By Krzysztof M. Ostaszewski
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Extra resources for Henstock Integration in the Plane
Sample text
Let I0 = [0,1] x [0,1], an = l - ^ 6n = i n for for n = 0 , 1 , 2 , . . ,a n ] x [ a n _ ! 75) Ln = {(uyv)eKn:v
Let f\nen^n D(x0,p{xo)), = {*(>}• There is an n 6 N such that In is contained in where a = ap and p G P. If A = A, or A = A* for t = 1,2, then we already get a contradiction, since {(zo>Jn)} is a partition of In which is contained in or. If A = Ai, then divide In into at most four subintervals, each of which has a vertex at z0. Let {J,} be those subintervals (j can run through any of the following sets: {1}, {1,2}, {1,2,3,4}). Then {(x0,Jj)} is a partition of In contained in a, and that is a contradiction.
X ^ y ' ) : » = 1 ' 2 ' - • • « *. }. 28) \U(p)-U(p)\