R is the reals. X is the square [0, 1] x [0, 1]. f : X → R is continuous. If ∫Y f(x, y) dx dy = 0 for all squares Y such that (1) Y ⊆ X, (2) Y has sides parallel to those of X, (3) at least one of Y's sides is contained in the boundary of X, is it true that f(x, y) = 0 for all x, y?
Solution
Answer: yes.
Given any square Z inside X with sides parallel to X, we can find squares Y1, Y2 satisfying the conditions in the question such that Z = Y1 ∩ Y2. Hence ∫Z f(x, y) dx dy = 0 for all Z.
If f(x, y) > 0 for any interior point (x, y) of X, then by continuity we can find a square Z such that f(x, y) > 0 on Z. Contradiction. Similarly for f(x, y) < 0 for an interior point of X. Hence f(x, y) = 0 on the interior of X and hence (by continuity) on the whole of X.
© John Scholes
jscholes@kalva.demon.co.uk
30 Nov 1999