Limit of measures is the measure of the intersection












1












$begingroup$


Suppose ${E_n}$ are measurable and suppose we have the following lemma:



If in addition, $E_n subseteq E_{n+1}$,



then lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcup_{n=1}^{infty} E_n)$.



Then, prove:



If $E_n$ measurable, $m(E_1) < + infty$ AND $E_{n+1} subseteq E_n$,



THEN, lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n=1}^{infty}E_n)$



So, Since the $E_n$ are measurable, we have monotonicity,



i.e.,



$E_{n+1} subseteq E_n$ $Rightarrow$ $m(E_{n+1}) leq m(E_n)$



but $m(E_1) < + infty$ gives us



$m(E_{n+1}) leq m(E_n) = k$ ; $forall n in mathbb{Z}^+$ and SOME $k in mathbb{R}$ (this denotes measure of $E_1$).



I get stuck on how to apply the Lemma, am I supposed to use the analogous version of the Lemma?



i.e., replacing $n$ with $n+1$?



and would this give



lim$_{n rightarrow infty}$ $m(E_{n+1})$ = $m(bigcup_{n=1}^{infty} E_{n+1})$??



or do I rewrite



$m(E_{n+1}$) = $m(E_n$) + $m(E_1)$



I highly doubt this, I never make assumptions in proofs I am not positive about and I never say I get something till it fully makes sense to me, please help, you can see where I am stuck. I apologize in advance if I am missing the obvious. I am merely a noob at measure theory.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Use Demorgan's law
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 0:14
















1












$begingroup$


Suppose ${E_n}$ are measurable and suppose we have the following lemma:



If in addition, $E_n subseteq E_{n+1}$,



then lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcup_{n=1}^{infty} E_n)$.



Then, prove:



If $E_n$ measurable, $m(E_1) < + infty$ AND $E_{n+1} subseteq E_n$,



THEN, lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n=1}^{infty}E_n)$



So, Since the $E_n$ are measurable, we have monotonicity,



i.e.,



$E_{n+1} subseteq E_n$ $Rightarrow$ $m(E_{n+1}) leq m(E_n)$



but $m(E_1) < + infty$ gives us



$m(E_{n+1}) leq m(E_n) = k$ ; $forall n in mathbb{Z}^+$ and SOME $k in mathbb{R}$ (this denotes measure of $E_1$).



I get stuck on how to apply the Lemma, am I supposed to use the analogous version of the Lemma?



i.e., replacing $n$ with $n+1$?



and would this give



lim$_{n rightarrow infty}$ $m(E_{n+1})$ = $m(bigcup_{n=1}^{infty} E_{n+1})$??



or do I rewrite



$m(E_{n+1}$) = $m(E_n$) + $m(E_1)$



I highly doubt this, I never make assumptions in proofs I am not positive about and I never say I get something till it fully makes sense to me, please help, you can see where I am stuck. I apologize in advance if I am missing the obvious. I am merely a noob at measure theory.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Use Demorgan's law
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 0:14














1












1








1





$begingroup$


Suppose ${E_n}$ are measurable and suppose we have the following lemma:



If in addition, $E_n subseteq E_{n+1}$,



then lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcup_{n=1}^{infty} E_n)$.



Then, prove:



If $E_n$ measurable, $m(E_1) < + infty$ AND $E_{n+1} subseteq E_n$,



THEN, lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n=1}^{infty}E_n)$



So, Since the $E_n$ are measurable, we have monotonicity,



i.e.,



$E_{n+1} subseteq E_n$ $Rightarrow$ $m(E_{n+1}) leq m(E_n)$



but $m(E_1) < + infty$ gives us



$m(E_{n+1}) leq m(E_n) = k$ ; $forall n in mathbb{Z}^+$ and SOME $k in mathbb{R}$ (this denotes measure of $E_1$).



I get stuck on how to apply the Lemma, am I supposed to use the analogous version of the Lemma?



i.e., replacing $n$ with $n+1$?



and would this give



lim$_{n rightarrow infty}$ $m(E_{n+1})$ = $m(bigcup_{n=1}^{infty} E_{n+1})$??



or do I rewrite



$m(E_{n+1}$) = $m(E_n$) + $m(E_1)$



I highly doubt this, I never make assumptions in proofs I am not positive about and I never say I get something till it fully makes sense to me, please help, you can see where I am stuck. I apologize in advance if I am missing the obvious. I am merely a noob at measure theory.






real-analysis measure-theory lebesgue-measure







share|cite|improve this question









$endgroup$




Suppose ${E_n}$ are measurable and suppose we have the following lemma:



If in addition, $E_n subseteq E_{n+1}$,



then lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcup_{n=1}^{infty} E_n)$.



Then, prove:



If $E_n$ measurable, $m(E_1) < + infty$ AND $E_{n+1} subseteq E_n$,



THEN, lim$_{n rightarrow infty}$ $m(E_n)$ = $m(bigcap_{n=1}^{infty}E_n)$



So, Since the $E_n$ are measurable, we have monotonicity,



i.e.,



$E_{n+1} subseteq E_n$ $Rightarrow$ $m(E_{n+1}) leq m(E_n)$



but $m(E_1) < + infty$ gives us



$m(E_{n+1}) leq m(E_n) = k$ ; $forall n in mathbb{Z}^+$ and SOME $k in mathbb{R}$ (this denotes measure of $E_1$).



I get stuck on how to apply the Lemma, am I supposed to use the analogous version of the Lemma?



i.e., replacing $n$ with $n+1$?



and would this give



lim$_{n rightarrow infty}$ $m(E_{n+1})$ = $m(bigcup_{n=1}^{infty} E_{n+1})$??



or do I rewrite



$m(E_{n+1}$) = $m(E_n$) + $m(E_1)$



I highly doubt this, I never make assumptions in proofs I am not positive about and I never say I get something till it fully makes sense to me, please help, you can see where I am stuck. I apologize in advance if I am missing the obvious. I am merely a noob at measure theory.






real-analysis measure-theory lebesgue-measure




real-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










asked Dec 24 '18 at 0:09









Hossien SahebjameHossien Sahebjame

1169




1169








  • 1




    $begingroup$
    Use Demorgan's law
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 0:14














  • 1




    $begingroup$
    Use Demorgan's law
    $endgroup$
    – rubikscube09
    Dec 24 '18 at 0:14








1




1




$begingroup$
Use Demorgan's law
$endgroup$
– rubikscube09
Dec 24 '18 at 0:14




$begingroup$
Use Demorgan's law
$endgroup$
– rubikscube09
Dec 24 '18 at 0:14










1 Answer
1






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oldest

votes


















2












$begingroup$

$E_1setminus E_n$ increases to $E_1setminus cap_n E_n$, so $m(E_1setminus E_n) to m(E_1setminus cap_n E_n)$. This gives $m(E_1)-m(E_n) to m(E_1)-m(cap_n E_n)$. Cancel $m(E_1)$ to get the result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
    $endgroup$
    – Hossien Sahebjame
    Jan 3 at 21:23












  • $begingroup$
    Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:22












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$begingroup$

$E_1setminus E_n$ increases to $E_1setminus cap_n E_n$, so $m(E_1setminus E_n) to m(E_1setminus cap_n E_n)$. This gives $m(E_1)-m(E_n) to m(E_1)-m(cap_n E_n)$. Cancel $m(E_1)$ to get the result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
    $endgroup$
    – Hossien Sahebjame
    Jan 3 at 21:23












  • $begingroup$
    Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:22
















2












$begingroup$

$E_1setminus E_n$ increases to $E_1setminus cap_n E_n$, so $m(E_1setminus E_n) to m(E_1setminus cap_n E_n)$. This gives $m(E_1)-m(E_n) to m(E_1)-m(cap_n E_n)$. Cancel $m(E_1)$ to get the result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
    $endgroup$
    – Hossien Sahebjame
    Jan 3 at 21:23












  • $begingroup$
    Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:22














2












2








2





$begingroup$

$E_1setminus E_n$ increases to $E_1setminus cap_n E_n$, so $m(E_1setminus E_n) to m(E_1setminus cap_n E_n)$. This gives $m(E_1)-m(E_n) to m(E_1)-m(cap_n E_n)$. Cancel $m(E_1)$ to get the result.






share|cite|improve this answer









$endgroup$



$E_1setminus E_n$ increases to $E_1setminus cap_n E_n$, so $m(E_1setminus E_n) to m(E_1setminus cap_n E_n)$. This gives $m(E_1)-m(E_n) to m(E_1)-m(cap_n E_n)$. Cancel $m(E_1)$ to get the result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 0:14









Kavi Rama MurthyKavi Rama Murthy

72k53170




72k53170












  • $begingroup$
    what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
    $endgroup$
    – Hossien Sahebjame
    Jan 3 at 21:23












  • $begingroup$
    Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:22


















  • $begingroup$
    what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
    $endgroup$
    – Hossien Sahebjame
    Jan 3 at 21:23












  • $begingroup$
    Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 23:22
















$begingroup$
what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
$endgroup$
– Hossien Sahebjame
Jan 3 at 21:23






$begingroup$
what do you mean by increases to ? as $n$ tends to $infty$? and is this because $m(E_1)$ is finite and by monotonicity $m(E_n)$ is also finite?
$endgroup$
– Hossien Sahebjame
Jan 3 at 21:23














$begingroup$
Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:22




$begingroup$
Increases means $E_1setminus E_n subset E_1setminus E_{n+1}$.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 23:22


















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