Mixing
Content $0$ implies boundary has content $0$
$begingroup$
Problem: If a set $C$ has content $0$, then the boundary must also have content $0$.
The solution say
Solution: Suppose a finite set of rectangle $ U_i = (a_{i1},b_{i1}) times dots times (a_{in}, b_{in})$ where $i in {1, dots, m }$ cover $C$ and have total volume less than $epsilon/ 2$ where $epsilon >0.$ Let $V_i = (a_{i1} - delta ,b_{i1} + delta) times dots times (a_{in} - delta , b_{in} + delta)$ where
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
Or equivalently,
Then the union of $V_i$ cover the boundary of $C$ and have total volume less than $epsilon$. Hence $partial C$ is also of content $0$.
This is a really simple question, but does $V_i - U_i$ cover $partial C$? $V_i - U_i$ means the part where we remove $U_i$ from $V_i$.
Also is the last step because of $$ sum_{i = 1}^m vol(cup( V_i - U_i)) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) - prod (b_{ij} - a_{aj} ) < sum_{i = 1}^m epsilon/2m = epsilon /2.$$
or equivalently,
$$ sum_{i = 1}^m vol(cup V_i ) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) < sum_{i = 1}^m epsilon/2m + prod (b_{ij} - a_{aj} ) < epsilon /2 + epsilon/2 = epsilon. $$
Am I also correct to understand the outline of this proof is
Use the finite cover that covers $C$, and take a $delta$ cover over that finite cover just enough to cover the boundary of $C$.
Using the finite cover of $C$, demand this cover so that it has volume less than $epsilon/2$. Pick $delta$ small enough so that we get
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
- Use finite subadditivitly to finish off the estimation.
real-analysis measure-theory
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
$endgroup$
|
show 1 more comment
$begingroup$
Problem: If a set $C$ has content $0$, then the boundary must also have content $0$.
The solution say
Solution: Suppose a finite set of rectangle $ U_i = (a_{i1},b_{i1}) times dots times (a_{in}, b_{in})$ where $i in {1, dots, m }$ cover $C$ and have total volume less than $epsilon/ 2$ where $epsilon >0.$ Let $V_i = (a_{i1} - delta ,b_{i1} + delta) times dots times (a_{in} - delta , b_{in} + delta)$ where
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
Or equivalently,
Then the union of $V_i$ cover the boundary of $C$ and have total volume less than $epsilon$. Hence $partial C$ is also of content $0$.
This is a really simple question, but does $V_i - U_i$ cover $partial C$? $V_i - U_i$ means the part where we remove $U_i$ from $V_i$.
Also is the last step because of $$ sum_{i = 1}^m vol(cup( V_i - U_i)) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) - prod (b_{ij} - a_{aj} ) < sum_{i = 1}^m epsilon/2m = epsilon /2.$$
or equivalently,
$$ sum_{i = 1}^m vol(cup V_i ) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) < sum_{i = 1}^m epsilon/2m + prod (b_{ij} - a_{aj} ) < epsilon /2 + epsilon/2 = epsilon. $$
Am I also correct to understand the outline of this proof is
Use the finite cover that covers $C$, and take a $delta$ cover over that finite cover just enough to cover the boundary of $C$.
Using the finite cover of $C$, demand this cover so that it has volume less than $epsilon/2$. Pick $delta$ small enough so that we get
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
- Use finite subadditivitly to finish off the estimation.
real-analysis measure-theory
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
$endgroup$
$begingroup$
Note that you can take the cover to be closed (with the same volume) and the closed cover also covers the closure of $C$, and hence the boundary.
$endgroup$
– copper.hat
May 6 '14 at 6:53
$begingroup$
Wait, why does the closed cover cover the closure of $C$? This seems like a rhetorical question as you are probably going to say "because it covers the boundary…"
$endgroup$
– Hawk
May 6 '14 at 7:10
$begingroup$
What is content of $Q^n$? What is content its boundary?
$endgroup$
– George
May 7 '14 at 13:12
$begingroup$
@GogiPantsulaia, where did you get $Q^n$ from?
$endgroup$
– Hawk
May 8 '14 at 1:43
$begingroup$
My question is the following: what does a "content" mean?
$endgroup$
– George
May 8 '14 at 7:21
|
show 1 more comment
$begingroup$
Problem: If a set $C$ has content $0$, then the boundary must also have content $0$.
The solution say
Solution: Suppose a finite set of rectangle $ U_i = (a_{i1},b_{i1}) times dots times (a_{in}, b_{in})$ where $i in {1, dots, m }$ cover $C$ and have total volume less than $epsilon/ 2$ where $epsilon >0.$ Let $V_i = (a_{i1} - delta ,b_{i1} + delta) times dots times (a_{in} - delta , b_{in} + delta)$ where
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
Or equivalently,
Then the union of $V_i$ cover the boundary of $C$ and have total volume less than $epsilon$. Hence $partial C$ is also of content $0$.
This is a really simple question, but does $V_i - U_i$ cover $partial C$? $V_i - U_i$ means the part where we remove $U_i$ from $V_i$.
Also is the last step because of $$ sum_{i = 1}^m vol(cup( V_i - U_i)) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) - prod (b_{ij} - a_{aj} ) < sum_{i = 1}^m epsilon/2m = epsilon /2.$$
or equivalently,
$$ sum_{i = 1}^m vol(cup V_i ) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) < sum_{i = 1}^m epsilon/2m + prod (b_{ij} - a_{aj} ) < epsilon /2 + epsilon/2 = epsilon. $$
Am I also correct to understand the outline of this proof is
Use the finite cover that covers $C$, and take a $delta$ cover over that finite cover just enough to cover the boundary of $C$.
Using the finite cover of $C$, demand this cover so that it has volume less than $epsilon/2$. Pick $delta$ small enough so that we get
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
- Use finite subadditivitly to finish off the estimation.
real-analysis measure-theory
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
$endgroup$
Problem: If a set $C$ has content $0$, then the boundary must also have content $0$.
The solution say
Solution: Suppose a finite set of rectangle $ U_i = (a_{i1},b_{i1}) times dots times (a_{in}, b_{in})$ where $i in {1, dots, m }$ cover $C$ and have total volume less than $epsilon/ 2$ where $epsilon >0.$ Let $V_i = (a_{i1} - delta ,b_{i1} + delta) times dots times (a_{in} - delta , b_{in} + delta)$ where
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
Or equivalently,
Then the union of $V_i$ cover the boundary of $C$ and have total volume less than $epsilon$. Hence $partial C$ is also of content $0$.
This is a really simple question, but does $V_i - U_i$ cover $partial C$? $V_i - U_i$ means the part where we remove $U_i$ from $V_i$.
Also is the last step because of $$ sum_{i = 1}^m vol(cup( V_i - U_i)) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) - prod (b_{ij} - a_{aj} ) < sum_{i = 1}^m epsilon/2m = epsilon /2.$$
or equivalently,
$$ sum_{i = 1}^m vol(cup V_i ) leq sum_{i = 1}^mprod (b_{ij} - a_{aj} + 2delta) < sum_{i = 1}^m epsilon/2m + prod (b_{ij} - a_{aj} ) < epsilon /2 + epsilon/2 = epsilon. $$
Am I also correct to understand the outline of this proof is
Use the finite cover that covers $C$, and take a $delta$ cover over that finite cover just enough to cover the boundary of $C$.
Using the finite cover of $C$, demand this cover so that it has volume less than $epsilon/2$. Pick $delta$ small enough so that we get
$$prod_{iin { 1,dots,n}} (b_{ij} - a_{ij} + 2delta) - prod_{iin { 1,dots,n}} (b_{ij} - a_{aj} ) < epsilon/2m .$$
- Use finite subadditivitly to finish off the estimation.
real-analysis measure-theory
real-analysis measure-theory
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
edited May 6 '14 at 7:06
Hawk
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
asked May 6 '14 at 6:45
HawkHawk
5,5651140109
5,5651140109
$begingroup$
Note that you can take the cover to be closed (with the same volume) and the closed cover also covers the closure of $C$, and hence the boundary.
$endgroup$
– copper.hat
May 6 '14 at 6:53
$begingroup$
Wait, why does the closed cover cover the closure of $C$? This seems like a rhetorical question as you are probably going to say "because it covers the boundary…"
$endgroup$
– Hawk
May 6 '14 at 7:10
$begingroup$
What is content of $Q^n$? What is content its boundary?
$endgroup$
– George
May 7 '14 at 13:12
$begingroup$
@GogiPantsulaia, where did you get $Q^n$ from?
$endgroup$
– Hawk
May 8 '14 at 1:43
$begingroup$
My question is the following: what does a "content" mean?
$endgroup$
– George
May 8 '14 at 7:21
|
show 1 more comment
$begingroup$
Note that you can take the cover to be closed (with the same volume) and the closed cover also covers the closure of $C$, and hence the boundary.
$endgroup$
– copper.hat
May 6 '14 at 6:53
$begingroup$
Wait, why does the closed cover cover the closure of $C$? This seems like a rhetorical question as you are probably going to say "because it covers the boundary…"
$endgroup$
– Hawk
May 6 '14 at 7:10
$begingroup$
What is content of $Q^n$? What is content its boundary?
$endgroup$
– George
May 7 '14 at 13:12
$begingroup$
@GogiPantsulaia, where did you get $Q^n$ from?
$endgroup$
– Hawk
May 8 '14 at 1:43
$begingroup$
My question is the following: what does a "content" mean?
$endgroup$
– George
May 8 '14 at 7:21
$begingroup$
Note that you can take the cover to be closed (with the same volume) and the closed cover also covers the closure of $C$, and hence the boundary.
$endgroup$
– copper.hat
May 6 '14 at 6:53
$begingroup$
Note that you can take the cover to be closed (with the same volume) and the closed cover also covers the closure of $C$, and hence the boundary.
$endgroup$
– copper.hat
May 6 '14 at 6:53
$begingroup$
Wait, why does the closed cover cover the closure of $C$? This seems like a rhetorical question as you are probably going to say "because it covers the boundary…"
$endgroup$
– Hawk
May 6 '14 at 7:10
$begingroup$
Wait, why does the closed cover cover the closure of $C$? This seems like a rhetorical question as you are probably going to say "because it covers the boundary…"
$endgroup$
– Hawk
May 6 '14 at 7:10
$begingroup$
What is content of $Q^n$? What is content its boundary?
$endgroup$
– George
May 7 '14 at 13:12
$begingroup$
What is content of $Q^n$? What is content its boundary?
$endgroup$
– George
May 7 '14 at 13:12
$begingroup$
@GogiPantsulaia, where did you get $Q^n$ from?
$endgroup$
– Hawk
May 8 '14 at 1:43
$begingroup$
@GogiPantsulaia, where did you get $Q^n$ from?
$endgroup$
– Hawk
May 8 '14 at 1:43
$begingroup$
My question is the following: what does a "content" mean?
$endgroup$
– George
May 8 '14 at 7:21
$begingroup$
My question is the following: what does a "content" mean?
$endgroup$
– George
May 8 '14 at 7:21
|
show 1 more comment
1 Answer
1
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oldest
votes
$begingroup$
$V_i-U_i$ will not cover $C$ in general. Note that $U_i$ is large enough to cover $C$, but it may be wasteful.
So the only trick is really to start with a sufficiently small finite cover for $C$, and blow it up slightly so that $V_isupset overline {U_i}$ and hence $bigcup V_isupset bigcup overline{U_i}=overline{bigcup U_i}supset overline Csupset partial C$, and thus obtain an arbitrarily small cover of $partial C$.
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
add a comment |
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$begingroup$
$V_i-U_i$ will not cover $C$ in general. Note that $U_i$ is large enough to cover $C$, but it may be wasteful.
So the only trick is really to start with a sufficiently small finite cover for $C$, and blow it up slightly so that $V_isupset overline {U_i}$ and hence $bigcup V_isupset bigcup overline{U_i}=overline{bigcup U_i}supset overline Csupset partial C$, and thus obtain an arbitrarily small cover of $partial C$.
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
add a comment |
$begingroup$
$V_i-U_i$ will not cover $C$ in general. Note that $U_i$ is large enough to cover $C$, but it may be wasteful.
So the only trick is really to start with a sufficiently small finite cover for $C$, and blow it up slightly so that $V_isupset overline {U_i}$ and hence $bigcup V_isupset bigcup overline{U_i}=overline{bigcup U_i}supset overline Csupset partial C$, and thus obtain an arbitrarily small cover of $partial C$.
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
add a comment |
$begingroup$
$V_i-U_i$ will not cover $C$ in general. Note that $U_i$ is large enough to cover $C$, but it may be wasteful.
So the only trick is really to start with a sufficiently small finite cover for $C$, and blow it up slightly so that $V_isupset overline {U_i}$ and hence $bigcup V_isupset bigcup overline{U_i}=overline{bigcup U_i}supset overline Csupset partial C$, and thus obtain an arbitrarily small cover of $partial C$.
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$V_i-U_i$ will not cover $C$ in general. Note that $U_i$ is large enough to cover $C$, but it may be wasteful.
So the only trick is really to start with a sufficiently small finite cover for $C$, and blow it up slightly so that $V_isupset overline {U_i}$ and hence $bigcup V_isupset bigcup overline{U_i}=overline{bigcup U_i}supset overline Csupset partial C$, and thus obtain an arbitrarily small cover of $partial C$.
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
answered May 6 '14 at 7:23
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
add a comment |
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
$begingroup$
Actually mentioning $V_i - U_i$ is red herring. It could possibly miss the entire set. Were my reasons for the outlines incorrect?
$endgroup$
– Hawk
May 6 '14 at 7:30
add a comment |
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$begingroup$
Note that you can take the cover to be closed (with the same volume) and the closed cover also covers the closure of $C$, and hence the boundary.
$endgroup$
– copper.hat
May 6 '14 at 6:53
$begingroup$
Wait, why does the closed cover cover the closure of $C$? This seems like a rhetorical question as you are probably going to say "because it covers the boundary…"
$endgroup$
– Hawk
May 6 '14 at 7:10
$begingroup$
What is content of $Q^n$? What is content its boundary?
$endgroup$
– George
May 7 '14 at 13:12
$begingroup$
@GogiPantsulaia, where did you get $Q^n$ from?
$endgroup$
– Hawk
May 8 '14 at 1:43
$begingroup$
My question is the following: what does a "content" mean?
$endgroup$
– George
May 8 '14 at 7:21