Talk:Utility functions on indivisible goods

Superadditivity

I believe the definition of super-additive set function is wrong since when $A=B$ we get $f(A)\geq 2f(A)$ which implies that $f(A)=0$ -- anonymous 2015-12-21

You are probably right. I corrected the definition here and in Superadditive set function. --Erel Segal (talk) 13:24, 21 December 2015 (UTC)