- 09 Feb '15 11:02 / 3 editsI 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? - 09 Feb '15 17:52

Because you can't add eggs and cheese!*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? - 09 Feb '15 21:37 / 13 edits

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*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*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. - 09 Feb '15 23:49

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).*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. - 10 Feb '15 00:18 / 1 edit

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.*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 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! - 10 Feb '15 00:27

Don't the sets have to be disjoint (not that that spoils your example much)?*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! - 10 Feb '15 01:00

We could also have:*Originally posted by Soothfast***Fixed.**

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 Feb '15 01:35

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.*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)