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In this paper the subadditivity of the Cantor function 0: [0, 1] [0, 1] is proved. Let us begin by recalling that a function f : R R is said to be subadditive, if it satisfies the inequality f (x + y) 1, f It is easy to check that each On is non-decreasing, that On (x) =0 for all x 1, and that the two lines in the definition of qn+1 agree in the overlap of their domains, both giving q0n+1(x) = when 1 y. Here we consider several cases. Case 1: y 1. In this case, O?n+1(X + y) <?1 +2 < On+?I(X) + On+?I(Y) Case 3: x < ?. As x, y and x + y are all < 2, we have 5n+1 (X + Y) n 2 (3x + 3y) On 2*?(3x) + 2 On (3Y) = On+I(X) + On+ I (Y) Received by the editors September 2, 1994 and, in revised form, May 11, 1995. 1991 Mathematics Subject Classification. Primary 26D15.

The standard Cantor function is subadditive