Mixing
Proof of Outer Regularity of Lebesgue Measure on $mathbb{R}$
$begingroup$
Let $E subseteq mathbb{R}$ be a measurable set, and $varepsilon > 0 $. Show that there exists an open set $G supset E$ such that $mu(G setminus E) < epsilon$.
By the definition of Lebesgue measure we can find a countable collection of open intervals $(A_{n})_{n in mathbb{N}}$ such that
$$
sumlimits_{n=1}^{infty} |A_n| leq mu(E) + epsilon.
$$
And I think setting $G = cup_{n in mathbb{N}}A_n$ gives us what we want. But what about if $E$ is of infinite measure?
measure-theory lebesgue-measure
edited Jan 10 at 17:14
Viktor Glombik
8211527
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
$endgroup$
add a comment |
$begingroup$
Let $E subseteq mathbb{R}$ be a measurable set, and $varepsilon > 0 $. Show that there exists an open set $G supset E$ such that $mu(G setminus E) < epsilon$.
By the definition of Lebesgue measure we can find a countable collection of open intervals $(A_{n})_{n in mathbb{N}}$ such that
$$
sumlimits_{n=1}^{infty} |A_n| leq mu(E) + epsilon.
$$
And I think setting $G = cup_{n in mathbb{N}}A_n$ gives us what we want. But what about if $E$ is of infinite measure?
measure-theory lebesgue-measure
edited Jan 10 at 17:14
Viktor Glombik
8211527
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
$endgroup$
$begingroup$
Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $mathbb{R}$. Approximate $E_N=Ecap B_N$ with an open set $G_N$ and then take the union.
$endgroup$
– Giuseppe Negro
Oct 23 '14 at 16:04
add a comment |
$begingroup$
Let $E subseteq mathbb{R}$ be a measurable set, and $varepsilon > 0 $. Show that there exists an open set $G supset E$ such that $mu(G setminus E) < epsilon$.
By the definition of Lebesgue measure we can find a countable collection of open intervals $(A_{n})_{n in mathbb{N}}$ such that
$$
sumlimits_{n=1}^{infty} |A_n| leq mu(E) + epsilon.
$$
And I think setting $G = cup_{n in mathbb{N}}A_n$ gives us what we want. But what about if $E$ is of infinite measure?
measure-theory lebesgue-measure
edited Jan 10 at 17:14
Viktor Glombik
8211527
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
$endgroup$
Let $E subseteq mathbb{R}$ be a measurable set, and $varepsilon > 0 $. Show that there exists an open set $G supset E$ such that $mu(G setminus E) < epsilon$.
By the definition of Lebesgue measure we can find a countable collection of open intervals $(A_{n})_{n in mathbb{N}}$ such that
$$
sumlimits_{n=1}^{infty} |A_n| leq mu(E) + epsilon.
$$
And I think setting $G = cup_{n in mathbb{N}}A_n$ gives us what we want. But what about if $E$ is of infinite measure?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 10 at 17:14
Viktor Glombik
8211527
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
edited Jan 10 at 17:14
Viktor Glombik
8211527
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
edited Jan 10 at 17:14
Viktor Glombik
8211527
edited Jan 10 at 17:14
Viktor Glombik
8211527
edited Jan 10 at 17:14
Viktor Glombik
8211527
8211527
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
asked Oct 23 '14 at 15:56
Tom OfferTom Offer
14712
14712
$begingroup$
Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $mathbb{R}$. Approximate $E_N=Ecap B_N$ with an open set $G_N$ and then take the union.
$endgroup$
– Giuseppe Negro
Oct 23 '14 at 16:04
add a comment |
$begingroup$
Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $mathbb{R}$. Approximate $E_N=Ecap B_N$ with an open set $G_N$ and then take the union.
$endgroup$
– Giuseppe Negro
Oct 23 '14 at 16:04
$begingroup$
Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $mathbb{R}$. Approximate $E_N=Ecap B_N$ with an open set $G_N$ and then take the union.
$endgroup$
– Giuseppe Negro
Oct 23 '14 at 16:04
$begingroup$
Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $mathbb{R}$. Approximate $E_N=Ecap B_N$ with an open set $G_N$ and then take the union.
$endgroup$
– Giuseppe Negro
Oct 23 '14 at 16:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The key element here is that $mathbb{R}$ is $sigma$-finite. The rest is nitty-gritty and a $sum_{n=1}^infty {1 over 2^n} = 1$ trick.
Suppose for each set $E$ of finite measure and each $epsilon >0$ you can find some open set $G$ containing $E$ with $mu(G setminus E) < varepsilon$.
Now let $E$ be and $varepsilon>0$ be arbitrary, and let $E_n = E cap (n, n+1]$. Now let choose open $G_n$ such that
$$
E_n subset G_n
qquad text{and}
qquad mu(G_n setminus E_n) < varepsilon {1 over 3cdot2^{|n|}}.
$$
Now let $G = bigcup_n G_n$, which is open.
By monotonicity, we have
$$G setminus E = bigcup_n G_n setminus E subset bigcup_n G_n setminus E_n
implies
mu(G setminus E) le sum_n mu (G_n setminus E_n ) < varepsilon.
$$
Elaboration:
$G setminus E = (bigcup_n G_n) setminus E = bigcup_n (G_n setminus E)$. Now, since $G_n setminus E subset G_n setminus E_n$, we have
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.
edited Jan 10 at 16:24
Viktor Glombik
8211527
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
1
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
|
show 1 more comment
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1 Answer
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1 Answer
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oldest
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votes
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oldest
votes
$begingroup$
The key element here is that $mathbb{R}$ is $sigma$-finite. The rest is nitty-gritty and a $sum_{n=1}^infty {1 over 2^n} = 1$ trick.
Suppose for each set $E$ of finite measure and each $epsilon >0$ you can find some open set $G$ containing $E$ with $mu(G setminus E) < varepsilon$.
Now let $E$ be and $varepsilon>0$ be arbitrary, and let $E_n = E cap (n, n+1]$. Now let choose open $G_n$ such that
$$
E_n subset G_n
qquad text{and}
qquad mu(G_n setminus E_n) < varepsilon {1 over 3cdot2^{|n|}}.
$$
Now let $G = bigcup_n G_n$, which is open.
By monotonicity, we have
$$G setminus E = bigcup_n G_n setminus E subset bigcup_n G_n setminus E_n
implies
mu(G setminus E) le sum_n mu (G_n setminus E_n ) < varepsilon.
$$
Elaboration:
$G setminus E = (bigcup_n G_n) setminus E = bigcup_n (G_n setminus E)$. Now, since $G_n setminus E subset G_n setminus E_n$, we have
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.
edited Jan 10 at 16:24
Viktor Glombik
8211527
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
1
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
|
show 1 more comment
$begingroup$
The key element here is that $mathbb{R}$ is $sigma$-finite. The rest is nitty-gritty and a $sum_{n=1}^infty {1 over 2^n} = 1$ trick.
Suppose for each set $E$ of finite measure and each $epsilon >0$ you can find some open set $G$ containing $E$ with $mu(G setminus E) < varepsilon$.
Now let $E$ be and $varepsilon>0$ be arbitrary, and let $E_n = E cap (n, n+1]$. Now let choose open $G_n$ such that
$$
E_n subset G_n
qquad text{and}
qquad mu(G_n setminus E_n) < varepsilon {1 over 3cdot2^{|n|}}.
$$
Now let $G = bigcup_n G_n$, which is open.
By monotonicity, we have
$$G setminus E = bigcup_n G_n setminus E subset bigcup_n G_n setminus E_n
implies
mu(G setminus E) le sum_n mu (G_n setminus E_n ) < varepsilon.
$$
Elaboration:
$G setminus E = (bigcup_n G_n) setminus E = bigcup_n (G_n setminus E)$. Now, since $G_n setminus E subset G_n setminus E_n$, we have
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.
edited Jan 10 at 16:24
Viktor Glombik
8211527
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
1
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
|
show 1 more comment
$begingroup$
The key element here is that $mathbb{R}$ is $sigma$-finite. The rest is nitty-gritty and a $sum_{n=1}^infty {1 over 2^n} = 1$ trick.
Suppose for each set $E$ of finite measure and each $epsilon >0$ you can find some open set $G$ containing $E$ with $mu(G setminus E) < varepsilon$.
Now let $E$ be and $varepsilon>0$ be arbitrary, and let $E_n = E cap (n, n+1]$. Now let choose open $G_n$ such that
$$
E_n subset G_n
qquad text{and}
qquad mu(G_n setminus E_n) < varepsilon {1 over 3cdot2^{|n|}}.
$$
Now let $G = bigcup_n G_n$, which is open.
By monotonicity, we have
$$G setminus E = bigcup_n G_n setminus E subset bigcup_n G_n setminus E_n
implies
mu(G setminus E) le sum_n mu (G_n setminus E_n ) < varepsilon.
$$
Elaboration:
$G setminus E = (bigcup_n G_n) setminus E = bigcup_n (G_n setminus E)$. Now, since $G_n setminus E subset G_n setminus E_n$, we have
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.
edited Jan 10 at 16:24
Viktor Glombik
8211527
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
$endgroup$
The key element here is that $mathbb{R}$ is $sigma$-finite. The rest is nitty-gritty and a $sum_{n=1}^infty {1 over 2^n} = 1$ trick.
Suppose for each set $E$ of finite measure and each $epsilon >0$ you can find some open set $G$ containing $E$ with $mu(G setminus E) < varepsilon$.
Now let $E$ be and $varepsilon>0$ be arbitrary, and let $E_n = E cap (n, n+1]$. Now let choose open $G_n$ such that
$$
E_n subset G_n
qquad text{and}
qquad mu(G_n setminus E_n) < varepsilon {1 over 3cdot2^{|n|}}.
$$
Now let $G = bigcup_n G_n$, which is open.
By monotonicity, we have
$$G setminus E = bigcup_n G_n setminus E subset bigcup_n G_n setminus E_n
implies
mu(G setminus E) le sum_n mu (G_n setminus E_n ) < varepsilon.
$$
Elaboration:
$G setminus E = (bigcup_n G_n) setminus E = bigcup_n (G_n setminus E)$. Now, since $G_n setminus E subset G_n setminus E_n$, we have
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.
edited Jan 10 at 16:24
Viktor Glombik
8211527
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
edited Jan 10 at 16:24
Viktor Glombik
8211527
edited Jan 10 at 16:24
Viktor Glombik
8211527
edited Jan 10 at 16:24
Viktor Glombik
8211527
8211527
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
answered Oct 23 '14 at 16:16
copper.hatcopper.hat
127k559160
127k559160
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
1
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
|
show 1 more comment
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
1
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
Thanks! This looks good. However I'm not sure about the final line. We have $ cup_n G_n setminus E subset (cup_n G_n) setminus E_n$, but surely not necessarily $ cup_n G_n setminus E subset cup_n (G_n setminus E_n)$ which is what we need for the conclusion?
$endgroup$
– Tom Offer
Oct 23 '14 at 21:11
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
I added an elaboration.
$endgroup$
– copper.hat
Oct 23 '14 at 21:27
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
$begingroup$
Oh yeah. Can Just take the set difference inside the union first... thanks! I had thought that we'd need to use some sort of covering of $mathbb{R}$ to do this for sets of infinite measure. The $E_n$ you suggest don't even cover $mathbb{R}^+$, but it seems to work. I didn't expect it to!
$endgroup$
– Tom Offer
Oct 23 '14 at 21:32
1
1
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Well $A setminus B = A cap B^c$, so $(cup_n A) setminus B = (cup_n A) cap B^c = cup_n (A cap B^c) = cap_n (A setminus B)$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:34
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
$begingroup$
Why do you care about a cover of $mathbb{R}^+$??? All you need is a collection of disjoint sets $E_n$ such that the measures are bounded and the union is $E$. Since $mathbb{R} = cup_n (n,n+1]$, you know that $E = cup_n E_n$.
$endgroup$
– copper.hat
Oct 23 '14 at 21:36
|
show 1 more comment
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$begingroup$
Try consider the family $B_N=[-N,N]$. This is a countable collection of measurable sets having finite measure and that cover $mathbb{R}$. Approximate $E_N=Ecap B_N$ with an open set $G_N$ and then take the union.
$endgroup$
– Giuseppe Negro
Oct 23 '14 at 16:04