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)