A question about the outer measure generated












0












$begingroup$


Let $mathcal{K}subseteq 2^{X}$ be a family of subset such that $emptysetin mathcal{K}$ and let $nucolonmathcal{K}to [0,+infty]$ an application such that $nu(emptyset)=0.$
We denote with $$mathcal{R}_{mathcal{K}}big(Ebig)=bigg{{I_k}_{kinmathbb{N}}subseteqmathcal{K};bigg|;Esubseteqbigcup_{kinmathbb{N}}I_kbigg}.$$ We define $$mu^{(mathcal{K},nu)}big(Ebig):=infbigg{sum_{kinmathbb{N}}nu(I_k);bigg|;{I_k}inmathcal{R}_{mathcal{K}}big(Ebig)bigg}quadtext{if}quadmathcal{R}_{mathcal{K}}big(Ebig)neemptyset$$ and $$mu^{(mathcal{K},nu)}big(Ebig):=+inftyquadtext{if}quad mathcal{R}_{mathcal{K}}big(Ebig)=emptyset.$$




Question. If I know that $mathcal{R}_{mathcal{K}}big(Ebig)ne emptyset$, can I conclude that $mu^{(mathcal{K},mu)}big(Ebig)<+infty?$




For me the answer is no. Because we do not know that every series is convergent. That is: we suppose that $mathcal{R}_{mathcal{K}}big(Ebig)neemptyset$, and we suppose that exists one ${I_k}inmathcal{R}_{mathcal{K}}big(Ebig)$, such that $sum_{kinmathbb{N}}nu(I_k)=+infty$, in this case what can we say about $mu^{(mathcal{K},nu)}big(Ebig)$?



Thanks!






measure-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You are correct. The answer is indeed:"no".
    $endgroup$
    – drhab
    Jan 21 at 14:21
















0












$begingroup$


Let $mathcal{K}subseteq 2^{X}$ be a family of subset such that $emptysetin mathcal{K}$ and let $nucolonmathcal{K}to [0,+infty]$ an application such that $nu(emptyset)=0.$
We denote with $$mathcal{R}_{mathcal{K}}big(Ebig)=bigg{{I_k}_{kinmathbb{N}}subseteqmathcal{K};bigg|;Esubseteqbigcup_{kinmathbb{N}}I_kbigg}.$$ We define $$mu^{(mathcal{K},nu)}big(Ebig):=infbigg{sum_{kinmathbb{N}}nu(I_k);bigg|;{I_k}inmathcal{R}_{mathcal{K}}big(Ebig)bigg}quadtext{if}quadmathcal{R}_{mathcal{K}}big(Ebig)neemptyset$$ and $$mu^{(mathcal{K},nu)}big(Ebig):=+inftyquadtext{if}quad mathcal{R}_{mathcal{K}}big(Ebig)=emptyset.$$




Question. If I know that $mathcal{R}_{mathcal{K}}big(Ebig)ne emptyset$, can I conclude that $mu^{(mathcal{K},mu)}big(Ebig)<+infty?$




For me the answer is no. Because we do not know that every series is convergent. That is: we suppose that $mathcal{R}_{mathcal{K}}big(Ebig)neemptyset$, and we suppose that exists one ${I_k}inmathcal{R}_{mathcal{K}}big(Ebig)$, such that $sum_{kinmathbb{N}}nu(I_k)=+infty$, in this case what can we say about $mu^{(mathcal{K},nu)}big(Ebig)$?



Thanks!






measure-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You are correct. The answer is indeed:"no".
    $endgroup$
    – drhab
    Jan 21 at 14:21














0












0








0





$begingroup$


Let $mathcal{K}subseteq 2^{X}$ be a family of subset such that $emptysetin mathcal{K}$ and let $nucolonmathcal{K}to [0,+infty]$ an application such that $nu(emptyset)=0.$
We denote with $$mathcal{R}_{mathcal{K}}big(Ebig)=bigg{{I_k}_{kinmathbb{N}}subseteqmathcal{K};bigg|;Esubseteqbigcup_{kinmathbb{N}}I_kbigg}.$$ We define $$mu^{(mathcal{K},nu)}big(Ebig):=infbigg{sum_{kinmathbb{N}}nu(I_k);bigg|;{I_k}inmathcal{R}_{mathcal{K}}big(Ebig)bigg}quadtext{if}quadmathcal{R}_{mathcal{K}}big(Ebig)neemptyset$$ and $$mu^{(mathcal{K},nu)}big(Ebig):=+inftyquadtext{if}quad mathcal{R}_{mathcal{K}}big(Ebig)=emptyset.$$




Question. If I know that $mathcal{R}_{mathcal{K}}big(Ebig)ne emptyset$, can I conclude that $mu^{(mathcal{K},mu)}big(Ebig)<+infty?$




For me the answer is no. Because we do not know that every series is convergent. That is: we suppose that $mathcal{R}_{mathcal{K}}big(Ebig)neemptyset$, and we suppose that exists one ${I_k}inmathcal{R}_{mathcal{K}}big(Ebig)$, such that $sum_{kinmathbb{N}}nu(I_k)=+infty$, in this case what can we say about $mu^{(mathcal{K},nu)}big(Ebig)$?



Thanks!






measure-theory







share|cite|improve this question









$endgroup$




Let $mathcal{K}subseteq 2^{X}$ be a family of subset such that $emptysetin mathcal{K}$ and let $nucolonmathcal{K}to [0,+infty]$ an application such that $nu(emptyset)=0.$
We denote with $$mathcal{R}_{mathcal{K}}big(Ebig)=bigg{{I_k}_{kinmathbb{N}}subseteqmathcal{K};bigg|;Esubseteqbigcup_{kinmathbb{N}}I_kbigg}.$$ We define $$mu^{(mathcal{K},nu)}big(Ebig):=infbigg{sum_{kinmathbb{N}}nu(I_k);bigg|;{I_k}inmathcal{R}_{mathcal{K}}big(Ebig)bigg}quadtext{if}quadmathcal{R}_{mathcal{K}}big(Ebig)neemptyset$$ and $$mu^{(mathcal{K},nu)}big(Ebig):=+inftyquadtext{if}quad mathcal{R}_{mathcal{K}}big(Ebig)=emptyset.$$




Question. If I know that $mathcal{R}_{mathcal{K}}big(Ebig)ne emptyset$, can I conclude that $mu^{(mathcal{K},mu)}big(Ebig)<+infty?$




For me the answer is no. Because we do not know that every series is convergent. That is: we suppose that $mathcal{R}_{mathcal{K}}big(Ebig)neemptyset$, and we suppose that exists one ${I_k}inmathcal{R}_{mathcal{K}}big(Ebig)$, such that $sum_{kinmathbb{N}}nu(I_k)=+infty$, in this case what can we say about $mu^{(mathcal{K},nu)}big(Ebig)$?



Thanks!






measure-theory




measure-theory






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 13:59









Jack J.Jack J.

4292419




4292419








  • 1




    $begingroup$
    You are correct. The answer is indeed:"no".
    $endgroup$
    – drhab
    Jan 21 at 14:21














  • 1




    $begingroup$
    You are correct. The answer is indeed:"no".
    $endgroup$
    – drhab
    Jan 21 at 14:21








1




1




$begingroup$
You are correct. The answer is indeed:"no".
$endgroup$
– drhab
Jan 21 at 14:21




$begingroup$
You are correct. The answer is indeed:"no".
$endgroup$
– drhab
Jan 21 at 14:21










1 Answer
1






active

oldest

votes


















1












$begingroup$

In general, the answer is indeed no, and you can see this with a very simple example. Just take $mathcal K = {emptyset, X}$ and $nu(emptyset)=0$, $nu(X)=+infty$. Then for any non-empty set $Esubset X$ we have $mathcal R_{mathcal K}(E)=big{ {X,X,ldots} big}neq emptyset$ and $mu^{(mathcal K, nu)}(E)=+infty$.



Note that in this case $mathcal K$ is a $sigma$-field and $nu$ is a measure, so there is a lot more structure than in your general setting, but the assertion is still false.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
    $endgroup$
    – Jack J.
    Jan 21 at 14:29










  • $begingroup$
    You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
    $endgroup$
    – Mars Plastic
    Jan 21 at 14:36













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$begingroup$

In general, the answer is indeed no, and you can see this with a very simple example. Just take $mathcal K = {emptyset, X}$ and $nu(emptyset)=0$, $nu(X)=+infty$. Then for any non-empty set $Esubset X$ we have $mathcal R_{mathcal K}(E)=big{ {X,X,ldots} big}neq emptyset$ and $mu^{(mathcal K, nu)}(E)=+infty$.



Note that in this case $mathcal K$ is a $sigma$-field and $nu$ is a measure, so there is a lot more structure than in your general setting, but the assertion is still false.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
    $endgroup$
    – Jack J.
    Jan 21 at 14:29










  • $begingroup$
    You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
    $endgroup$
    – Mars Plastic
    Jan 21 at 14:36


















1












$begingroup$

In general, the answer is indeed no, and you can see this with a very simple example. Just take $mathcal K = {emptyset, X}$ and $nu(emptyset)=0$, $nu(X)=+infty$. Then for any non-empty set $Esubset X$ we have $mathcal R_{mathcal K}(E)=big{ {X,X,ldots} big}neq emptyset$ and $mu^{(mathcal K, nu)}(E)=+infty$.



Note that in this case $mathcal K$ is a $sigma$-field and $nu$ is a measure, so there is a lot more structure than in your general setting, but the assertion is still false.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
    $endgroup$
    – Jack J.
    Jan 21 at 14:29










  • $begingroup$
    You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
    $endgroup$
    – Mars Plastic
    Jan 21 at 14:36
















1












1








1





$begingroup$

In general, the answer is indeed no, and you can see this with a very simple example. Just take $mathcal K = {emptyset, X}$ and $nu(emptyset)=0$, $nu(X)=+infty$. Then for any non-empty set $Esubset X$ we have $mathcal R_{mathcal K}(E)=big{ {X,X,ldots} big}neq emptyset$ and $mu^{(mathcal K, nu)}(E)=+infty$.



Note that in this case $mathcal K$ is a $sigma$-field and $nu$ is a measure, so there is a lot more structure than in your general setting, but the assertion is still false.






share|cite|improve this answer









$endgroup$



In general, the answer is indeed no, and you can see this with a very simple example. Just take $mathcal K = {emptyset, X}$ and $nu(emptyset)=0$, $nu(X)=+infty$. Then for any non-empty set $Esubset X$ we have $mathcal R_{mathcal K}(E)=big{ {X,X,ldots} big}neq emptyset$ and $mu^{(mathcal K, nu)}(E)=+infty$.



Note that in this case $mathcal K$ is a $sigma$-field and $nu$ is a measure, so there is a lot more structure than in your general setting, but the assertion is still false.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 14:24









Mars PlasticMars Plastic

64118




64118












  • $begingroup$
    Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
    $endgroup$
    – Jack J.
    Jan 21 at 14:29










  • $begingroup$
    You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
    $endgroup$
    – Mars Plastic
    Jan 21 at 14:36




















  • $begingroup$
    Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
    $endgroup$
    – Jack J.
    Jan 21 at 14:29










  • $begingroup$
    You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
    $endgroup$
    – Mars Plastic
    Jan 21 at 14:36


















$begingroup$
Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
$endgroup$
– Jack J.
Jan 21 at 14:29




$begingroup$
Thanks for your precise answer. So, It's correct to say that $inf{+infty}=+infty$?
$endgroup$
– Jack J.
Jan 21 at 14:29












$begingroup$
You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
$endgroup$
– Mars Plastic
Jan 21 at 14:36






$begingroup$
You're welcome! And yes, that is also correct. The set $mathbb R cup {+infty,-infty}$ remains a totally ordered set, so the definition of infimum and supremum can be applied as usual.
$endgroup$
– Mars Plastic
Jan 21 at 14:36




















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