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  1. 09 Feb '15 11:02 / 3 edits
    I have been reading this:

    http://en.wikipedia.org/wiki/Sigma_additivity#Additive_.28or_finitely_additive.29_set_functions

    and, in the main, understanding it although there are some aspects of it that I still struggle with.

    But, I really want to better understand it is at least just a couple of different examples of mathematical functions u defined on an algebra of sets A with values in [−∞, +∞] that are NOT additive complete with explanations of why they are not additive if that why isn't self-evident.

    Anyone?
  2. Standard member DeepThought
    Losing the Thread
    09 Feb '15 17:22
    What's wrong with the example they gave?

    http://en.wikipedia.org/wiki/Sigma_additivity#An_additive_function_which_is_not_.CF.83-additive
  3. Standard member Duncan Clarke
    Student
    09 Feb '15 17:52
    Originally posted by humy
    I have been reading this:

    http://en.wikipedia.org/wiki/Sigma_additivity#Additive_.28or_finitely_additive.29_set_functions

    and, in the main, understanding it although there are some aspects of it that I still struggle with.

    But, I really want to better understand it is at least just a couple of different examples of mathematical functions u defined on a ...[text shortened]... planations of why they are not additive if that why isn't self-evident.

    Anyone?
    Because you can't add eggs and cheese!
  4. 09 Feb '15 21:37 / 13 edits
    Originally posted by DeepThought
    What's wrong with the example they gave?

    http://en.wikipedia.org/wiki/Sigma_additivity#An_additive_function_which_is_not_.CF.83-additive
    That isn't an example of a NONE additive function. That is an example of an additive function which is not σ-additive, but is still additive because it is finitely additive. I am asking for example/examples of a function that can take the same input (an algebra of sets A with values in [−∞, +∞] ) and has the same type of output (i.e. either real numbers or a subset of real numbers) but is NOT additive i.e. both not finitely additive and not σ-additive.

    I tried googling this but so far, although I see some hints that such a function may exist, I have failed to find a single clear example of one. This makes me a little unsure if one exists.
  5. Standard member DeepThought
    Losing the Thread
    09 Feb '15 23:49
    Originally posted by humy
    That isn't an example of a NONE additive function. That is an example of an additive function which is not σ-additive, but is still additive because it is finitely additive. I am asking for example/examples of a function that can take the same input (an algebra of sets A with values in [−∞, +∞] ) and has the same type of output (i.e. either real numbers ...[text shortened]... have failed to find a single clear example of one. This makes me a little unsure if one exists.
    With you. Consider the function µ(A) = λ(A)², where λ(A) is the Lebesgue measure of A. Let B and C be disjoint sets and let D = B∪C. µ is not additive since µ(D) = λ(D)² = λ(B∪C)² = (λ(B) + λ(C))² = λ(B)² + λ(C)² + 2λ(B)λ(C) ≥ µ(B) + µ(C).
  6. Standard member Soothfast
    0,1,1,2,3,5,8,13,21,
    10 Feb '15 00:18 / 1 edit
    Originally posted by humy
    That isn't an example of a NONE additive function. That is an example of an additive function which is not σ-additive, but is still additive because it is finitely additive. I am asking for example/examples of a function that can take the same input (an algebra of sets A with values in [−∞, +∞] ) and has the same type of output (i.e. either real numbers ...[text shortened]... have failed to find a single clear example of one. This makes me a little unsure if one exists.
    Let A be the algebra that is the power set of the integers (i.e. the set of all subsets of the integers). Define a set function µ on A as follows: for any element S in A (so S is a subset of the integers) let µ(S)=10 if S is an infinite set, let µ(S)=5 if S is finite and has an even number of elements, let µ(S)=7 if S is finite and has an odd number of elements. Also let µ({ })=0, which takes care of the empty set. This makes µ a legitimate set function. So let's see how this works out in the following example.

    Let S_1={1,2,3} and S_2={4,5,6}, so that S_1∪S_2={1,2,3,4,5,6}

    Then µ(S_1∪S_2) = 5, but µ(S_1) + µ(S_2) = 7+7 = 14.

    There you go!
  7. Standard member DeepThought
    Losing the Thread
    10 Feb '15 00:27
    Originally posted by Soothfast
    Let A be the algebra that is the power set of the integers (i.e. the set of all subsets of the integers). Define a set function µ on A as follows: for any element S in A (so S is a subset of the integers) let µ(S)=10 if S is an infinite set, let µ(S)=5 if S is finite and has an even number of elements, let µ(S)=7 if S is finite and has an odd number of el ...[text shortened]... that S_1∪S_2={1,2,3,4}

    Then µ(S_1∪S_2) = 5, but µ(S_1) + µ(S_2) = 7+5 = 12.

    There you go!
    Don't the sets have to be disjoint (not that that spoils your example much)?
  8. Standard member Soothfast
    0,1,1,2,3,5,8,13,21,
    10 Feb '15 00:33
    Originally posted by DeepThought
    Don't the sets have to be disjoint (not that that spoils your example much)?
    Fixed.
  9. Standard member DeepThought
    Losing the Thread
    10 Feb '15 01:00
    Originally posted by Soothfast
    Fixed.
    We could also have:

    A = {1,3,5,···}
    B = {0,2,4,···}
    A∪B = {0,1,2,···} = ℕ

    Both sets are infinite so µ(A) = µ(B) = 10, they are also disjoint.

    µ(A∪B) = µ({1, 2, 3, ···}) = µ(ℕ) = 10
    µ(A) + µ(B) = 10 + 10 = 20

    I got my head about how to get unicode symbols on linux (ctrl + shift + u)
  10. Standard member Soothfast
    0,1,1,2,3,5,8,13,21,
    10 Feb '15 01:35
    Originally posted by DeepThought
    We could also have:

    A = {1,3,5,···}
    B = {0,2,4,···}
    A∪B = {0,1,2,···} = ℕ

    Both sets are infinite so µ(A) = µ(B) = 10, they are also disjoint.

    µ(A∪B) = µ({1, 2, 3, ···}) = µ(ℕ) = 10
    µ(A) + µ(B) = 10 + 10 = 20

    I got my head about how to get unicode symbols on linux (ctrl + shift + u)
    Right. So basically it's really easy to come up with set functions that are not additive. I would guess the reason they're not in wide currency in cyberspace is because they tend to be not very interesting or useful. In my own experience set functions are usually given to be measures, so additivity is built in.
  11. 10 Feb '15 08:11 / 1 edit
    Thanks all for the examples
    I will gradually mull through them in due course.