(i) Let X ( a.s. on a set A with P(A) > 0, and suppose that (A X dP = 0. Then show that X = 0 a.s. on A.
(ii) Let X ( 0, integrable, and X = 0 a.s. on a set A ( B (( A), with P (A) > 0. Then show that εB X = 0 a.s. on A.
(i) With C = (X > 0), we have 0 = (A X dP = (D X dP, D = A ( C. So, (D X dP = 0 and X > 0 on D. Show that P (D) = 0 which would be equivalent to saying that X = 0 a.s. On A. Do it by going through the four familiar steps.
(ii) Use the fact that (B X dP = (B εB X dPB for all B ( B, replace B by B ( A, and conclude that IAεB X = 0 a.s. This would imply that εB X = 0 a.s. on A.