Let be a continuous function on the interval and assume
for every function which is continuous on the interval . Prove that for all .
Proof. Since must hold for every function that is continuous on , it must hold for itself (since is continuous on by hypothesis). Therefore we must have,
However, for all . We know from the previous exercise (Section 3.20, #7) that a non-negative function whose integral is zero on an interval must be zero at every point at which it is continuous. By hypothesis is continuous at every point of ; hence, is also continuous at every point of (since the product of continuous functions is continuous). Therefore,