Question 8.T.4: (Darboux) Suppose the function f : [a, b] → R is bounded. Gi...

Let ε > 0 be given. From the definition of U (f ) , there is a partition P_{0} = \left\{u_{0}, u_{1}, . . . u_{r}\right\} such that

U (f, P_{0}) < U (f) + ε/2.        (8.9)

Since f is bounded, there is a number K such that |f(x)| ≤ K  for all x ∈ [a, b] . Take δ_{1} = ε/8rK, and suppose P = \left\{x_{0}, x_{1}, x_{2}, …, x_{n}\right\} is any partition satisfying \left\|P\right\| < δ_{1}. Set P_{1} = P_{0} ∪ P and let us calculate U (f, P ) − U (f, P_{1}) by calculating the contribution of each subinterval \left[x_{i}, x_{i+1}\right] to this difference.

If the subinterval does not contain any member of P_{0}, then its contribution to U (f, P ) equals its contribution to U (f, P_{1}) and thereby contributes nothing to the difference. Suppose the subinterval contains an element of P_{0}. Since its contribution to U (f, P ) is at most K (x_{i+1} − x_{i}) , while its contribution to U (f, P_{1}) is at least −K(x_{i+1} − x_{i}), we see that its contribution to the difference does not exceed 2K (x_{i+1} − x_{i}). Hence

U (f, P ) − U (f, P_{1}) ≤ \sum\limits_{i∈J}{2K \left(x_{i+1} − x_{i}\right) ,}

where J = \left\{i : [x_{i}, x_{i+1}] ∩ P_{0} ≠ ∅\right\} .

Since each member of P_{0} can belong to at most two subintervals of P, J cannot have more than 2r members. Consequently

U (f, P ) − U (f, P_{1}) ≤ \sum\limits_{i∈J}{2K \left\|P\right\| ≤ 4Kr \left\|P\right\|< ε/2.}

With U (f, P_{1}) ≤ U (f, P_{0}) , we obtain

U (f, P ) − U (f, P_{0}) < ε/2

and deduce from (8.9) that

U (f, P ) − U (f) = [U (f, P ) − U (f, P_{0})] + [U (f, P_{0}) − U (f )] < ε.

In a similar fashion we can find a δ_{2} > 0 such that

\left\|P\right\| < δ_{2} ⇒ L (f ) − L (f, P ) < ε.

Now the number δ = \min \left\{δ_{1}, δ_{2}\right\} meets the requirements.