[1] Show that ∫$\_$0/$^$1/ [∫$\_$0/$^$1/ f($chi$,y)dy] d$chi$ = ∫$\_$0/$^$1/[∫$\_$0/$^$1/ f($chi$,y)d$chi$] Counterexample: If pk denotes the k-th prime number, let S(pk) = (equation omitted), let S = ∪$\_$k=1/$^$$infty$/ S(pk), and let Q = [0, 1]${ imes}$[0, 1]. Define f on Q as follows; f($chi$, y) = 0 ($chi$, y)$in$S, f($chi$, y) = 1 ($chi$, y)$in$Q - S.(omitted)