Given rectangle $L subseteq mathbb{R}^n$ st. $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$...
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Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?
Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.
real-analysis measure-theory
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
$endgroup$
add a comment |
$begingroup$
Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?
Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.
real-analysis measure-theory
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
$endgroup$
$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59
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@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00
$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03
add a comment |
$begingroup$
Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?
Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.
real-analysis measure-theory
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
$endgroup$
Consider a rectangle $L = [a_1, b_1 ] times dots times [a_n, b_n]subseteq mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $sum_{i=1}^{infty}v(Q_i) < r$?
Note that $v(L) = (b_1 - a_1)dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.
real-analysis measure-theory
real-analysis measure-theory
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
asked Dec 13 '18 at 5:55
PerturbativePerturbative
4,35811552
4,35811552
$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59
$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00
$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03
add a comment |
$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59
$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00
$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03
$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59
$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59
$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00
$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00
$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03
$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, of course this follows from the theory of Lebesgue measure.
If you don't want to appeal to that, then consider this. One can expand
the $Q_i$ slightly so they become open but still the sum of their measures
is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
number of the $Q_i$ cover $L$. Now it is quite elementary that
$sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
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add a comment |
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I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
$$
v(L) le vleft(bigcup_{i=1}^infty Q_iright)
$$
since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
$$
r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
$$
This is a contradiction.
Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
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add a comment |
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2 Answers
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$begingroup$
No, of course this follows from the theory of Lebesgue measure.
If you don't want to appeal to that, then consider this. One can expand
the $Q_i$ slightly so they become open but still the sum of their measures
is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
number of the $Q_i$ cover $L$. Now it is quite elementary that
$sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
No, of course this follows from the theory of Lebesgue measure.
If you don't want to appeal to that, then consider this. One can expand
the $Q_i$ slightly so they become open but still the sum of their measures
is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
number of the $Q_i$ cover $L$. Now it is quite elementary that
$sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
No, of course this follows from the theory of Lebesgue measure.
If you don't want to appeal to that, then consider this. One can expand
the $Q_i$ slightly so they become open but still the sum of their measures
is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
number of the $Q_i$ cover $L$. Now it is quite elementary that
$sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
No, of course this follows from the theory of Lebesgue measure.
If you don't want to appeal to that, then consider this. One can expand
the $Q_i$ slightly so they become open but still the sum of their measures
is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite
number of the $Q_i$ cover $L$. Now it is quite elementary that
$sum_{i=1}^n v(Q_i)ge v(L)$ for a finite covering of rectangles.
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 13 '18 at 6:00
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
add a comment |
add a comment |
$begingroup$
I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
$$
v(L) le vleft(bigcup_{i=1}^infty Q_iright)
$$
since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
$$
r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
$$
This is a contradiction.
Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
$$
v(L) le vleft(bigcup_{i=1}^infty Q_iright)
$$
since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
$$
r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
$$
This is a contradiction.
Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
$$
v(L) le vleft(bigcup_{i=1}^infty Q_iright)
$$
since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
$$
r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
$$
This is a contradiction.
Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
$endgroup$
I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then
$$
v(L) le vleft(bigcup_{i=1}^infty Q_iright)
$$
since $v$ is monotonic and $Lsubset cup_{i=1}^infty Q_i$. However, $v$ is also subadditive, i.e. $vleft(bigcup_{i=1}^infty A_iright) le sum_{i=1}^infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence
$$
r=v(L) le vleft(bigcup_{i=1}^infty Q_iright) le sum_{i=1}^infty v(Q_i) <r.
$$
This is a contradiction.
Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
edited Dec 13 '18 at 17:09
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
answered Dec 13 '18 at 6:10
BigbearZzzBigbearZzz
8,72621652
8,72621652
add a comment |
add a comment |
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$begingroup$
What do you mean by covering?
$endgroup$
– BigbearZzz
Dec 13 '18 at 5:59
$begingroup$
@BigbearZzz I mean that $L subseteq bigcup_{i=1}^{infty} Q_i$
$endgroup$
– Perturbative
Dec 13 '18 at 6:00
$begingroup$
I presume that by $v(A)$ you mean the Lebesgue measure of $A$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 6:03