the outer measure of cantor like set

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Let $D subset [0,1]$ be a cantor like set. Then, the outer measure, $mu^*(D) = varepsilon$ for $varepsilon>0.$



Attempt: Suppose that $a^k$ is removed from the middle interval at $k$th stage, and $D_k$ be the $k$th stage of a construction of $D$. Then, we have $$mu^*(D_k) = 1-asum_n=0^k-1 (2a)^n = 1-afrac1-(2a)^k1-2a. $$
Then, for $k to infty,$ we have $$mu^*(D)=1-fraca1-2a$$ for $a<1/2.$ This shows that if $a = 1/3$, then $mu^*(D)=0.$

Of course, this does not answer my question that $mu^*(D)=varepsilon$, but this is what I know. So, could you give some help to tackle this question?



Thank you in advance.




measure-theory




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  • Does "$mu^*(D)=varepsilon$ for $varepsilongt0$" mean "$mu^*(D)=varepsilon$ for SOME $varepsilongt0$" or "$mu^*(D)=varepsilon$ for ALL $varepsilongt0$"? If the former, it is equivalent to asking "is $mu^*(D)gt0,$" why make it more complicated? If the latter, the answer is no; if $varepsilon_1nevarepsilon_2,$ it's impossible to have both $mu^*(D)=varepsilon_1$ and $mu^*(D)=varepsilon_2.$
    – bof
    Aug 18 at 2:05











  • In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^k-1$ intervals obtained at the stage $k-1$. The length of each of these intervals is $frac13^k-1$, so at the stage $k$ you remove a set whose measure is $frac2^k-13^k$. The measure of the union of the sets removed up to stage $k$ is $sum_i=1^k frac2^i-13^i$. Taking the limit, you get $mu(D)=mu(1)-lim_krightarrowinftysum_i=1^k frac2^i-13^i = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure.
    – Emanuele Bottazzi
    Aug 18 at 6:50










  • If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want.
    – DanielWainfleet
    Aug 18 at 13:36














up vote
0
down vote

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Let $D subset [0,1]$ be a cantor like set. Then, the outer measure, $mu^*(D) = varepsilon$ for $varepsilon>0.$



Attempt: Suppose that $a^k$ is removed from the middle interval at $k$th stage, and $D_k$ be the $k$th stage of a construction of $D$. Then, we have $$mu^*(D_k) = 1-asum_n=0^k-1 (2a)^n = 1-afrac1-(2a)^k1-2a. $$
Then, for $k to infty,$ we have $$mu^*(D)=1-fraca1-2a$$ for $a<1/2.$ This shows that if $a = 1/3$, then $mu^*(D)=0.$

Of course, this does not answer my question that $mu^*(D)=varepsilon$, but this is what I know. So, could you give some help to tackle this question?



Thank you in advance.




measure-theory




share|cite|improve this question




















  • Does "$mu^*(D)=varepsilon$ for $varepsilongt0$" mean "$mu^*(D)=varepsilon$ for SOME $varepsilongt0$" or "$mu^*(D)=varepsilon$ for ALL $varepsilongt0$"? If the former, it is equivalent to asking "is $mu^*(D)gt0,$" why make it more complicated? If the latter, the answer is no; if $varepsilon_1nevarepsilon_2,$ it's impossible to have both $mu^*(D)=varepsilon_1$ and $mu^*(D)=varepsilon_2.$
    – bof
    Aug 18 at 2:05











  • In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^k-1$ intervals obtained at the stage $k-1$. The length of each of these intervals is $frac13^k-1$, so at the stage $k$ you remove a set whose measure is $frac2^k-13^k$. The measure of the union of the sets removed up to stage $k$ is $sum_i=1^k frac2^i-13^i$. Taking the limit, you get $mu(D)=mu(1)-lim_krightarrowinftysum_i=1^k frac2^i-13^i = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure.
    – Emanuele Bottazzi
    Aug 18 at 6:50










  • If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want.
    – DanielWainfleet
    Aug 18 at 13:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $D subset [0,1]$ be a cantor like set. Then, the outer measure, $mu^*(D) = varepsilon$ for $varepsilon>0.$



Attempt: Suppose that $a^k$ is removed from the middle interval at $k$th stage, and $D_k$ be the $k$th stage of a construction of $D$. Then, we have $$mu^*(D_k) = 1-asum_n=0^k-1 (2a)^n = 1-afrac1-(2a)^k1-2a. $$
Then, for $k to infty,$ we have $$mu^*(D)=1-fraca1-2a$$ for $a<1/2.$ This shows that if $a = 1/3$, then $mu^*(D)=0.$

Of course, this does not answer my question that $mu^*(D)=varepsilon$, but this is what I know. So, could you give some help to tackle this question?



Thank you in advance.




measure-theory




share|cite|improve this question












Let $D subset [0,1]$ be a cantor like set. Then, the outer measure, $mu^*(D) = varepsilon$ for $varepsilon>0.$



Attempt: Suppose that $a^k$ is removed from the middle interval at $k$th stage, and $D_k$ be the $k$th stage of a construction of $D$. Then, we have $$mu^*(D_k) = 1-asum_n=0^k-1 (2a)^n = 1-afrac1-(2a)^k1-2a. $$
Then, for $k to infty,$ we have $$mu^*(D)=1-fraca1-2a$$ for $a<1/2.$ This shows that if $a = 1/3$, then $mu^*(D)=0.$

Of course, this does not answer my question that $mu^*(D)=varepsilon$, but this is what I know. So, could you give some help to tackle this question?



Thank you in advance.





measure-theory





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asked Aug 18 at 1:53









Sihyun Kim

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  • Does "$mu^*(D)=varepsilon$ for $varepsilongt0$" mean "$mu^*(D)=varepsilon$ for SOME $varepsilongt0$" or "$mu^*(D)=varepsilon$ for ALL $varepsilongt0$"? If the former, it is equivalent to asking "is $mu^*(D)gt0,$" why make it more complicated? If the latter, the answer is no; if $varepsilon_1nevarepsilon_2,$ it's impossible to have both $mu^*(D)=varepsilon_1$ and $mu^*(D)=varepsilon_2.$
    – bof
    Aug 18 at 2:05











  • In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^k-1$ intervals obtained at the stage $k-1$. The length of each of these intervals is $frac13^k-1$, so at the stage $k$ you remove a set whose measure is $frac2^k-13^k$. The measure of the union of the sets removed up to stage $k$ is $sum_i=1^k frac2^i-13^i$. Taking the limit, you get $mu(D)=mu(1)-lim_krightarrowinftysum_i=1^k frac2^i-13^i = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure.
    – Emanuele Bottazzi
    Aug 18 at 6:50










  • If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want.
    – DanielWainfleet
    Aug 18 at 13:36
















  • Does "$mu^*(D)=varepsilon$ for $varepsilongt0$" mean "$mu^*(D)=varepsilon$ for SOME $varepsilongt0$" or "$mu^*(D)=varepsilon$ for ALL $varepsilongt0$"? If the former, it is equivalent to asking "is $mu^*(D)gt0,$" why make it more complicated? If the latter, the answer is no; if $varepsilon_1nevarepsilon_2,$ it's impossible to have both $mu^*(D)=varepsilon_1$ and $mu^*(D)=varepsilon_2.$
    – bof
    Aug 18 at 2:05











  • In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^k-1$ intervals obtained at the stage $k-1$. The length of each of these intervals is $frac13^k-1$, so at the stage $k$ you remove a set whose measure is $frac2^k-13^k$. The measure of the union of the sets removed up to stage $k$ is $sum_i=1^k frac2^i-13^i$. Taking the limit, you get $mu(D)=mu(1)-lim_krightarrowinftysum_i=1^k frac2^i-13^i = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure.
    – Emanuele Bottazzi
    Aug 18 at 6:50










  • If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want.
    – DanielWainfleet
    Aug 18 at 13:36















Does "$mu^*(D)=varepsilon$ for $varepsilongt0$" mean "$mu^*(D)=varepsilon$ for SOME $varepsilongt0$" or "$mu^*(D)=varepsilon$ for ALL $varepsilongt0$"? If the former, it is equivalent to asking "is $mu^*(D)gt0,$" why make it more complicated? If the latter, the answer is no; if $varepsilon_1nevarepsilon_2,$ it's impossible to have both $mu^*(D)=varepsilon_1$ and $mu^*(D)=varepsilon_2.$
– bof
Aug 18 at 2:05





Does "$mu^*(D)=varepsilon$ for $varepsilongt0$" mean "$mu^*(D)=varepsilon$ for SOME $varepsilongt0$" or "$mu^*(D)=varepsilon$ for ALL $varepsilongt0$"? If the former, it is equivalent to asking "is $mu^*(D)gt0,$" why make it more complicated? If the latter, the answer is no; if $varepsilon_1nevarepsilon_2,$ it's impossible to have both $mu^*(D)=varepsilon_1$ and $mu^*(D)=varepsilon_2.$
– bof
Aug 18 at 2:05













In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^k-1$ intervals obtained at the stage $k-1$. The length of each of these intervals is $frac13^k-1$, so at the stage $k$ you remove a set whose measure is $frac2^k-13^k$. The measure of the union of the sets removed up to stage $k$ is $sum_i=1^k frac2^i-13^i$. Taking the limit, you get $mu(D)=mu(1)-lim_krightarrowinftysum_i=1^k frac2^i-13^i = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure.
– Emanuele Bottazzi
Aug 18 at 6:50




In the construction of the Cantor set, at the stage $k$ you remove the middle third to each of the $2^k-1$ intervals obtained at the stage $k-1$. The length of each of these intervals is $frac13^k-1$, so at the stage $k$ you remove a set whose measure is $frac2^k-13^k$. The measure of the union of the sets removed up to stage $k$ is $sum_i=1^k frac2^i-13^i$. Taking the limit, you get $mu(D)=mu(1)-lim_krightarrowinftysum_i=1^k frac2^i-13^i = 0$. As a consequence the Cantor set is a null set and its inner measure coincides with its outer measure.
– Emanuele Bottazzi
Aug 18 at 6:50












If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want.
– DanielWainfleet
Aug 18 at 13:36




If $0<a<1/3$ then $1-a/(1-2a)$ can be any member of $(0,1)$ that you want.
– DanielWainfleet
Aug 18 at 13:36















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