(A_n)≤nint_X|f|dmu<infty$.
answered Dec 3 at 12:55
UserS
1,506112
edited Dec 3 at 13:08
answered Dec 3 at 12:55
UserS
1,506112
answered Dec 3 at 12:55
UserS
1,506112
answered Dec 3 at 12:55
UserS
1,506112
1,506112
add a comment |
add a comment |
up vote
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Ahhh, thank you! That makes sense!
But what is about the following? How can I use the upper statement?
Let s ≥ 0 be a real number and f : Rn → [0,∞) a Borel- measurable mapping. Suppose
0 < ∫fdH^s above R^n <∞
Proof that s is the Hausdorff-Dimension of Rn f−1(0).
answered Dec 3 at 16:12
Luuuuu
1
add a comment |
up vote
0
down vote
Ahhh, thank you! That makes sense!
But what is about the following? How can I use the upper statement?
Let s ≥ 0 be a real number and f : Rn → [0,∞) a Borel- measurable mapping. Suppose
0 < ∫fdH^s above R^n <∞
Proof that s is the Hausdorff-Dimension of Rn f−1(0).
answered Dec 3 at 16:12
Luuuuu
1
add a comment |
up vote
0
down vote
up vote
0
down vote
Ahhh, thank you! That makes sense!
But what is about the following? How can I use the upper statement?
Let s ≥ 0 be a real number and f : Rn → [0,∞) a Borel- measurable mapping. Suppose
0 < ∫fdH^s above R^n <∞
Proof that s is the Hausdorff-Dimension of Rn f−1(0).
answered Dec 3 at 16:12
Luuuuu
1
Ahhh, thank you! That makes sense!
But what is about the following? How can I use the upper statement?
Let s ≥ 0 be a real number and f : Rn → [0,∞) a Borel- measurable mapping. Suppose
0 < ∫fdH^s above R^n <∞
Proof that s is the Hausdorff-Dimension of Rn f−1(0).
answered Dec 3 at 16:12
Luuuuu
1
answered Dec 3 at 16:12
Luuuuu
1
answered Dec 3 at 16:12
Luuuuu
1
answered Dec 3 at 16:12
Luuuuu
1
1
add a comment |
add a comment |
