Show that the open interval (a, b) is Lebesgue measurable












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 20:11









postmortes

1,91721117




1,91721117










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




















  • $begingroup$
    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












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 $|Acap I|+|Acap I^c|le |A|+epsilon.$






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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 $|Acap I|+|Acap I^c|le |A|+epsilon.$






    share|cite|improve this answer









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      0












      $begingroup$

      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 $|Acap I|+|Acap I^c|le |A|+epsilon.$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $|Acap I|+|Acap I^c|le |A|+epsilon.$






        share|cite|improve this answer









        $endgroup$



        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 $|Acap I|+|Acap I^c|le |A|+epsilon.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 21:48









        MatematletaMatematleta

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