$f:[a,b] rightarrow Bbb R$ measurable function, $fgeq0$ a.e. If $f^{-1}((0,infty))$ has measure $>0$ ,...
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This is exercise #8, p.103 if Gail S. Nelson's A User-Friendly Guide to Lebesgue Measure and Integration. Here is my attempt: For each $n$ let $E_n:=f^{-1}((1/n,infty))$. Note that $f$ is measurable, so $E_n$ is measurable for all $n$. Then $f^{-1}((0,infty))=bigcup_{n=1}^{infty}E_n$ (the reverse inclusion is clear, the forward inclusion follows from the Archimedean Property.) Then $$m(f^{-1}((0,infty))) = m(bigcup_{n=1}^{infty}E_n) leq sum_{n=1}^{infty} m(E_n)$$
by subadditivity. So if every $E_n$ had zero measure, so would $f^{-1}((0,infty))$, which proves the contrapositive. I'm fairly sure this is wrong, since I haven't used that $f$ is nonnegative almost everywhere; I'm just not sure where I went wrong. Any help would be greatly appreciated!
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 16 at 23:21
Alex Sanger
675
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This is exercise #8, p.103 if Gail S. Nelson's A User-Friendly Guide to Lebesgue Measure and Integration. Here is my attempt: For each $n$ let $E_n:=f^{-1}((1/n,infty))$. Note that $f$ is measurable, so $E_n$ is measurable for all $n$. Then $f^{-1}((0,infty))=bigcup_{n=1}^{infty}E_n$ (the reverse inclusion is clear, the forward inclusion follows from the Archimedean Property.) Then $$m(f^{-1}((0,infty))) = m(bigcup_{n=1}^{infty}E_n) leq sum_{n=1}^{infty} m(E_n)$$
by subadditivity. So if every $E_n$ had zero measure, so would $f^{-1}((0,infty))$, which proves the contrapositive. I'm fairly sure this is wrong, since I haven't used that $f$ is nonnegative almost everywhere; I'm just not sure where I went wrong. Any help would be greatly appreciated!
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 16 at 23:21
Alex Sanger
675
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is exercise #8, p.103 if Gail S. Nelson's A User-Friendly Guide to Lebesgue Measure and Integration. Here is my attempt: For each $n$ let $E_n:=f^{-1}((1/n,infty))$. Note that $f$ is measurable, so $E_n$ is measurable for all $n$. Then $f^{-1}((0,infty))=bigcup_{n=1}^{infty}E_n$ (the reverse inclusion is clear, the forward inclusion follows from the Archimedean Property.) Then $$m(f^{-1}((0,infty))) = m(bigcup_{n=1}^{infty}E_n) leq sum_{n=1}^{infty} m(E_n)$$
by subadditivity. So if every $E_n$ had zero measure, so would $f^{-1}((0,infty))$, which proves the contrapositive. I'm fairly sure this is wrong, since I haven't used that $f$ is nonnegative almost everywhere; I'm just not sure where I went wrong. Any help would be greatly appreciated!
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 16 at 23:21
Alex Sanger
675
This is exercise #8, p.103 if Gail S. Nelson's A User-Friendly Guide to Lebesgue Measure and Integration. Here is my attempt: For each $n$ let $E_n:=f^{-1}((1/n,infty))$. Note that $f$ is measurable, so $E_n$ is measurable for all $n$. Then $f^{-1}((0,infty))=bigcup_{n=1}^{infty}E_n$ (the reverse inclusion is clear, the forward inclusion follows from the Archimedean Property.) Then $$m(f^{-1}((0,infty))) = m(bigcup_{n=1}^{infty}E_n) leq sum_{n=1}^{infty} m(E_n)$$
by subadditivity. So if every $E_n$ had zero measure, so would $f^{-1}((0,infty))$, which proves the contrapositive. I'm fairly sure this is wrong, since I haven't used that $f$ is nonnegative almost everywhere; I'm just not sure where I went wrong. Any help would be greatly appreciated!
real-analysis measure-theory proof-verification lebesgue-measure
real-analysis measure-theory proof-verification lebesgue-measure
asked Nov 16 at 23:21
Alex Sanger
675
asked Nov 16 at 23:21
Alex Sanger
675
asked Nov 16 at 23:21
Alex Sanger
675
asked Nov 16 at 23:21
Alex Sanger
675
asked Nov 16 at 23:21
Alex Sanger
675
675
add a comment |
add a comment |
2 Answers
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In general, subadditivity shows that if $E_n$ has measure zero each, then so does $cup E_n.$ Hence the exercise. Q.E.D.
answered Nov 16 at 23:23
Will M.
1,969213
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
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Prove by contradiction. If $f^{-1}((frac 1 n , infty)$ has measure $0$ for each $n$ then the union of these sets has measure $0$. The union is exactly $f^{-1}((0 , infty)$
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In general, subadditivity shows that if $E_n$ has measure zero each, then so does $cup E_n.$ Hence the exercise. Q.E.D.
answered Nov 16 at 23:23
Will M.
1,969213
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
add a comment |
up vote
1
down vote
accepted
In general, subadditivity shows that if $E_n$ has measure zero each, then so does $cup E_n.$ Hence the exercise. Q.E.D.
answered Nov 16 at 23:23
Will M.
1,969213
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In general, subadditivity shows that if $E_n$ has measure zero each, then so does $cup E_n.$ Hence the exercise. Q.E.D.
answered Nov 16 at 23:23
Will M.
1,969213
In general, subadditivity shows that if $E_n$ has measure zero each, then so does $cup E_n.$ Hence the exercise. Q.E.D.
answered Nov 16 at 23:23
Will M.
1,969213
answered Nov 16 at 23:23
Will M.
1,969213
answered Nov 16 at 23:23
Will M.
1,969213
answered Nov 16 at 23:23
Will M.
1,969213
1,969213
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
add a comment |
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem?
– Alex Sanger
Nov 16 at 23:46
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Either $f^{-1}((0, infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e.
– Will M.
Nov 18 at 6:19
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
Ah, I see! Thanks for your replies!
– Alex Sanger
Nov 18 at 22:43
add a comment |
up vote
0
down vote
Prove by contradiction. If $f^{-1}((frac 1 n , infty)$ has measure $0$ for each $n$ then the union of these sets has measure $0$. The union is exactly $f^{-1}((0 , infty)$
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
add a comment |
up vote
0
down vote
Prove by contradiction. If $f^{-1}((frac 1 n , infty)$ has measure $0$ for each $n$ then the union of these sets has measure $0$. The union is exactly $f^{-1}((0 , infty)$
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
add a comment |
up vote
0
down vote
up vote
0
down vote
Prove by contradiction. If $f^{-1}((frac 1 n , infty)$ has measure $0$ for each $n$ then the union of these sets has measure $0$. The union is exactly $f^{-1}((0 , infty)$
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
Prove by contradiction. If $f^{-1}((frac 1 n , infty)$ has measure $0$ for each $n$ then the union of these sets has measure $0$. The union is exactly $f^{-1}((0 , infty)$
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
answered Nov 16 at 23:23
Kavi Rama Murthy
41.4k31751
41.4k31751
add a comment |
add a comment |
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