Example of countably additive function from a Boolean algebra to [0,infinity] that is not finitely additive The Next CEO of Stack OverflowFinitely but not countably additive set functionFinite additivity, atomlessness and countable additivityFrom finitely additive to countably additiveMotivation for and differences between properties of measure, outer measureWhy does not the Euclidean space support a countably additive measure defined for all subsets?Bounded finely additive signed measure is a signed measure.Is there a unique finitely additive extension of a countably additive product probability measure to the product $sigma$-algebra for finite products?a subset of a zero Lebesgue measure set is measurable?Show that the additive group of rational numbers does not have a Haar measure.Does the existence of a merely finitely additive probability on a Boolean algebra require axioms beyond ZF?
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Example of countably additive function from a Boolean algebra to [0,infinity] that is not finitely additive
The Next CEO of Stack OverflowFinitely but not countably additive set functionFinite additivity, atomlessness and countable additivityFrom finitely additive to countably additiveMotivation for and differences between properties of measure, outer measureWhy does not the Euclidean space support a countably additive measure defined for all subsets?Bounded finely additive signed measure is a signed measure.Is there a unique finitely additive extension of a countably additive product probability measure to the product $sigma$-algebra for finite products?a subset of a zero Lebesgue measure set is measurable?Show that the additive group of rational numbers does not have a Haar measure.Does the existence of a merely finitely additive probability on a Boolean algebra require axioms beyond ZF?
$begingroup$
This example is implied to exist by Tao’s undergrad measure theory text, exercise 1.7.4, with the claim that further assuming that the empty set getting mapped to 0 prevents such a function from existing. As far as I can tell, padding out the countable union with empty sets forces any other set to be mapped to infinity if the image of the empty set is nonzero (and thus infinite). But the constant infinity map is finitely additive, right? What am I missing here?
real-analysis measure-theory outer-measure
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
$endgroup$
|
show 3 more comments
$begingroup$
This example is implied to exist by Tao’s undergrad measure theory text, exercise 1.7.4, with the claim that further assuming that the empty set getting mapped to 0 prevents such a function from existing. As far as I can tell, padding out the countable union with empty sets forces any other set to be mapped to infinity if the image of the empty set is nonzero (and thus infinite). But the constant infinity map is finitely additive, right? What am I missing here?
real-analysis measure-theory outer-measure
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
$endgroup$
1
$begingroup$
What are the exact definitions of "countably additive" and "finitely additive"?
$endgroup$
– Eric Wofsey
Mar 19 at 20:36
$begingroup$
On my phone now so I mean that the function maps disjoint unions to sums.
$endgroup$
– Zach Boyd
Mar 19 at 20:37
$begingroup$
Either finite or Countably additive respectively.
$endgroup$
– Zach Boyd
Mar 19 at 20:38
$begingroup$
I'm not asking what the definition is roughly, but what it is exactly. I'm guessing the answer to your question has to do with some subtlety in the precise definitions Tao uses.
$endgroup$
– Eric Wofsey
Mar 19 at 20:38
$begingroup$
Ok I will have to wait until I am at a computer.
$endgroup$
– Zach Boyd
Mar 19 at 20:39
|
show 3 more comments
$begingroup$
This example is implied to exist by Tao’s undergrad measure theory text, exercise 1.7.4, with the claim that further assuming that the empty set getting mapped to 0 prevents such a function from existing. As far as I can tell, padding out the countable union with empty sets forces any other set to be mapped to infinity if the image of the empty set is nonzero (and thus infinite). But the constant infinity map is finitely additive, right? What am I missing here?
real-analysis measure-theory outer-measure
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
$endgroup$
This example is implied to exist by Tao’s undergrad measure theory text, exercise 1.7.4, with the claim that further assuming that the empty set getting mapped to 0 prevents such a function from existing. As far as I can tell, padding out the countable union with empty sets forces any other set to be mapped to infinity if the image of the empty set is nonzero (and thus infinite). But the constant infinity map is finitely additive, right? What am I missing here?
real-analysis measure-theory outer-measure
real-analysis measure-theory outer-measure
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
asked Mar 19 at 20:35
Zach BoydZach Boyd
7932518
7932518
1
$begingroup$
What are the exact definitions of "countably additive" and "finitely additive"?
$endgroup$
– Eric Wofsey
Mar 19 at 20:36
$begingroup$
On my phone now so I mean that the function maps disjoint unions to sums.
$endgroup$
– Zach Boyd
Mar 19 at 20:37
$begingroup$
Either finite or Countably additive respectively.
$endgroup$
– Zach Boyd
Mar 19 at 20:38
$begingroup$
I'm not asking what the definition is roughly, but what it is exactly. I'm guessing the answer to your question has to do with some subtlety in the precise definitions Tao uses.
$endgroup$
– Eric Wofsey
Mar 19 at 20:38
$begingroup$
Ok I will have to wait until I am at a computer.
$endgroup$
– Zach Boyd
Mar 19 at 20:39
|
show 3 more comments
1
$begingroup$
What are the exact definitions of "countably additive" and "finitely additive"?
$endgroup$
– Eric Wofsey
Mar 19 at 20:36
$begingroup$
On my phone now so I mean that the function maps disjoint unions to sums.
$endgroup$
– Zach Boyd
Mar 19 at 20:37
$begingroup$
Either finite or Countably additive respectively.
$endgroup$
– Zach Boyd
Mar 19 at 20:38
$begingroup$
I'm not asking what the definition is roughly, but what it is exactly. I'm guessing the answer to your question has to do with some subtlety in the precise definitions Tao uses.
$endgroup$
– Eric Wofsey
Mar 19 at 20:38
$begingroup$
Ok I will have to wait until I am at a computer.
$endgroup$
– Zach Boyd
Mar 19 at 20:39
1
1
$begingroup$
What are the exact definitions of "countably additive" and "finitely additive"?
$endgroup$
– Eric Wofsey
Mar 19 at 20:36
$begingroup$
What are the exact definitions of "countably additive" and "finitely additive"?
$endgroup$
– Eric Wofsey
Mar 19 at 20:36
$begingroup$
On my phone now so I mean that the function maps disjoint unions to sums.
$endgroup$
– Zach Boyd
Mar 19 at 20:37
$begingroup$
On my phone now so I mean that the function maps disjoint unions to sums.
$endgroup$
– Zach Boyd
Mar 19 at 20:37
$begingroup$
Either finite or Countably additive respectively.
$endgroup$
– Zach Boyd
Mar 19 at 20:38
$begingroup$
Either finite or Countably additive respectively.
$endgroup$
– Zach Boyd
Mar 19 at 20:38
$begingroup$
I'm not asking what the definition is roughly, but what it is exactly. I'm guessing the answer to your question has to do with some subtlety in the precise definitions Tao uses.
$endgroup$
– Eric Wofsey
Mar 19 at 20:38
$begingroup$
I'm not asking what the definition is roughly, but what it is exactly. I'm guessing the answer to your question has to do with some subtlety in the precise definitions Tao uses.
$endgroup$
– Eric Wofsey
Mar 19 at 20:38
$begingroup$
Ok I will have to wait until I am at a computer.
$endgroup$
– Zach Boyd
Mar 19 at 20:39
$begingroup$
Ok I will have to wait until I am at a computer.
$endgroup$
– Zach Boyd
Mar 19 at 20:39
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
That book is online:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwj_yb6Xlo_hAhWT51QKHU1YAHIQFjABegQIBBAC&url=https%3A%2F%2Fterrytao.files.wordpress.com%2F2012%2F12%2Fgsm-126-tao5-measure-book.pdf&usg=AOvVaw0GnO1oyFyno__7TnnfBVCl
The Exercise 1.7.4 does not imply existence of function that is countably additive but not finitely additive. Eric's comment on the precise definition of finitely additive is the key here.
Finitely additive
Definition 1.4.19 gives: With $mathcalB$ a Boolean algebra, a map $mu:mathcalBrightarrow [0,infty]$ is finitely additive if
(Axiom 1) $mu(phi)=0$.
(Axiom 2) $mu(A cup B) = mu(A) + mu(B)$ whenever $A, B$ are disjoint and in $mathcalB$.
Countably additive
The defintion of countably additive (Definition 1.4.27) keeps Axiom 1 but changes Axiom 2 to treat countable unions of disjoint sets (it also assumes $mathcalB$ is a sigma-algebra rather than an algebra).
Exercise 1.7.4a
The exercise 1.7.4a just wants you to show that if $mathcalB$ is a Boolean algebra and if $mu:mathcalB rightarrow [0,infty]$ a function, then $mu$ is finitely additive whenever the following two properties hold:
(i) $mu(phi)=0$.
(ii) $mu(cup_i=1^inftyA_i)= sum_i=1^inftymu(A_i)$ whenever $A_i_i=1^infty$ are disjoint sets in the Boolean algebra such that $cup_i=1^infty A_i$ is also in the Boolean algebra.
That exercise was not trying to imagine a scenario where $mu(phi)>0$ (that would not fit the definition of either finitely additive or countably additive).
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
$endgroup$
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
add a comment |
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1 Answer
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1 Answer
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$begingroup$
That book is online:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwj_yb6Xlo_hAhWT51QKHU1YAHIQFjABegQIBBAC&url=https%3A%2F%2Fterrytao.files.wordpress.com%2F2012%2F12%2Fgsm-126-tao5-measure-book.pdf&usg=AOvVaw0GnO1oyFyno__7TnnfBVCl
The Exercise 1.7.4 does not imply existence of function that is countably additive but not finitely additive. Eric's comment on the precise definition of finitely additive is the key here.
Finitely additive
Definition 1.4.19 gives: With $mathcalB$ a Boolean algebra, a map $mu:mathcalBrightarrow [0,infty]$ is finitely additive if
(Axiom 1) $mu(phi)=0$.
(Axiom 2) $mu(A cup B) = mu(A) + mu(B)$ whenever $A, B$ are disjoint and in $mathcalB$.
Countably additive
The defintion of countably additive (Definition 1.4.27) keeps Axiom 1 but changes Axiom 2 to treat countable unions of disjoint sets (it also assumes $mathcalB$ is a sigma-algebra rather than an algebra).
Exercise 1.7.4a
The exercise 1.7.4a just wants you to show that if $mathcalB$ is a Boolean algebra and if $mu:mathcalB rightarrow [0,infty]$ a function, then $mu$ is finitely additive whenever the following two properties hold:
(i) $mu(phi)=0$.
(ii) $mu(cup_i=1^inftyA_i)= sum_i=1^inftymu(A_i)$ whenever $A_i_i=1^infty$ are disjoint sets in the Boolean algebra such that $cup_i=1^infty A_i$ is also in the Boolean algebra.
That exercise was not trying to imagine a scenario where $mu(phi)>0$ (that would not fit the definition of either finitely additive or countably additive).
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
$endgroup$
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
add a comment |
$begingroup$
That book is online:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwj_yb6Xlo_hAhWT51QKHU1YAHIQFjABegQIBBAC&url=https%3A%2F%2Fterrytao.files.wordpress.com%2F2012%2F12%2Fgsm-126-tao5-measure-book.pdf&usg=AOvVaw0GnO1oyFyno__7TnnfBVCl
The Exercise 1.7.4 does not imply existence of function that is countably additive but not finitely additive. Eric's comment on the precise definition of finitely additive is the key here.
Finitely additive
Definition 1.4.19 gives: With $mathcalB$ a Boolean algebra, a map $mu:mathcalBrightarrow [0,infty]$ is finitely additive if
(Axiom 1) $mu(phi)=0$.
(Axiom 2) $mu(A cup B) = mu(A) + mu(B)$ whenever $A, B$ are disjoint and in $mathcalB$.
Countably additive
The defintion of countably additive (Definition 1.4.27) keeps Axiom 1 but changes Axiom 2 to treat countable unions of disjoint sets (it also assumes $mathcalB$ is a sigma-algebra rather than an algebra).
Exercise 1.7.4a
The exercise 1.7.4a just wants you to show that if $mathcalB$ is a Boolean algebra and if $mu:mathcalB rightarrow [0,infty]$ a function, then $mu$ is finitely additive whenever the following two properties hold:
(i) $mu(phi)=0$.
(ii) $mu(cup_i=1^inftyA_i)= sum_i=1^inftymu(A_i)$ whenever $A_i_i=1^infty$ are disjoint sets in the Boolean algebra such that $cup_i=1^infty A_i$ is also in the Boolean algebra.
That exercise was not trying to imagine a scenario where $mu(phi)>0$ (that would not fit the definition of either finitely additive or countably additive).
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
$endgroup$
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
add a comment |
$begingroup$
That book is online:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwj_yb6Xlo_hAhWT51QKHU1YAHIQFjABegQIBBAC&url=https%3A%2F%2Fterrytao.files.wordpress.com%2F2012%2F12%2Fgsm-126-tao5-measure-book.pdf&usg=AOvVaw0GnO1oyFyno__7TnnfBVCl
The Exercise 1.7.4 does not imply existence of function that is countably additive but not finitely additive. Eric's comment on the precise definition of finitely additive is the key here.
Finitely additive
Definition 1.4.19 gives: With $mathcalB$ a Boolean algebra, a map $mu:mathcalBrightarrow [0,infty]$ is finitely additive if
(Axiom 1) $mu(phi)=0$.
(Axiom 2) $mu(A cup B) = mu(A) + mu(B)$ whenever $A, B$ are disjoint and in $mathcalB$.
Countably additive
The defintion of countably additive (Definition 1.4.27) keeps Axiom 1 but changes Axiom 2 to treat countable unions of disjoint sets (it also assumes $mathcalB$ is a sigma-algebra rather than an algebra).
Exercise 1.7.4a
The exercise 1.7.4a just wants you to show that if $mathcalB$ is a Boolean algebra and if $mu:mathcalB rightarrow [0,infty]$ a function, then $mu$ is finitely additive whenever the following two properties hold:
(i) $mu(phi)=0$.
(ii) $mu(cup_i=1^inftyA_i)= sum_i=1^inftymu(A_i)$ whenever $A_i_i=1^infty$ are disjoint sets in the Boolean algebra such that $cup_i=1^infty A_i$ is also in the Boolean algebra.
That exercise was not trying to imagine a scenario where $mu(phi)>0$ (that would not fit the definition of either finitely additive or countably additive).
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
$endgroup$
That book is online:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwj_yb6Xlo_hAhWT51QKHU1YAHIQFjABegQIBBAC&url=https%3A%2F%2Fterrytao.files.wordpress.com%2F2012%2F12%2Fgsm-126-tao5-measure-book.pdf&usg=AOvVaw0GnO1oyFyno__7TnnfBVCl
The Exercise 1.7.4 does not imply existence of function that is countably additive but not finitely additive. Eric's comment on the precise definition of finitely additive is the key here.
Finitely additive
Definition 1.4.19 gives: With $mathcalB$ a Boolean algebra, a map $mu:mathcalBrightarrow [0,infty]$ is finitely additive if
(Axiom 1) $mu(phi)=0$.
(Axiom 2) $mu(A cup B) = mu(A) + mu(B)$ whenever $A, B$ are disjoint and in $mathcalB$.
Countably additive
The defintion of countably additive (Definition 1.4.27) keeps Axiom 1 but changes Axiom 2 to treat countable unions of disjoint sets (it also assumes $mathcalB$ is a sigma-algebra rather than an algebra).
Exercise 1.7.4a
The exercise 1.7.4a just wants you to show that if $mathcalB$ is a Boolean algebra and if $mu:mathcalB rightarrow [0,infty]$ a function, then $mu$ is finitely additive whenever the following two properties hold:
(i) $mu(phi)=0$.
(ii) $mu(cup_i=1^inftyA_i)= sum_i=1^inftymu(A_i)$ whenever $A_i_i=1^infty$ are disjoint sets in the Boolean algebra such that $cup_i=1^infty A_i$ is also in the Boolean algebra.
That exercise was not trying to imagine a scenario where $mu(phi)>0$ (that would not fit the definition of either finitely additive or countably additive).
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
edited Mar 20 at 6:07
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
answered Mar 19 at 21:57
MichaelMichael
13.2k11429
13.2k11429
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
add a comment |
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
Exercise 1.7.4. is not in that linked text. What is its statement exactly?
$endgroup$
– Henno Brandsma
Mar 19 at 22:23
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
It is there on page 185.
$endgroup$
– Michael
Mar 20 at 3:38
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
$begingroup$
Thank you Michael—this matches the rest of the exercise as well.
$endgroup$
– Zach Boyd
Mar 20 at 10:21
add a comment |
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1
$begingroup$
What are the exact definitions of "countably additive" and "finitely additive"?
$endgroup$
– Eric Wofsey
Mar 19 at 20:36
$begingroup$
On my phone now so I mean that the function maps disjoint unions to sums.
$endgroup$
– Zach Boyd
Mar 19 at 20:37
$begingroup$
Either finite or Countably additive respectively.
$endgroup$
– Zach Boyd
Mar 19 at 20:38
$begingroup$
I'm not asking what the definition is roughly, but what it is exactly. I'm guessing the answer to your question has to do with some subtlety in the precise definitions Tao uses.
$endgroup$
– Eric Wofsey
Mar 19 at 20:38
$begingroup$
Ok I will have to wait until I am at a computer.
$endgroup$
– Zach Boyd
Mar 19 at 20:39