Show that the open interval (a, b) is Lebesgue measurable
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I have to show that an open interval in the form $(a,b)$, where $a,b in {mathbb R}$ and $a < b$ is Lebesgue measurable.
I think I'm supposed to show, that the subset $(a,b)$ is Lebesgue measurable, if and only if:
$$m(A) = m(A ∩ S) + m(A ∩ S^c)$$
where $S subseteq {mathbb R}^n$ and $S^c$ is the complement of $S$.
But how do I actually prove that the open interval $(a,b)$ is Lebesgue measurable?
analysis measure-theory lebesgue-measure
edited Dec 9 '18 at 20:11
postmortes
1,89521117
asked Dec 9 '18 at 19:43
RHARHA
11
$endgroup$
add a comment |
$begingroup$
I have to show that an open interval in the form $(a,b)$, where $a,b in {mathbb R}$ and $a < b$ is Lebesgue measurable.
I think I'm supposed to show, that the subset $(a,b)$ is Lebesgue measurable, if and only if:
$$m(A) = m(A ∩ S) + m(A ∩ S^c)$$
where $S subseteq {mathbb R}^n$ and $S^c$ is the complement of $S$.
But how do I actually prove that the open interval $(a,b)$ is Lebesgue measurable?
analysis measure-theory lebesgue-measure
edited Dec 9 '18 at 20:11
postmortes
1,89521117
asked Dec 9 '18 at 19:43
RHARHA
11
$endgroup$
$begingroup$
What is your definition of being measurable? Are you using outer measure?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:31
$begingroup$
@mathcounterexamples.net Yes, I think OP is using the Carathéodory's criterion as definition.
$endgroup$
– Alex Vong
Dec 9 '18 at 20:34
$begingroup$
But at lest OP should precise what is $A$ and $S$ as he is looking to prove the measurability of $(a,b)$.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:36
$begingroup$
The purpose of the exercise is to proove that any interval in ℝ is Lebesgue measurable by showing that any open interval in the form (a,b) is Lebesgue measurable. S in this exercise is the subset (a,b) and A is any S⊆ℝn. We use the outer measure as a definition of being measurable.
$endgroup$
– RHA
Dec 9 '18 at 21:11
add a comment |
$begingroup$
I have to show that an open interval in the form $(a,b)$, where $a,b in {mathbb R}$ and $a < b$ is Lebesgue measurable.
I think I'm supposed to show, that the subset $(a,b)$ is Lebesgue measurable, if and only if:
$$m(A) = m(A ∩ S) + m(A ∩ S^c)$$
where $S subseteq {mathbb R}^n$ and $S^c$ is the complement of $S$.
But how do I actually prove that the open interval $(a,b)$ is Lebesgue measurable?
analysis measure-theory lebesgue-measure
edited Dec 9 '18 at 20:11
postmortes
1,89521117
asked Dec 9 '18 at 19:43
RHARHA
11
$endgroup$
I have to show that an open interval in the form $(a,b)$, where $a,b in {mathbb R}$ and $a < b$ is Lebesgue measurable.
I think I'm supposed to show, that the subset $(a,b)$ is Lebesgue measurable, if and only if:
$$m(A) = m(A ∩ S) + m(A ∩ S^c)$$
where $S subseteq {mathbb R}^n$ and $S^c$ is the complement of $S$.
But how do I actually prove that the open interval $(a,b)$ is Lebesgue measurable?
analysis measure-theory lebesgue-measure
analysis measure-theory lebesgue-measure
edited Dec 9 '18 at 20:11
postmortes
1,89521117
asked Dec 9 '18 at 19:43
RHARHA
11
edited Dec 9 '18 at 20:11
postmortes
1,89521117
asked Dec 9 '18 at 19:43
RHARHA
11
edited Dec 9 '18 at 20:11
postmortes
1,89521117
edited Dec 9 '18 at 20:11
postmortes
1,89521117
edited Dec 9 '18 at 20:11
postmortes
1,89521117
1,89521117
asked Dec 9 '18 at 19:43
RHARHA
11
asked Dec 9 '18 at 19:43
RHARHA
11
asked Dec 9 '18 at 19:43
RHARHA
11
11
$begingroup$
What is your definition of being measurable? Are you using outer measure?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:31
$begingroup$
@mathcounterexamples.net Yes, I think OP is using the Carathéodory's criterion as definition.
$endgroup$
– Alex Vong
Dec 9 '18 at 20:34
$begingroup$
But at lest OP should precise what is $A$ and $S$ as he is looking to prove the measurability of $(a,b)$.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:36
$begingroup$
The purpose of the exercise is to proove that any interval in ℝ is Lebesgue measurable by showing that any open interval in the form (a,b) is Lebesgue measurable. S in this exercise is the subset (a,b) and A is any S⊆ℝn. We use the outer measure as a definition of being measurable.
$endgroup$
– RHA
Dec 9 '18 at 21:11
add a comment |
$begingroup$
What is your definition of being measurable? Are you using outer measure?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:31
$begingroup$
@mathcounterexamples.net Yes, I think OP is using the Carathéodory's criterion as definition.
$endgroup$
– Alex Vong
Dec 9 '18 at 20:34
$begingroup$
But at lest OP should precise what is $A$ and $S$ as he is looking to prove the measurability of $(a,b)$.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:36
$begingroup$
The purpose of the exercise is to proove that any interval in ℝ is Lebesgue measurable by showing that any open interval in the form (a,b) is Lebesgue measurable. S in this exercise is the subset (a,b) and A is any S⊆ℝn. We use the outer measure as a definition of being measurable.
$endgroup$
– RHA
Dec 9 '18 at 21:11
$begingroup$
What is your definition of being measurable? Are you using outer measure?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:31
$begingroup$
What is your definition of being measurable? Are you using outer measure?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:31
$begingroup$
@mathcounterexamples.net Yes, I think OP is using the Carathéodory's criterion as definition.
$endgroup$
– Alex Vong
Dec 9 '18 at 20:34
$begingroup$
@mathcounterexamples.net Yes, I think OP is using the Carathéodory's criterion as definition.
$endgroup$
– Alex Vong
Dec 9 '18 at 20:34
$begingroup$
But at lest OP should precise what is $A$ and $S$ as he is looking to prove the measurability of $(a,b)$.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:36
$begingroup$
But at lest OP should precise what is $A$ and $S$ as he is looking to prove the measurability of $(a,b)$.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:36
$begingroup$
The purpose of the exercise is to proove that any interval in ℝ is Lebesgue measurable by showing that any open interval in the form (a,b) is Lebesgue measurable. S in this exercise is the subset (a,b) and A is any S⊆ℝn. We use the outer measure as a definition of being measurable.
$endgroup$
– RHA
Dec 9 '18 at 21:11
$begingroup$
The purpose of the exercise is to proove that any interval in ℝ is Lebesgue measurable by showing that any open interval in the form (a,b) is Lebesgue measurable. S in this exercise is the subset (a,b) and A is any S⊆ℝn. We use the outer measure as a definition of being measurable.
$endgroup$
– RHA
Dec 9 '18 at 21:11
add a comment |
1 Answer
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I assume you are considering sets in $mathbb R. $ Let $Ain mathcal P (mathbb R), I=(a,b)$ and $ epsilon>0. $ For convenience, denote both the outer measure and length of intervals by $|cdot|.$
There is a sequence $(I_n)$ of intervals such that $bigcup I_nsupseteq A$ and $sum |I_n|<|A|+epsilon. $ Set $J_n=Icap I_n; J_n'=I^ccap I_n. $ Some of these may be empty, but that's ok.
Then,
$|I_n|=|J_n|+|J_n'|, A cap I subseteq bigcup_{n=1}^{infty} J_n, A cap I^c subseteq bigcup_{n=1}^{infty} J'_n, $ and these facts imply that $|Acap I|+|Acap I^c|le |A|+epsilon.$
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
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1 Answer
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1 Answer
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$begingroup$
I assume you are considering sets in $mathbb R. $ Let $Ain mathcal P (mathbb R), I=(a,b)$ and $ epsilon>0. $ For convenience, denote both the outer measure and length of intervals by $|cdot|.$
There is a sequence $(I_n)$ of intervals such that $bigcup I_nsupseteq A$ and $sum |I_n|<|A|+epsilon. $ Set $J_n=Icap I_n; J_n'=I^ccap I_n. $ Some of these may be empty, but that's ok.
Then,
$|I_n|=|J_n|+|J_n'|, A cap I subseteq bigcup_{n=1}^{infty} J_n, A cap I^c subseteq bigcup_{n=1}^{infty} J'_n, $ and these facts imply that $|Acap I|+|Acap I^c|le |A|+epsilon.$
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
$endgroup$
add a comment |
$begingroup$
I assume you are considering sets in $mathbb R. $ Let $Ain mathcal P (mathbb R), I=(a,b)$ and $ epsilon>0. $ For convenience, denote both the outer measure and length of intervals by $|cdot|.$
There is a sequence $(I_n)$ of intervals such that $bigcup I_nsupseteq A$ and $sum |I_n|<|A|+epsilon. $ Set $J_n=Icap I_n; J_n'=I^ccap I_n. $ Some of these may be empty, but that's ok.
Then,
$|I_n|=|J_n|+|J_n'|, A cap I subseteq bigcup_{n=1}^{infty} J_n, A cap I^c subseteq bigcup_{n=1}^{infty} J'_n, $ and these facts imply that $|Acap I|+|Acap I^c|le |A|+epsilon.$
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
$endgroup$
add a comment |
$begingroup$
I assume you are considering sets in $mathbb R. $ Let $Ain mathcal P (mathbb R), I=(a,b)$ and $ epsilon>0. $ For convenience, denote both the outer measure and length of intervals by $|cdot|.$
There is a sequence $(I_n)$ of intervals such that $bigcup I_nsupseteq A$ and $sum |I_n|<|A|+epsilon. $ Set $J_n=Icap I_n; J_n'=I^ccap I_n. $ Some of these may be empty, but that's ok.
Then,
$|I_n|=|J_n|+|J_n'|, A cap I subseteq bigcup_{n=1}^{infty} J_n, A cap I^c subseteq bigcup_{n=1}^{infty} J'_n, $ and these facts imply that $|Acap I|+|Acap I^c|le |A|+epsilon.$
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
$endgroup$
I assume you are considering sets in $mathbb R. $ Let $Ain mathcal P (mathbb R), I=(a,b)$ and $ epsilon>0. $ For convenience, denote both the outer measure and length of intervals by $|cdot|.$
There is a sequence $(I_n)$ of intervals such that $bigcup I_nsupseteq A$ and $sum |I_n|<|A|+epsilon. $ Set $J_n=Icap I_n; J_n'=I^ccap I_n. $ Some of these may be empty, but that's ok.
Then,
$|I_n|=|J_n|+|J_n'|, A cap I subseteq bigcup_{n=1}^{infty} J_n, A cap I^c subseteq bigcup_{n=1}^{infty} J'_n, $ and these facts imply that $|Acap I|+|Acap I^c|le |A|+epsilon.$
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
answered Dec 9 '18 at 21:48
MatematletaMatematleta
10.5k2918
10.5k2918
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$begingroup$
What is your definition of being measurable? Are you using outer measure?
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:31
$begingroup$
@mathcounterexamples.net Yes, I think OP is using the Carathéodory's criterion as definition.
$endgroup$
– Alex Vong
Dec 9 '18 at 20:34
$begingroup$
But at lest OP should precise what is $A$ and $S$ as he is looking to prove the measurability of $(a,b)$.
$endgroup$
– mathcounterexamples.net
Dec 9 '18 at 20:36
$begingroup$
The purpose of the exercise is to proove that any interval in ℝ is Lebesgue measurable by showing that any open interval in the form (a,b) is Lebesgue measurable. S in this exercise is the subset (a,b) and A is any S⊆ℝn. We use the outer measure as a definition of being measurable.
$endgroup$
– RHA
Dec 9 '18 at 21:11