Theorem. Suppose that f and a are both continuous and non-decreasing, then

Proof. First we observe both integrals exist by our previous results. Let P be a partition

and set x_i = x_i-1,  x¢_i = x_i, then

This follows from the ``summation by parts'' equation

We set u_j = f(x_j) and v_j = a(x_j) and chose a partition P so that both the left hand side is arbitrarily close to ò_a^b f  da and the Riemann-Stieltjes sum on the right hand side is arbitrarily close to ò_a^b adf.   ^[¯]

Theorem.  If a is monotone increasing on [a,b] and f is bounded, then f is Riemann-Stieltjes integrable with respect to a on[a,b] if and only if f a¢ is Riemann integrable. In this case,

Pf. Suppose e > 0 is arbitrary. Since a¢ is Riemann integrable, pick a partition P such that U(P,a¢)-L(P,a¢) < e. For each subinterval I_i: = [x_i,x_i-1], by the Mean Value Theorem there is a t_i in the interval so that Da_i = a¢(t_i) Dx_i. If s_i is arbitrary in I_i, then

By varying s_i, this implies

Using the same inequality, this estimate also holds for U(P;f a¢) - U(P;f,a). A similar argument establishes the corresponding estimate for lower sums.   ^[¯]

Theorem.  Suppose that f has range [m,M] and f is continuous on [m,M]. If f is Riemann-Stieltjes integrable with respect to a, then F = f_°f is Riemann-Stiletjes integrable with respect to a.

Pf. Let e > 0 be arbitary. Since f is uniformly continuous, there is a d > 0 (which we may as well assume is smaller than e) such that |f(u)-f(v)|<e if |u-v| < d. For this d pick a partition P so that

Let M_j,  m_j denote sup and inf of f respectively over the subinterval [x_j-1,x_j]. Similarly, define M_j^*,  m_j^* as the sup and inf of F over same subinterval. Let G be defined as the index set of ``good'' intervals where M_j - m_j < d and B as the remainder where M_j - m_j ³ d. For j Î G we have M_j^*-m_j^* < e, while for j Î B there holds å_{j Î B} Da_j < d. The last estimate follows since d å_{j Î B} Da_j £ å_{j Î B} (M_i-m_i) Da_j £ d². Now we can estimate the difference between the upper and lower Riemann-Stieltjes sums of F:

Corollary.  If f is Riemann-Stieltjes integrable with respect to a, then so is f². Further, if g is also Riemann-Stieltjes integrable with respect to a, then the product f g is as well.

Pf. Apply the above theorem with f(y) = y² to establish the first statement. To establish the second, use the identity

and note that each term of the sum on the right hand side is Riemann-Stieltjes integrable with respect to a.   ^[¯]

Corollary.  If f is Riemann-Stieltjes integrable with respect to a, then so is |f|.

Pf. Apply the above theorem with f(y) = |y|.   ^[¯]

Defn.  A function g is said to be of  bounded variation  if for any partition P of the interval [a,b],

is finite where t(P,g) : = å_{j = 1}ⁿ |Dg_j|.

Theorem. A function g is of bounded variation if and only if it can be decomposed as a difference (g = b-a) of two monotone nondecreasing functions.

Pf. First define u⁺ = max(u,0) and u^- = min(u,0) , then u = u⁺ +u^- and |u| = u⁺ -u^-. For a partition P of [a,x] define

p(P,x) = å_{j = 1}ⁿ |Dg_j|⁺

and

n(P,x) = å_{j = 1}ⁿ |Dg_j|^-.

It is clear that b(x) : = sup_{partitions P of [a,x]} p(P,x) and a(x) = sup_{partitions P of [a,x]} -n(P,x) are nondecreasing functions. Finish by showing that g = b-a.   ^[¯]

Note. Using this decomposition, one may now extend both the definition of the Riemann-Stieltjes integral and its properties from g monotone nondecreasing to g being a function of bounded variation:

All the properties given in the previous lectures have their obvious analogues. We may also easily extend to the case of complex and vector valued functions with corresponding results.