Mixing
The measure of a sequence of open sets
$begingroup$
Another fun qualifying exam practice question!
Suppose $E$ is a given set and $O_n$ is the open set defined as $O_n = left{x: operatorname{dist}(x,E) < frac1nright}$
a) Show that if $E$ is compact then $m(E) = lim_{n to infty} m(O_n)$
What I have so far:
Define $B_k = bigcap^k_{i=0}O_n = O_k$
Then $B_{k+1} subseteq B_k$
Also $lim_{n rightarrow infty}B_k = bigcap^infty_{i=0}O_n = E$
(Since $E$ is closed and bounded? I'm not really sure why this is true!)
So $B_k$ converges downward to $E$. Since measures are monotone, $m(B_k)$ converges downward to $m(E)$
So $lim_{k rightarrow infty}m(B_k)
=lim_{k rightarrow infty}m(bigcap^k_{n=1}O_n)
=lim_{k rightarrow infty}m(O_k)=m(E)$
Basically I'm wondering how compactness plays into this argument.
b) Show that the conclusion may be false for $E$ closed and unbounded or open and bounded.
Since I'm not sure how compactness plays into the above argument I'm having a hard time coming up with a counter example.
general-topology measure-theory
edited Jan 5 at 14:55
Henrik
5,99792030
asked Jan 5 at 14:45
Math LadyMath Lady
1196
$endgroup$
add a comment |
$begingroup$
Another fun qualifying exam practice question!
Suppose $E$ is a given set and $O_n$ is the open set defined as $O_n = left{x: operatorname{dist}(x,E) < frac1nright}$
a) Show that if $E$ is compact then $m(E) = lim_{n to infty} m(O_n)$
What I have so far:
Define $B_k = bigcap^k_{i=0}O_n = O_k$
Then $B_{k+1} subseteq B_k$
Also $lim_{n rightarrow infty}B_k = bigcap^infty_{i=0}O_n = E$
(Since $E$ is closed and bounded? I'm not really sure why this is true!)
So $B_k$ converges downward to $E$. Since measures are monotone, $m(B_k)$ converges downward to $m(E)$
So $lim_{k rightarrow infty}m(B_k)
=lim_{k rightarrow infty}m(bigcap^k_{n=1}O_n)
=lim_{k rightarrow infty}m(O_k)=m(E)$
Basically I'm wondering how compactness plays into this argument.
b) Show that the conclusion may be false for $E$ closed and unbounded or open and bounded.
Since I'm not sure how compactness plays into the above argument I'm having a hard time coming up with a counter example.
general-topology measure-theory
edited Jan 5 at 14:55
Henrik
5,99792030
asked Jan 5 at 14:45
Math LadyMath Lady
1196
$endgroup$
$begingroup$
Your argument works if $m(O_1)$ is finite and $bigcap O_i = E$, so...
$endgroup$
– Rigel
Jan 5 at 14:53
add a comment |
$begingroup$
Another fun qualifying exam practice question!
Suppose $E$ is a given set and $O_n$ is the open set defined as $O_n = left{x: operatorname{dist}(x,E) < frac1nright}$
a) Show that if $E$ is compact then $m(E) = lim_{n to infty} m(O_n)$
What I have so far:
Define $B_k = bigcap^k_{i=0}O_n = O_k$
Then $B_{k+1} subseteq B_k$
Also $lim_{n rightarrow infty}B_k = bigcap^infty_{i=0}O_n = E$
(Since $E$ is closed and bounded? I'm not really sure why this is true!)
So $B_k$ converges downward to $E$. Since measures are monotone, $m(B_k)$ converges downward to $m(E)$
So $lim_{k rightarrow infty}m(B_k)
=lim_{k rightarrow infty}m(bigcap^k_{n=1}O_n)
=lim_{k rightarrow infty}m(O_k)=m(E)$
Basically I'm wondering how compactness plays into this argument.
b) Show that the conclusion may be false for $E$ closed and unbounded or open and bounded.
Since I'm not sure how compactness plays into the above argument I'm having a hard time coming up with a counter example.
general-topology measure-theory
edited Jan 5 at 14:55
Henrik
5,99792030
asked Jan 5 at 14:45
Math LadyMath Lady
1196
$endgroup$
Another fun qualifying exam practice question!
Suppose $E$ is a given set and $O_n$ is the open set defined as $O_n = left{x: operatorname{dist}(x,E) < frac1nright}$
a) Show that if $E$ is compact then $m(E) = lim_{n to infty} m(O_n)$
What I have so far:
Define $B_k = bigcap^k_{i=0}O_n = O_k$
Then $B_{k+1} subseteq B_k$
Also $lim_{n rightarrow infty}B_k = bigcap^infty_{i=0}O_n = E$
(Since $E$ is closed and bounded? I'm not really sure why this is true!)
So $B_k$ converges downward to $E$. Since measures are monotone, $m(B_k)$ converges downward to $m(E)$
So $lim_{k rightarrow infty}m(B_k)
=lim_{k rightarrow infty}m(bigcap^k_{n=1}O_n)
=lim_{k rightarrow infty}m(O_k)=m(E)$
Basically I'm wondering how compactness plays into this argument.
b) Show that the conclusion may be false for $E$ closed and unbounded or open and bounded.
Since I'm not sure how compactness plays into the above argument I'm having a hard time coming up with a counter example.
general-topology measure-theory
general-topology measure-theory
edited Jan 5 at 14:55
Henrik
5,99792030
asked Jan 5 at 14:45
Math LadyMath Lady
1196
edited Jan 5 at 14:55
Henrik
5,99792030
asked Jan 5 at 14:45
Math LadyMath Lady
1196
edited Jan 5 at 14:55
Henrik
5,99792030
edited Jan 5 at 14:55
Henrik
5,99792030
edited Jan 5 at 14:55
Henrik
5,99792030
5,99792030
asked Jan 5 at 14:45
Math LadyMath Lady
1196
asked Jan 5 at 14:45
Math LadyMath Lady
1196
asked Jan 5 at 14:45
Math LadyMath Lady
1196
1196
$begingroup$
Your argument works if $m(O_1)$ is finite and $bigcap O_i = E$, so...
$endgroup$
– Rigel
Jan 5 at 14:53
add a comment |
$begingroup$
Your argument works if $m(O_1)$ is finite and $bigcap O_i = E$, so...
$endgroup$
– Rigel
Jan 5 at 14:53
$begingroup$
Your argument works if $m(O_1)$ is finite and $bigcap O_i = E$, so...
$endgroup$
– Rigel
Jan 5 at 14:53
$begingroup$
Your argument works if $m(O_1)$ is finite and $bigcap O_i = E$, so...
$endgroup$
– Rigel
Jan 5 at 14:53
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If $xinbigcap_{n=1}^{infty}O_n$ then $d(x,E)<frac1n$ for every positive integer $n$ and this can only be true iff $d(x,E)=0$ or equivalently $xinoverline E$.
$E$ is compact hence closed so that $E=overline E$, so we have $E=bigcap_{n=1}^{infty}O_n$.
Actually here we only need that $E$ is closed, but the condition of boundedness comes in to ensure that $m(O_n)<infty$ for some $n$.
Under that extra condition your solution works.
If we would e.g. take $E=mathbb Rtimes{0}subseteqmathbb R^2$ then $m(O_n)=infty$ for every $n$.
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
$endgroup$
add a comment |
$begingroup$
You need closedness for $bigcap_n O_n = E$ to hold: if $x$ lies in the intersection it means that $d(x,E)=0$ which is equivalent to $x in overline{E}$, so $x in E$ if $E$ is closed. So $bigcap_n O_n = overline{E}$ holds in general. We do not need boundedness of $E$ at all, that I can see.
The continuity argument for $m$ works when we have finite measure for $O_1$ (And hence all $O_n$) and it could fail for infinite measures.
The $B_k$ are unnecessary as the $O_n$ are already decreasing themselves: $x in O_{n+1}$ or $d(x, E) < frac{1}{n+1}$ trivially implies $d(x,E) < frac{1}{n}$ or $x in O_n$, as $frac{1}{n+1} < frac{1}{n}$, so $O_{n+1} subseteq O_n$. It's the standard way to write a closed set as the intersection of countably many open sets in a metric space.
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
So far none of the answers address part (b) of your question fully, so I will give you two examples below:
1) To see why being bounded is crucial to the argument, consider $E=mathbb{Z}subsetmathbb{R}$.
2) To see why being closed is crucial consider an enumeration of the rationals inside the unit interval, say $(q_k)$, and place an open interval on top of each $q_k$ of length $epsiloncdot2^{-k}$ in order to construct your set $E$.
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
$endgroup$
add a comment |
$begingroup$
I'm assuming you mean that $Esubseteqmathbb{R}$.
(a) Note that $bigcap_{n=1}^infty O_n={x:d(x,E)=0}$, which is just the closure of $E$, which since $E$ is compact is just $E$ itself. By continuity of Lebesgue measure together with $O_nsupseteq O_{n+1}$ we now have
$$m(E)=mleft(bigcap_{n=1}^infty O_nright)=lim_{ntoinfty}m(O_n).$$
(Note that the second equality holds if and only if the measures are finite.)
(b) For closed and unbounded $E$, take $E=mathbb{N}$. For open and bounded $E$, take the complement of the fat Cantor set in $[0,1]$.
Good luck with the Quals! That brings back memories.
answered Jan 5 at 15:07
Ben WBen W
2,234615
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $xinbigcap_{n=1}^{infty}O_n$ then $d(x,E)<frac1n$ for every positive integer $n$ and this can only be true iff $d(x,E)=0$ or equivalently $xinoverline E$.
$E$ is compact hence closed so that $E=overline E$, so we have $E=bigcap_{n=1}^{infty}O_n$.
Actually here we only need that $E$ is closed, but the condition of boundedness comes in to ensure that $m(O_n)<infty$ for some $n$.
Under that extra condition your solution works.
If we would e.g. take $E=mathbb Rtimes{0}subseteqmathbb R^2$ then $m(O_n)=infty$ for every $n$.
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
$endgroup$
add a comment |
$begingroup$
If $xinbigcap_{n=1}^{infty}O_n$ then $d(x,E)<frac1n$ for every positive integer $n$ and this can only be true iff $d(x,E)=0$ or equivalently $xinoverline E$.
$E$ is compact hence closed so that $E=overline E$, so we have $E=bigcap_{n=1}^{infty}O_n$.
Actually here we only need that $E$ is closed, but the condition of boundedness comes in to ensure that $m(O_n)<infty$ for some $n$.
Under that extra condition your solution works.
If we would e.g. take $E=mathbb Rtimes{0}subseteqmathbb R^2$ then $m(O_n)=infty$ for every $n$.
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
$endgroup$
add a comment |
$begingroup$
If $xinbigcap_{n=1}^{infty}O_n$ then $d(x,E)<frac1n$ for every positive integer $n$ and this can only be true iff $d(x,E)=0$ or equivalently $xinoverline E$.
$E$ is compact hence closed so that $E=overline E$, so we have $E=bigcap_{n=1}^{infty}O_n$.
Actually here we only need that $E$ is closed, but the condition of boundedness comes in to ensure that $m(O_n)<infty$ for some $n$.
Under that extra condition your solution works.
If we would e.g. take $E=mathbb Rtimes{0}subseteqmathbb R^2$ then $m(O_n)=infty$ for every $n$.
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
$endgroup$
If $xinbigcap_{n=1}^{infty}O_n$ then $d(x,E)<frac1n$ for every positive integer $n$ and this can only be true iff $d(x,E)=0$ or equivalently $xinoverline E$.
$E$ is compact hence closed so that $E=overline E$, so we have $E=bigcap_{n=1}^{infty}O_n$.
Actually here we only need that $E$ is closed, but the condition of boundedness comes in to ensure that $m(O_n)<infty$ for some $n$.
Under that extra condition your solution works.
If we would e.g. take $E=mathbb Rtimes{0}subseteqmathbb R^2$ then $m(O_n)=infty$ for every $n$.
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
answered Jan 5 at 14:57
drhabdrhab
99.5k544130
99.5k544130
add a comment |
add a comment |
$begingroup$
You need closedness for $bigcap_n O_n = E$ to hold: if $x$ lies in the intersection it means that $d(x,E)=0$ which is equivalent to $x in overline{E}$, so $x in E$ if $E$ is closed. So $bigcap_n O_n = overline{E}$ holds in general. We do not need boundedness of $E$ at all, that I can see.
The continuity argument for $m$ works when we have finite measure for $O_1$ (And hence all $O_n$) and it could fail for infinite measures.
The $B_k$ are unnecessary as the $O_n$ are already decreasing themselves: $x in O_{n+1}$ or $d(x, E) < frac{1}{n+1}$ trivially implies $d(x,E) < frac{1}{n}$ or $x in O_n$, as $frac{1}{n+1} < frac{1}{n}$, so $O_{n+1} subseteq O_n$. It's the standard way to write a closed set as the intersection of countably many open sets in a metric space.
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
You need closedness for $bigcap_n O_n = E$ to hold: if $x$ lies in the intersection it means that $d(x,E)=0$ which is equivalent to $x in overline{E}$, so $x in E$ if $E$ is closed. So $bigcap_n O_n = overline{E}$ holds in general. We do not need boundedness of $E$ at all, that I can see.
The continuity argument for $m$ works when we have finite measure for $O_1$ (And hence all $O_n$) and it could fail for infinite measures.
The $B_k$ are unnecessary as the $O_n$ are already decreasing themselves: $x in O_{n+1}$ or $d(x, E) < frac{1}{n+1}$ trivially implies $d(x,E) < frac{1}{n}$ or $x in O_n$, as $frac{1}{n+1} < frac{1}{n}$, so $O_{n+1} subseteq O_n$. It's the standard way to write a closed set as the intersection of countably many open sets in a metric space.
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
You need closedness for $bigcap_n O_n = E$ to hold: if $x$ lies in the intersection it means that $d(x,E)=0$ which is equivalent to $x in overline{E}$, so $x in E$ if $E$ is closed. So $bigcap_n O_n = overline{E}$ holds in general. We do not need boundedness of $E$ at all, that I can see.
The continuity argument for $m$ works when we have finite measure for $O_1$ (And hence all $O_n$) and it could fail for infinite measures.
The $B_k$ are unnecessary as the $O_n$ are already decreasing themselves: $x in O_{n+1}$ or $d(x, E) < frac{1}{n+1}$ trivially implies $d(x,E) < frac{1}{n}$ or $x in O_n$, as $frac{1}{n+1} < frac{1}{n}$, so $O_{n+1} subseteq O_n$. It's the standard way to write a closed set as the intersection of countably many open sets in a metric space.
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
You need closedness for $bigcap_n O_n = E$ to hold: if $x$ lies in the intersection it means that $d(x,E)=0$ which is equivalent to $x in overline{E}$, so $x in E$ if $E$ is closed. So $bigcap_n O_n = overline{E}$ holds in general. We do not need boundedness of $E$ at all, that I can see.
The continuity argument for $m$ works when we have finite measure for $O_1$ (And hence all $O_n$) and it could fail for infinite measures.
The $B_k$ are unnecessary as the $O_n$ are already decreasing themselves: $x in O_{n+1}$ or $d(x, E) < frac{1}{n+1}$ trivially implies $d(x,E) < frac{1}{n}$ or $x in O_n$, as $frac{1}{n+1} < frac{1}{n}$, so $O_{n+1} subseteq O_n$. It's the standard way to write a closed set as the intersection of countably many open sets in a metric space.
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 5 at 14:58
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
$begingroup$
So far none of the answers address part (b) of your question fully, so I will give you two examples below:
1) To see why being bounded is crucial to the argument, consider $E=mathbb{Z}subsetmathbb{R}$.
2) To see why being closed is crucial consider an enumeration of the rationals inside the unit interval, say $(q_k)$, and place an open interval on top of each $q_k$ of length $epsiloncdot2^{-k}$ in order to construct your set $E$.
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
$endgroup$
add a comment |
$begingroup$
So far none of the answers address part (b) of your question fully, so I will give you two examples below:
1) To see why being bounded is crucial to the argument, consider $E=mathbb{Z}subsetmathbb{R}$.
2) To see why being closed is crucial consider an enumeration of the rationals inside the unit interval, say $(q_k)$, and place an open interval on top of each $q_k$ of length $epsiloncdot2^{-k}$ in order to construct your set $E$.
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
$endgroup$
add a comment |
$begingroup$
So far none of the answers address part (b) of your question fully, so I will give you two examples below:
1) To see why being bounded is crucial to the argument, consider $E=mathbb{Z}subsetmathbb{R}$.
2) To see why being closed is crucial consider an enumeration of the rationals inside the unit interval, say $(q_k)$, and place an open interval on top of each $q_k$ of length $epsiloncdot2^{-k}$ in order to construct your set $E$.
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
$endgroup$
So far none of the answers address part (b) of your question fully, so I will give you two examples below:
1) To see why being bounded is crucial to the argument, consider $E=mathbb{Z}subsetmathbb{R}$.
2) To see why being closed is crucial consider an enumeration of the rationals inside the unit interval, say $(q_k)$, and place an open interval on top of each $q_k$ of length $epsiloncdot2^{-k}$ in order to construct your set $E$.
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
answered Jan 5 at 15:05
ItsJustAMeasureBroItsJustAMeasureBro
342
342
add a comment |
add a comment |
$begingroup$
I'm assuming you mean that $Esubseteqmathbb{R}$.
(a) Note that $bigcap_{n=1}^infty O_n={x:d(x,E)=0}$, which is just the closure of $E$, which since $E$ is compact is just $E$ itself. By continuity of Lebesgue measure together with $O_nsupseteq O_{n+1}$ we now have
$$m(E)=mleft(bigcap_{n=1}^infty O_nright)=lim_{ntoinfty}m(O_n).$$
(Note that the second equality holds if and only if the measures are finite.)
(b) For closed and unbounded $E$, take $E=mathbb{N}$. For open and bounded $E$, take the complement of the fat Cantor set in $[0,1]$.
Good luck with the Quals! That brings back memories.
answered Jan 5 at 15:07
Ben WBen W
2,234615
$endgroup$
add a comment |
$begingroup$
I'm assuming you mean that $Esubseteqmathbb{R}$.
(a) Note that $bigcap_{n=1}^infty O_n={x:d(x,E)=0}$, which is just the closure of $E$, which since $E$ is compact is just $E$ itself. By continuity of Lebesgue measure together with $O_nsupseteq O_{n+1}$ we now have
$$m(E)=mleft(bigcap_{n=1}^infty O_nright)=lim_{ntoinfty}m(O_n).$$
(Note that the second equality holds if and only if the measures are finite.)
(b) For closed and unbounded $E$, take $E=mathbb{N}$. For open and bounded $E$, take the complement of the fat Cantor set in $[0,1]$.
Good luck with the Quals! That brings back memories.
answered Jan 5 at 15:07
Ben WBen W
2,234615
$endgroup$
add a comment |
$begingroup$
I'm assuming you mean that $Esubseteqmathbb{R}$.
(a) Note that $bigcap_{n=1}^infty O_n={x:d(x,E)=0}$, which is just the closure of $E$, which since $E$ is compact is just $E$ itself. By continuity of Lebesgue measure together with $O_nsupseteq O_{n+1}$ we now have
$$m(E)=mleft(bigcap_{n=1}^infty O_nright)=lim_{ntoinfty}m(O_n).$$
(Note that the second equality holds if and only if the measures are finite.)
(b) For closed and unbounded $E$, take $E=mathbb{N}$. For open and bounded $E$, take the complement of the fat Cantor set in $[0,1]$.
Good luck with the Quals! That brings back memories.
answered Jan 5 at 15:07
Ben WBen W
2,234615
$endgroup$
I'm assuming you mean that $Esubseteqmathbb{R}$.
(a) Note that $bigcap_{n=1}^infty O_n={x:d(x,E)=0}$, which is just the closure of $E$, which since $E$ is compact is just $E$ itself. By continuity of Lebesgue measure together with $O_nsupseteq O_{n+1}$ we now have
$$m(E)=mleft(bigcap_{n=1}^infty O_nright)=lim_{ntoinfty}m(O_n).$$
(Note that the second equality holds if and only if the measures are finite.)
(b) For closed and unbounded $E$, take $E=mathbb{N}$. For open and bounded $E$, take the complement of the fat Cantor set in $[0,1]$.
Good luck with the Quals! That brings back memories.
answered Jan 5 at 15:07
Ben WBen W
2,234615
edited Jan 5 at 15:12
answered Jan 5 at 15:07
Ben WBen W
2,234615
answered Jan 5 at 15:07
Ben WBen W
2,234615
answered Jan 5 at 15:07
Ben WBen W
2,234615
2,234615
add a comment |
add a comment |
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$begingroup$
Your argument works if $m(O_1)$ is finite and $bigcap O_i = E$, so...
$endgroup$
– Rigel
Jan 5 at 14:53