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Set and Set Complement of Uniform Distribution on [0,1]

A question regarding Borel sets, the Lebesgue measure, and Cantor-type setsUnion of Sets Dense in $[0,1]$For set of positive measure $E$, $alpha in (0, 1)$, there is interval $I$ such that $m(E cap I) > alpha , m(I)$Measure Theory - working with unusual measures and set functionsShowing $chi_GnotinmathcalR[0,1]$ where: $G$ open, $mathbbQcap[0,1] subset G,$ and $m(G)<frac12$How can I construct such a Borel subset?How to prove complement of generalized Cantor set is dense in $[0,1]$A discontinuous function at every point in $[0,1]$Asking about a hint: constructing a cantor-like setCantor-Like Sets

0

$begingroup$

This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^C in [0,1]$ $S$ and $S^C$ both have measure $frac12$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_S = mu (S : cap : (a,b))=mu (S^C : cap : (a,b)) = mu_S^C$ ? Just to clarify $S^C$ is the complement of $S$ on the unit interval, or put differently $S^C = S^complement : cap : [0,1]$

real-analysis measure-theory

share|cite|improve this question

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

asked Mar 22 at 1:16

David G.David G.

156

$endgroup$

add a comment | 

0

$begingroup$

This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^C in [0,1]$ $S$ and $S^C$ both have measure $frac12$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_S = mu (S : cap : (a,b))=mu (S^C : cap : (a,b)) = mu_S^C$ ? Just to clarify $S^C$ is the complement of $S$ on the unit interval, or put differently $S^C = S^complement : cap : [0,1]$

real-analysis measure-theory

share|cite|improve this question

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

asked Mar 22 at 1:16

David G.David G.

156

$endgroup$

add a comment | 

$begingroup$

This questions first came to mind a few years ago when I was taking a course on real analysis as an undergraduate. I posed it to my instructor but he did not know a means of solving my inquiry. But to the point, does there exist a set (or is there a means of constructing a set) such that for a set of points $S in [0,1]$ and $S^C in [0,1]$ $S$ and $S^C$ both have measure $frac12$ and for any arbitrary sub-interval $(a,b)$ of $[0,1]$ the value $mu_S = mu (S : cap : (a,b))=mu (S^C : cap : (a,b)) = mu_S^C$ ? Just to clarify $S^C$ is the complement of $S$ on the unit interval, or put differently $S^C = S^complement : cap : [0,1]$

real-analysis measure-theory

share|cite|improve this question

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

asked Mar 22 at 1:16

David G.David G.

156

$endgroup$

share|cite|improve this question

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

asked Mar 22 at 1:16

David G.David G.

156

share|cite|improve this question

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

asked Mar 22 at 1:16

David G.David G.

156

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

edited Mar 22 at 3:05

Andrés E. Caicedo

65.9k8160252

asked Mar 22 at 1:16

David G.David G.

156

asked Mar 22 at 1:16

David G.David G.

156

2 Answers
2

active

oldest

votes

1

$begingroup$

No, there is no such measurable set. If there were you can get a contradiction as follows.

Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.

share|cite|improve this answer

edited Mar 22 at 11:38

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_S$. Could you please expand?
  $endgroup$
  – David G.
  Mar 22 at 2:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG. I have edited my answer. I hope it makes it clearer.
  $endgroup$
  – kimchi lover
  Mar 22 at 11:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
  $endgroup$
  – David G.
  Mar 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
  $endgroup$
  – kimchi lover
  Mar 22 at 15:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you clarify why $mu_S ll mu$? It seems to me that the relation should be $mu_s = frac12 mu$ which still holds when $mu (A) = 0$
  $endgroup$
  – David G.
  Mar 22 at 16:11
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

By a standard argument $mu(Scap (a,b))=mu(S^ccap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^ccap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.

share|cite|improve this answer

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
  $endgroup$
  – David G.
  Mar 22 at 6:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you extend the equality from intervals to measurable sets you can then take $A=S$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 22 at 6:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
  $endgroup$
  – David G.
  Mar 22 at 7:24
  

  
  
  
  

add a comment | 

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1

$begingroup$

No, there is no such measurable set. If there were you can get a contradiction as follows.

Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.

share|cite|improve this answer

edited Mar 22 at 11:38

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_S$. Could you please expand?
  $endgroup$
  – David G.
  Mar 22 at 2:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG. I have edited my answer. I hope it makes it clearer.
  $endgroup$
  – kimchi lover
  Mar 22 at 11:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
  $endgroup$
  – David G.
  Mar 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
  $endgroup$
  – kimchi lover
  Mar 22 at 15:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you clarify why $mu_S ll mu$? It seems to me that the relation should be $mu_s = frac12 mu$ which still holds when $mu (A) = 0$
  $endgroup$
  – David G.
  Mar 22 at 16:11
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

No, there is no such measurable set. If there were you can get a contradiction as follows.

Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.

share|cite|improve this answer

edited Mar 22 at 11:38

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So unfortunately, I don't quite understand why the RN derivative is equal to the indicator function $chi_S$. Could you please expand?
  $endgroup$
  – David G.
  Mar 22 at 2:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG. I have edited my answer. I hope it makes it clearer.
  $endgroup$
  – kimchi lover
  Mar 22 at 11:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't see why $mu (A) = 0$ since $A = (a,b)$ wouldn't it necessarily be the case that $mu (A) = b-a$?
  $endgroup$
  – David G.
  Mar 22 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The condition for $mu_Sllmu$ is that whenever $mu(A)=0$ we also have $mu_S(A)=0$, too. Which is satisfied by your $mu_S$.
  $endgroup$
  – kimchi lover
  Mar 22 at 15:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you clarify why $mu_S ll mu$? It seems to me that the relation should be $mu_s = frac12 mu$ which still holds when $mu (A) = 0$
  $endgroup$
  – David G.
  Mar 22 at 16:11
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

No, there is no such measurable set. If there were you can get a contradiction as follows.

Note that $mu_Sllmu$ because $mu(A)=0$ implies $mu_S(A)=mu(Acap S)=0$, so the Radon Nikodym theorem applies: there exists a function $f$ such that $mu_S(A)=int_A f dx$, unique up to measure $0$ modifications. Calculate this Radon Nikodym derivative two different ways. On the one hand, $f=chi_S$, the indicator function of $S$, since $mu_S(A)=int_A chi_S(x)dx$. On the other, it is the constant function $1/2$. (Evaluate $F(x)=mu_S([0,x]) = x/2$, and so on.)
These two formulas do not agree almost everywhere.

share|cite|improve this answer

edited Mar 22 at 11:38

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

$endgroup$

share|cite|improve this answer

edited Mar 22 at 11:38

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

answered Mar 22 at 1:46

kimchi loverkimchi lover

11.6k31229

1

$begingroup$

By a standard argument $mu(Scap (a,b))=mu(S^ccap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^ccap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.

share|cite|improve this answer

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
  $endgroup$
  – David G.
  Mar 22 at 6:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you extend the equality from intervals to measurable sets you can then take $A=S$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 22 at 6:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
  $endgroup$
  – David G.
  Mar 22 at 7:24
  

  
  
  
  

add a comment | 

1

$begingroup$

By a standard argument $mu(Scap (a,b))=mu(S^ccap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^ccap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.

share|cite|improve this answer

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand your point, however, I'm inclined to believe that $S$ cannot equal $A$ as I don't see how $S$ is an interval.
  $endgroup$
  – David G.
  Mar 22 at 6:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you extend the equality from intervals to measurable sets you can then take $A=S$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 22 at 6:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Though this is true, that breaks the statement of the question and circumvents the basis of the question. The subintervals are there to illustrated uniform distribution of points across the unit interval
  $endgroup$
  – David G.
  Mar 22 at 7:24
  

  
  
  
  

add a comment | 

$begingroup$

By a standard argument $mu(Scap (a,b))=mu(S^ccap (a,b))$ for every subinterval $(a,b)$ implies $mu(Scap A)=mu(S^ccap A)$ for every Bore set $A$. Taking $A=S$ we get a contradiction.

share|cite|improve this answer

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

share|cite|improve this answer

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

answered Mar 22 at 6:06

Kavi Rama MurthyKavi Rama Murthy

72.6k53170