Reimann Sum and Induction

Question

let f be a continuous positive valued function defined on an interval [a, b]. Take a ’partition’ of the interval consisting of only the interval [a, b] itself. Let U[a,b] be the upper Riemann sum of f on that partition. Create a refinement of the partition by adding another point c so that the partition is now [a, c, b] and let S[a,c,b] be any Riemann sum on the finer partition. Show that S[a,c,b] ≤ U[a,b] . Show by induction that this holds for any refinement of the partition by adding any number of points to the partition.

Answer 1

The original Riemann sum is f(z)(b -a), where z is between an and b. The Riemann sum of the new partition is f(w)(c - a) + f(r)(b - c), where w is between a and c and r is between c and b. The Upper sum U(a,b) was M(b -a), where M is the maximum value of the function f on [a,b]. All the terms involved are nonnegative and so

f(w) is less than or equal to M as is f(r).

We then have S(a,c,b) = f(w)(c - a) = f(r)(b - c) \s less than or equal to M(c - a) + M(b - c) = M(b -a) = U(a, b).