Proof of Outer Regularity of Lebesgue Measure on $mathbb{R}$












3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 17:14









Viktor Glombik

772527




772527










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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













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1 Answer
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1 Answer
1






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active

oldest

votes









1












$begingroup$

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
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.






share|cite|improve this answer











$endgroup$



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
$bigcup_n (G_n setminus E) subset bigcup_n (G_n setminus E_n)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 16:24









Viktor Glombik

772527




772527










answered Oct 23 '14 at 16:16









copper.hatcopper.hat

126k559160




126k559160












  • $begingroup$
    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












  • $begingroup$
    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




















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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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