$lim_{k to infty}v(E_k) = v(E)$
$begingroup$
$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.
In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$
We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.
I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.
We are working with jordan measure here (as in, boundary is negligible)
calculus measure-theory volume
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
$endgroup$
add a comment |
$begingroup$
$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.
In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$
We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.
I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.
We are working with jordan measure here (as in, boundary is negligible)
calculus measure-theory volume
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
$endgroup$
1
$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11
$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20
$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21
$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24
1
$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30
add a comment |
$begingroup$
$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.
In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$
We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.
I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.
We are working with jordan measure here (as in, boundary is negligible)
calculus measure-theory volume
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
$endgroup$
$E_1,E_2, dots$ are measurable sets, $E subset E_i$ for all $i$ is also a measurable set and bounded.
In addition we are given that for all $k in mathbb N$ and for all $x in E_k$ there is a $y in E$ such that $|x-y| < frac{1}{k}$
We are asked to prove that $lim_{k to infty}v(E_k) = v(E)$, where $v$ stands for volume.
I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.
We are working with jordan measure here (as in, boundary is negligible)
calculus measure-theory volume
calculus measure-theory volume
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
edited Jan 18 at 22:31
Rick Joker
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
asked Jan 18 at 21:56
Rick JokerRick Joker
285214
285214
1
$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11
$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20
$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21
$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24
1
$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30
add a comment |
1
$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11
$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20
$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21
$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24
1
$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30
1
1
$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11
$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11
$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20
$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20
$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21
$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21
$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24
$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24
1
1
$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30
$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I believe I may have found a solution.
I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.
(*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.
Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.
Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.
This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
$endgroup$
add a comment |
$begingroup$
Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe I may have found a solution.
I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.
(*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.
Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.
Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.
This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
$endgroup$
add a comment |
$begingroup$
I believe I may have found a solution.
I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.
(*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.
Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.
Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.
This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
$endgroup$
add a comment |
$begingroup$
I believe I may have found a solution.
I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.
(*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.
Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.
Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.
This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
$endgroup$
I believe I may have found a solution.
I will attempt to prove that $lim_{k to infty}v(E_ksetminus E) = 0$ which is identical.
(*) Consider $x in E_k setminus E$. The closest point $y in E$ to it can't be an interior point, so infact $y in partial E$ and $|x-y| < frac{1}{k}$.
Let $epsilon > 0$, and $k$ such that $(frac{2}{k})^n < epsilon$.
Since $partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $partial E$ by boxes with side length $frac{2}{k}$.
This is also a cover of $E_ksetminus E$ because of (*), and each box has volume $(frac{2}{k})^n < epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k setminus E$, so it's negligible, which means it has zero volume.
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
answered Jan 18 at 23:24
Rick JokerRick Joker
285214
285214
add a comment |
add a comment |
$begingroup$
Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
add a comment |
$begingroup$
Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
add a comment |
$begingroup$
Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
Let $E^{(k)}$ denote ${x:|x-y| <frac 1 k text {for some}, y in E}$. Since the boundary of $E$ has measure $0$ it follows that $nu (E^{(k)}) to nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $cap_k E^{(k)}$ lies between $E$ and $overline {E}$]. Since $E_k subset E^{(k)}$ we get $nu (E_k) leq nu( E^{(k)})$ so $lim sup nu (E_k) leq nu(E)$. Since $E subset E^{(k)}$ for all $k$ we are done.
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Jan 18 at 23:28
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
76.4k53370
add a comment |
add a comment |
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1
$begingroup$
I guess you are missing some condition. Take for example $$ E_k=(-1-1/k; 1+1/k)$$ then both $[-1;1]$ and $ [-1;1]cap mathbb{Q}$ are valide candidates for $E$, but they do not have the same measure.
$endgroup$
– Severin Schraven
Jan 18 at 22:11
$begingroup$
Maybe you want $E$ to be open? Then you could squeeze $$nu(E)leq nu(E_k) leq nu(E^{(1/k)})$$ where $$E^{(1/k)}={ x : d(x, E)leq 1/k}$$
$endgroup$
– Severin Schraven
Jan 18 at 22:20
$begingroup$
Nope, it's exactly as I wrote it. $E$ is not known to be anything but bounded and jordan measurable.
$endgroup$
– Rick Joker
Jan 18 at 22:21
$begingroup$
Why must $E$ be open to use squeeze? and why is the volume of $E^{(1/k)}$ approach zero?
$endgroup$
– Rick Joker
Jan 18 at 22:24
1
$begingroup$
@RickJoker You should be aware that when people say "measurable" they tend to mean "Lebesgue measurable" or measurable in an abstract measure space. Jordan measure isn't a measure in the modern sense of the word, and so if you ask a question about Jordan measurable sets, you need to specify that (and not just say measurable).
$endgroup$
– Aaron
Jan 18 at 22:30