Talk:Additive Function is Strongly Additive

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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)