Any countable set has measurable zero












1














To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory







share|cite|improve this question


















  • 1




    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    – rubikscube09
    Dec 10 '18 at 2:05










  • Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    Thanks, Rubik! :)
    – Rafael Vergnaud
    Dec 10 '18 at 2:14
















1














To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory







share|cite|improve this question


















  • 1




    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    – rubikscube09
    Dec 10 '18 at 2:05










  • Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    Thanks, Rubik! :)
    – Rafael Vergnaud
    Dec 10 '18 at 2:14














1












1








1


1





To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory







share|cite|improve this question













To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory




real-analysis general-topology elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 1:49









Rafael Vergnaud

318116




318116








  • 1




    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    – rubikscube09
    Dec 10 '18 at 2:05










  • Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    Thanks, Rubik! :)
    – Rafael Vergnaud
    Dec 10 '18 at 2:14














  • 1




    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    – rubikscube09
    Dec 10 '18 at 2:05










  • Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    Thanks, Rubik! :)
    – Rafael Vergnaud
    Dec 10 '18 at 2:14








1




1




Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
– carmichael561
Dec 10 '18 at 2:04




Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
– carmichael561
Dec 10 '18 at 2:04




1




1




No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
– rubikscube09
Dec 10 '18 at 2:05




No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
– rubikscube09
Dec 10 '18 at 2:05












Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
– Rafael Vergnaud
Dec 10 '18 at 2:06






Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
– Rafael Vergnaud
Dec 10 '18 at 2:06






1




1




Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
– rubikscube09
Dec 10 '18 at 2:14






Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
– rubikscube09
Dec 10 '18 at 2:14






1




1




Thanks, Rubik! :)
– Rafael Vergnaud
Dec 10 '18 at 2:14




Thanks, Rubik! :)
– Rafael Vergnaud
Dec 10 '18 at 2:14










1 Answer
1






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2














(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer























  • Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    Right. Got it! Thanks a lot Holo!
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    – Holo
    Dec 10 '18 at 2:29













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2














(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer























  • Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    Right. Got it! Thanks a lot Holo!
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    – Holo
    Dec 10 '18 at 2:29


















2














(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer























  • Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    Right. Got it! Thanks a lot Holo!
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    – Holo
    Dec 10 '18 at 2:29
















2












2








2






(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer














(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 2:29

























answered Dec 10 '18 at 2:10









Holo

5,5302930




5,5302930












  • Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    Right. Got it! Thanks a lot Holo!
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    – Holo
    Dec 10 '18 at 2:29




















  • Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    Right. Got it! Thanks a lot Holo!
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    – Holo
    Dec 10 '18 at 2:29


















Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
– Rafael Vergnaud
Dec 10 '18 at 2:11






Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
– Rafael Vergnaud
Dec 10 '18 at 2:11














@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
– Holo
Dec 10 '18 at 2:14






@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
– Holo
Dec 10 '18 at 2:14






1




1




Right. Got it! Thanks a lot Holo!
– Rafael Vergnaud
Dec 10 '18 at 2:15




Right. Got it! Thanks a lot Holo!
– Rafael Vergnaud
Dec 10 '18 at 2:15




1




1




@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
– Holo
Dec 10 '18 at 2:29






@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
– Holo
Dec 10 '18 at 2:29




















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