# Talk:Additive Function is Strongly Additive

When the codomain is $\overline{\R}$, subtraction is tricky business; the page needs rephrasing to avoid minus signs. --Lord_Farin 11:59, 16 March 2012 (EDT)

- What is proven is $f\left({A}\right) + f\left({B}\right) = f\left({A \cup B}\right) + f\left({A \cap B}\right)$. How about changing it to that? Abcxyz 12:22, 16 March 2012 (EDT)

- That particular statement is called that '$f$ is strongly additive'; I put it up a sec ago for measures: Measure is Strongly Additive. So I suggest this page to be rephrased and moved to Additive Function is Strongly Additive. Anyone disagree? --Lord_Farin 12:30, 16 March 2012 (EDT)

- I think that's a good idea. Anyway, maybe we should have a definition page for a strongly additive function? Abcxyz 12:38, 16 March 2012 (EDT)

Yes, haven't gotten round to writing it, but there's a link already on Measure is Strongly Additive. --Lord_Farin 15:29, 16 March 2012 (EDT)

Moved it as suggested; I have a doubt concerning the algebra of sets requirement, since it seems not to imply that $\mathcal S$ is $\cap$-closed. --Lord_Farin 14:07, 21 May 2012 (EDT)

- Is it not? I understood that $\cap$-closed was a given. --prime mover 15:29, 21 May 2012 (EDT)

- Mm, yes, never mind; I took a look at the wrong def'n. Maybe good to put up explicitly what a ring of sets is, so that it is immediately clear that it is not only $\cup$, but also $\cap$-closed. --Lord_Farin 16:00, 21 May 2012 (EDT)

About the asking for explanation by Usagiop : It is kind of baked into the definition of additive functions that $\map f A + \map f B$ is always defined. This follows from the definition if $A,B$ are disjoint, and if they are not disjoint, we can construct a short proof by examining $\map f { A \cap B }$. I could add the proof as a Lemma if needed. Anghel (talk) 15:07, 13 August 2022 (UTC)

- Yes, if $A\cap B \ne \O$, it is not obvious that $\map f A + \map f B$ is well-defined. Maybe it is easier to proof backwards:

As:

- $\map f {A\cup B} = \map f {A \symdif B} + \map f {A \cap B}$

it must satisfy:

- $\struct { \map f {A\cup B}, \map f {A \cap B} } \not \in \set { \struct {+\infty, -\infty}, \struct {-\infty, +\infty} } $

so that:

- $\map f {A\cup B} + \map f {A \cap B} $

is well-defined, for example. --Usagiop (talk) 18:36, 13 August 2022 (UTC)

- To Anghel: Please add the missing lemma, if you like. I could add an alternative proof.