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complement of a set and its measure

0

$begingroup$

if $E = [0,1] bigcap mathbbQ $, what is $E^complement$ with respect to $[0,1]$ and its outer measure $m^ast(E^complement)$?

The outer measure is defined thus:

$m^ast(E^complement):= inf_P m(P)$

where $P$ ranges over the set of multi-intervals containing $E$: $E subseteq P$

measure-theory

share|cite|improve this question

edited Mar 26 at 10:50

Henno Brandsma

116k349127

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the properties of $m^ast$, not its definition. "multi-intervals" is not a standard term, you mean at most countable union of intervals?
  $endgroup$
  – Henno Brandsma
  Mar 26 at 10:48
  
  

  
  
  
  

add a comment | 

0

$begingroup$

if $E = [0,1] bigcap mathbbQ $, what is $E^complement$ with respect to $[0,1]$ and its outer measure $m^ast(E^complement)$?

The outer measure is defined thus:

$m^ast(E^complement):= inf_P m(P)$

where $P$ ranges over the set of multi-intervals containing $E$: $E subseteq P$

measure-theory

share|cite|improve this question

edited Mar 26 at 10:50

Henno Brandsma

116k349127

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the properties of $m^ast$, not its definition. "multi-intervals" is not a standard term, you mean at most countable union of intervals?
  $endgroup$
  – Henno Brandsma
  Mar 26 at 10:48
  
  

  
  
  
  

add a comment | 

$begingroup$

if $E = [0,1] bigcap mathbbQ $, what is $E^complement$ with respect to $[0,1]$ and its outer measure $m^ast(E^complement)$?

The outer measure is defined thus:

$m^ast(E^complement):= inf_P m(P)$

where $P$ ranges over the set of multi-intervals containing $E$: $E subseteq P$

measure-theory

share|cite|improve this question

edited Mar 26 at 10:50

Henno Brandsma

116k349127

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

$endgroup$

share|cite|improve this question

edited Mar 26 at 10:50

Henno Brandsma

116k349127

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

share|cite|improve this question

edited Mar 26 at 10:50

Henno Brandsma

116k349127

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

edited Mar 26 at 10:50

Henno Brandsma

116k349127

edited Mar 26 at 10:50

Henno Brandsma

116k349127

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

asked Mar 26 at 10:06

Giuliano MalatestaGiuliano Malatesta

347

2 Answers
2

active

oldest

votes

3

$begingroup$

$E^complement = x in [0,1]: x notin mathbbQ$, the irrational points of $[0,1]$, when we take the complement in $[0,1]$, as $E$ consists of the rational points (so is countable and hence has $m^ast(E)=0$).

$m^ast(E^complement)=1$ as is easy to see. Note that subadditivity and monotonicity of $m^ast$ already tell us that

$$1=m^ast([0,1]) le m^ast(E) + m^ast(E^complement) = 0 + m^ast(E^complement)= m^ast(E^complement) le m^ast([0,1])=1$$

share|cite|improve this answer

edited Mar 29 at 9:36

supinf

6,7571129

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to use De Morgan's Law: The complement of an intersection is the union of complements $left(Acap Bright)^c=A^ccup B^c$ which in this case should be: $[0,1]^c cup mathbbQ^C =emptyset cup [0,1]$ but it is wrong
  $endgroup$
  – Giuliano Malatesta
  Mar 26 at 10:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta inside $[0,1]$ the complement of $[0,1]$ is $emptyset$, indeed but the complement of the rationals (even inside $[0,1]$) is not $[0,1]$, but like I said, the irationals in that interval.
  $endgroup$
  – Henno Brandsma
  Mar 26 at 10:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta It's formulaic expression is unimportant for the computation ofits outer measure. I don't use it. Just that $E$ is countable and thus has outer measure $0$; that's all that matters.
  $endgroup$
  – Henno Brandsma
  Mar 26 at 11:02
  

  
  
  
  

add a comment | 

1

$begingroup$

The outer measure is clearly less than or equal to $1$. Suppose $m^*(E^c) =r<1$. Cover the rational numbers in $[0,1]$ by intervals with total length $t <1-r$ and cover $E$ by intervals with total length less than $r$. Putting these together we can cover $[0,1]$ by intervals with total length $t+r <1$ which contradicts the fact that measure of $[0,1]$ is $1$. Hence $m^*(E^c)=1$

share|cite|improve this answer

edited Mar 26 at 10:19

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the expression for $E^c$ ?
  $endgroup$
  – Giuliano Malatesta
  Mar 26 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $E^c=[0,1]setminus E$ the set of all irrational numbers in $[0,1]$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 11:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if $E^c$ is an irrational set why are you talking of covering the rationals in $[0,1]$?
  $endgroup$
  – Giuliano Malatesta
  Mar 29 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta The idea is to cover both $E$ and $E^c$ to get a cover for $[0,1]$ with total length less than $1$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 29 at 8:49
  

  
  
  
  

add a comment | 

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3

$begingroup$

$E^complement = x in [0,1]: x notin mathbbQ$, the irrational points of $[0,1]$, when we take the complement in $[0,1]$, as $E$ consists of the rational points (so is countable and hence has $m^ast(E)=0$).

$m^ast(E^complement)=1$ as is easy to see. Note that subadditivity and monotonicity of $m^ast$ already tell us that

$$1=m^ast([0,1]) le m^ast(E) + m^ast(E^complement) = 0 + m^ast(E^complement)= m^ast(E^complement) le m^ast([0,1])=1$$

share|cite|improve this answer

edited Mar 29 at 9:36

supinf

6,7571129

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to use De Morgan's Law: The complement of an intersection is the union of complements $left(Acap Bright)^c=A^ccup B^c$ which in this case should be: $[0,1]^c cup mathbbQ^C =emptyset cup [0,1]$ but it is wrong
  $endgroup$
  – Giuliano Malatesta
  Mar 26 at 10:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta inside $[0,1]$ the complement of $[0,1]$ is $emptyset$, indeed but the complement of the rationals (even inside $[0,1]$) is not $[0,1]$, but like I said, the irationals in that interval.
  $endgroup$
  – Henno Brandsma
  Mar 26 at 10:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta It's formulaic expression is unimportant for the computation ofits outer measure. I don't use it. Just that $E$ is countable and thus has outer measure $0$; that's all that matters.
  $endgroup$
  – Henno Brandsma
  Mar 26 at 11:02
  

  
  
  
  

add a comment | 

3

$begingroup$

$E^complement = x in [0,1]: x notin mathbbQ$, the irrational points of $[0,1]$, when we take the complement in $[0,1]$, as $E$ consists of the rational points (so is countable and hence has $m^ast(E)=0$).

$m^ast(E^complement)=1$ as is easy to see. Note that subadditivity and monotonicity of $m^ast$ already tell us that

$$1=m^ast([0,1]) le m^ast(E) + m^ast(E^complement) = 0 + m^ast(E^complement)= m^ast(E^complement) le m^ast([0,1])=1$$

share|cite|improve this answer

edited Mar 29 at 9:36

supinf

6,7571129

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried to use De Morgan's Law: The complement of an intersection is the union of complements $left(Acap Bright)^c=A^ccup B^c$ which in this case should be: $[0,1]^c cup mathbbQ^C =emptyset cup [0,1]$ but it is wrong
  $endgroup$
  – Giuliano Malatesta
  Mar 26 at 10:57
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta inside $[0,1]$ the complement of $[0,1]$ is $emptyset$, indeed but the complement of the rationals (even inside $[0,1]$) is not $[0,1]$, but like I said, the irationals in that interval.
  $endgroup$
  – Henno Brandsma
  Mar 26 at 10:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta It's formulaic expression is unimportant for the computation ofits outer measure. I don't use it. Just that $E$ is countable and thus has outer measure $0$; that's all that matters.
  $endgroup$
  – Henno Brandsma
  Mar 26 at 11:02
  

  
  
  
  

add a comment | 

$begingroup$

$E^complement = x in [0,1]: x notin mathbbQ$, the irrational points of $[0,1]$, when we take the complement in $[0,1]$, as $E$ consists of the rational points (so is countable and hence has $m^ast(E)=0$).

$m^ast(E^complement)=1$ as is easy to see. Note that subadditivity and monotonicity of $m^ast$ already tell us that

$$1=m^ast([0,1]) le m^ast(E) + m^ast(E^complement) = 0 + m^ast(E^complement)= m^ast(E^complement) le m^ast([0,1])=1$$

share|cite|improve this answer

edited Mar 29 at 9:36

supinf

6,7571129

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

share|cite|improve this answer

edited Mar 29 at 9:36

supinf

6,7571129

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

edited Mar 29 at 9:36

supinf

6,7571129

edited Mar 29 at 9:36

supinf

6,7571129

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

answered Mar 26 at 10:47

Henno BrandsmaHenno Brandsma

116k349127

1

$begingroup$

The outer measure is clearly less than or equal to $1$. Suppose $m^*(E^c) =r<1$. Cover the rational numbers in $[0,1]$ by intervals with total length $t <1-r$ and cover $E$ by intervals with total length less than $r$. Putting these together we can cover $[0,1]$ by intervals with total length $t+r <1$ which contradicts the fact that measure of $[0,1]$ is $1$. Hence $m^*(E^c)=1$

share|cite|improve this answer

edited Mar 26 at 10:19

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the expression for $E^c$ ?
  $endgroup$
  – Giuliano Malatesta
  Mar 26 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $E^c=[0,1]setminus E$ the set of all irrational numbers in $[0,1]$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 11:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if $E^c$ is an irrational set why are you talking of covering the rationals in $[0,1]$?
  $endgroup$
  – Giuliano Malatesta
  Mar 29 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta The idea is to cover both $E$ and $E^c$ to get a cover for $[0,1]$ with total length less than $1$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 29 at 8:49
  

  
  
  
  

add a comment | 

1

$begingroup$

The outer measure is clearly less than or equal to $1$. Suppose $m^*(E^c) =r<1$. Cover the rational numbers in $[0,1]$ by intervals with total length $t <1-r$ and cover $E$ by intervals with total length less than $r$. Putting these together we can cover $[0,1]$ by intervals with total length $t+r <1$ which contradicts the fact that measure of $[0,1]$ is $1$. Hence $m^*(E^c)=1$

share|cite|improve this answer

edited Mar 26 at 10:19

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the expression for $E^c$ ?
  $endgroup$
  – Giuliano Malatesta
  Mar 26 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $E^c=[0,1]setminus E$ the set of all irrational numbers in $[0,1]$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 26 at 11:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  if $E^c$ is an irrational set why are you talking of covering the rationals in $[0,1]$?
  $endgroup$
  – Giuliano Malatesta
  Mar 29 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GiulianoMalatesta The idea is to cover both $E$ and $E^c$ to get a cover for $[0,1]$ with total length less than $1$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 29 at 8:49
  

  
  
  
  

add a comment | 

$begingroup$

The outer measure is clearly less than or equal to $1$. Suppose $m^*(E^c) =r<1$. Cover the rational numbers in $[0,1]$ by intervals with total length $t <1-r$ and cover $E$ by intervals with total length less than $r$. Putting these together we can cover $[0,1]$ by intervals with total length $t+r <1$ which contradicts the fact that measure of $[0,1]$ is $1$. Hence $m^*(E^c)=1$

share|cite|improve this answer

edited Mar 26 at 10:19

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

$endgroup$

share|cite|improve this answer

edited Mar 26 at 10:19

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270

answered Mar 26 at 10:09

Kavi Rama MurthyKavi Rama Murthy

74.9k53270