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: |
inf{ f(x), xj-1where tj is an arbitrary number in [xj-1, xj] and j = 1, 2, ..., n. That immediately implies thatx
xj }
f(tj)
sup{ f(x), xj-1
x
xj }
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), xand the upper sum for P' would be[a, b] } × (b - a)
U(f, P') = sup{ f(x), xGeometrically, 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.[a, x0] } × (x0 - a) + sup{ f(x), x
[x0, b] } × (b - x0)
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Let's show this mathematically, in case one additional point t0 is added to a particular subinterval [xj-1, xj]. Let:
Aj
and
cj
Bj
so that
cj (xj - xj-1) = cj (xj - t0 + t0 - xj-1) = cj (xj - t0) + cj (t0 - xj-1)That shows that if P = {x0, ... xj-1, xj, ..., xn} and P' = {x0, ... xj-1, t0, xj, ..., xn} then U(f, P)
Bj (xj - t0) + Aj (x0 - tj-1)
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.