Measure of a compact subset of $mathbb{R}^d$
Let $K subset mathbb{R}^d$ be closed and bounded.
Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$
wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.
ATTEMPT:
So I wanted to use double sided inequality to show the equality holds.
NOTE: $E_1 supset E_2 supset E_2 ..$
I have a lemma stating: if additionally $m(E_1) < + infty$ then
lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **
(is ** lemma applicable her?)
using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!
Also, I have an observation, I would like to know is true:
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
real-analysis measure-theory lebesgue-measure
asked Jan 3 at 22:09
eyeheartmath
657
add a comment |
Let $K subset mathbb{R}^d$ be closed and bounded.
Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$
wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.
ATTEMPT:
So I wanted to use double sided inequality to show the equality holds.
NOTE: $E_1 supset E_2 supset E_2 ..$
I have a lemma stating: if additionally $m(E_1) < + infty$ then
lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **
(is ** lemma applicable her?)
using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!
Also, I have an observation, I would like to know is true:
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
real-analysis measure-theory lebesgue-measure
asked Jan 3 at 22:09
eyeheartmath
657
2
If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19
oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26
add a comment |
Let $K subset mathbb{R}^d$ be closed and bounded.
Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$
wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.
ATTEMPT:
So I wanted to use double sided inequality to show the equality holds.
NOTE: $E_1 supset E_2 supset E_2 ..$
I have a lemma stating: if additionally $m(E_1) < + infty$ then
lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **
(is ** lemma applicable her?)
using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!
Also, I have an observation, I would like to know is true:
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
real-analysis measure-theory lebesgue-measure
asked Jan 3 at 22:09
eyeheartmath
657
Let $K subset mathbb{R}^d$ be closed and bounded.
Define $E_n = {x in mathbb{R}^d vert exists y in K$, $ vert x - y vert < frac{1}{n} }$
wish to show lim$_{n rightarrow infty}$ $m(E_n)$ = $m(K)$.
ATTEMPT:
So I wanted to use double sided inequality to show the equality holds.
NOTE: $E_1 supset E_2 supset E_2 ..$
I have a lemma stating: if additionally $m(E_1) < + infty$ then
lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n = 1}^{infty} E_n$) **
(is ** lemma applicable her?)
using this, If I can make a claim stating $K = bigcap_{n=1}^{infty} E_n $ ( I doubt this is true) then I am done. But I feel like I am missing a subtlety.
Sorry for being a noob, any help is greatly appreciated, I am new to measure theory and trying to learn!
Also, I have an observation, I would like to know is true:
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 3 at 22:09
eyeheartmath
657
asked Jan 3 at 22:09
eyeheartmath
657
asked Jan 3 at 22:09
eyeheartmath
657
asked Jan 3 at 22:09
eyeheartmath
657
asked Jan 3 at 22:09
eyeheartmath
657
657
2
If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19
oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26
add a comment |
2
If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19
oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26
2
2
If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19
If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19
oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26
oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26
add a comment |
2 Answers
2
active
oldest
votes
For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.
To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
add a comment |
I doubt this is true
It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.
answered Jan 3 at 22:20
user3482749
2,738414
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
add a comment |
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2 Answers
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active
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votes
2 Answers
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For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.
To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
add a comment |
For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.
To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
add a comment |
For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.
To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
For each $xinmathbb{R}^{d}$ and $r>0$, let $B(x,r)={ymid d(x,y)<r}$
be the open ball centered at $x$ with radius $r$. Note that
$$
E_{n}=bigcup_{xin K}B(x,frac{1}{n})
$$
which is a bounded (because $K$ is bounded) open set. In particular
$m(E_{n})<infty$.
To show that $K=cap_{n}E_{n}$. Clearly $E_{1}supseteq E_{2}supseteqcdotssupseteq K$.
We only need to show that $cap_{n}E_{n}subseteq K$. Let $xincap_{n}E_{n}$.
For each $n$, $xin E_{n}Rightarrowexists y_{n}in K$ such that
$xin B(y_{n},frac{1}{n})$. Since $K$ is compact, for the sequence
$(y_{n})$, there exists a convergent subsequence $(y_{n_{k}})$ and
$yin K$ such that $y_{n_{k}}rightarrow y$ as $krightarrowinfty$.
Now
begin{eqnarray*}
d(x,y) & leq & d(x,y_{n_{k}})+d(y_{n_{k}},y)\
& < & frac{1}{n_{k}}+d(y_{n_{k}},y)\
& rightarrow & 0,
end{eqnarray*}
as $krightarrowinfty$. Therefore, $d(x,y)=0$ and hence $x=yin K$.
This shows that $cap_{n}E_{n}subseteq K$.
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
edited Jan 3 at 22:30
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
answered Jan 3 at 22:28
Danny Pak-Keung Chan
2,19038
2,19038
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
add a comment |
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
got it thanks so much!!!
– eyeheartmath
Jan 3 at 22:29
add a comment |
I doubt this is true
It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.
answered Jan 3 at 22:20
user3482749
2,738414
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
add a comment |
I doubt this is true
It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.
answered Jan 3 at 22:20
user3482749
2,738414
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
add a comment |
I doubt this is true
It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.
answered Jan 3 at 22:20
user3482749
2,738414
I doubt this is true
It is, and your method is fine, once you've also checked that $m(E_1) < +infty$.
Do all compact subsets of $mathbb{R}^d$ have finite (Lebesgue) measure.
Yes. This follows from the Heine-Borel Theorem: all compact subsets of $mathbb{R}^d$ are bounded, so are contained in some product of closed, bounded intervals, hence have measure bounded above by the measure of that product of closed, bounded intervals, which is finite. There's extra work to do to check that all compact subsets of $mathbb{R}^d$ are measurable, but indeed they are.
answered Jan 3 at 22:20
user3482749
2,738414
answered Jan 3 at 22:20
user3482749
2,738414
answered Jan 3 at 22:20
user3482749
2,738414
answered Jan 3 at 22:20
user3482749
2,738414
2,738414
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
add a comment |
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
Awesome!! thank you so much!! I am always very very hesitant on making claims unless I'm 100000% certain. I thought the equality was true but wanted to be extra certain, and yes Heine-Borel, luckily enough my point-set topology is relatively strong specially in comparison to my measure theory. again, thanks!!!!
– eyeheartmath
Jan 3 at 22:23
add a comment |
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2
If you have $x in bigcap_{n=1}^{infty}E_{n}$, then $d(x, K) = 0$, right? So since $K$ is closed, $x in K$. So $bigcap_{n=1}^{infty}E_{n} subseteq K$. Also, since $K subseteq E_{n}$ for all $n in mathbb{N}$, you also have $K subseteq bigcap_{n=1}^{infty}E_{n}$, so $K = bigcap_{n=1}^{infty}E_{n}$. Try to work out the details for yourself.
– Matija Sreckovic
Jan 3 at 22:19
oh yes true, this I can workout, its a simple analysis/point set topology problem. thank you so much!!
– eyeheartmath
Jan 3 at 22:26