7.1. Riemann Integral

Proposition 7.1.7: Size of Riemann Sums

Suppose P = { x0, x1, x2, ..., xn} is a partition of the closed interval [a, b], f a bounded function defined on that interval. Then we have:
  • The lower sum is increasing with respect to refinements of partitions, i.e. L(f, P') L(f, P) for every refinement P' of the partition P
  • The upper sum is decreasing with respect to refinements of partitions, i.e. U(f, P') U(f,P) for every refinement P' of the partition P
  • L(f, P) R(f, P) U(f, P) for every partition P


The last statement is simple to prove: take any partition P = {x0, x1, ..., xn}. Then
inf{ f(x), xj-1 x xj } f(tj) sup{ f(x), xj-1 x xj }
where tj is an arbitrary number in [xj-1, xj] and j = 1, 2, ..., n. That immediately implies that
L(f, P) R(f, P) U(f, P)

The other statements are somewhat trickier. Let's first find out why they should be true. To make it simple, let's say that P = {a, b} and P' = {a, x0, b}. Then

U(f, P) = sup{ f(x), x [a, b] } × (b - a)
and the upper sum for P' would be
U(f, P') = sup{ f(x), x [a, x0] } × (x0 - a) + sup{ f(x), x [x0, b] } × (b - x0)
Geometrically, the upper sum for P corresponds to one large rectangle, the one for P' to two smaller rectangles, where the smaller rectangles fit into the larger one but do not cover it.

U(f, P) = 1.089

U(f, P') = 0.86
Since the area covered by the two rectangles is smaller than that covered by the first one, we have U(f, P) > U(f, P').

Let's show this mathematically, in case one additional point t0 is added to a particular subinterval [xj-1, xj]. Let:

Then cj Aj and cj Bj so that
cj (xj - xj-1) = cj (xj - t0 + t0 - xj-1) = cj (xj - t0) + cj (t0 - xj-1)
      Bj (xj - t0) + Aj (x0 - tj-1)
That shows that if P = {x0, ... xj-1, xj, ..., xn} and P' = {x0, ... xj-1, t0, xj, ..., xn} then U(f, P) U(f, P').

The proof for a general refinement P' of P uses the same idea plus some confusing indexing scheme. No more details should be necessary.

The proof for the statement regarding the lower sum is analogous.

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