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I believe the definition of super-additive set function is wrong since when ${\displaystyle A=B}$ we get ${\displaystyle f(A)\geq 2f(A)}$ which implies that ${\displaystyle f(A)=0}$ -- 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) reply