In the previous section, we saw that it was possible for a to be discontinuous but for the Reiemann-Stieltjes integral of f to still exist. The following example shows that the integral may not exist however, if both f and a are discontinuous at a point.

Example.  Let f = a where f(x) is one for nonnegative x and zero otherwise. In this case, if P is any partition, U(P;f,a) = 1, while L(P;f,a) = 0. This shows that the Riemann-Stieltjes integral for this pair does not exist.

Theorem.  A necessary and sufficient condition for f to be Riemann-Stieltjes integrable with respect to a is for each given e > 0, that one can obtain a partition P of [a,b] such that

(*)

Pf.  First we show that (*) is a sufficient condition. This follows immediately, since for each e > 0 that there is a partition P such that (*) holds,

(U) ó õ b a  f(x) d a(x)  -  (L) ó õ b a  f(x)  d a(x)  £  U(P;f,a) - L(P;f,a)  <  e.

Since e > 0 was arbitrary, then the upper and lower Riemann-Stieltjes integrals of f must coincide.

To prove that (*) is a necessary condition for f to be Riemann integrable, we let e > 0. By the definition of the upper Riemann-Stieltjes integral as a infimum of upper sums, we can find a partition P₁ of [a,b] such that

Similarly, we have

Let P be a common refinement of P₁ and P₂, then subtracting the two previous inequalities implies,

Theorem.  If f is continuous on [a,b], then f is Riemann-Stieltjes integrable with respect to a on[a,b].

Pf. We use the condition (*) to establish the proof. If e > 0, we set e₀ : = e/(1+a(b)-a(a)). Since f is continuous on [a,b], f is uniformly continuous. Hence there is a d > 0 such that |f(y)-f(x)| < e₀ if |y-x| < d. Suppose that ||P || < d, then it follows that |M_i - m_i| < e₀   (1 £ i £ n). Hence

Theorem.  If f is monotone and a is continuous on [a,b], then f is Riemann-Stieltjes integrable with respect to a on [a,b].

Pf.  We prove the case for f monotone increasing and note that the case for monotone decreasing is similiar. We again use the condition (*) to prove the theorem. If e > 0, we set e₀ : = e/(1+f(b)-f(a)), Since a is continuous and [a,b] is compact, a is uniformly continuous. So for e₀ we can determine a d > 0, so that if P is a partition with ||P|| < d, then Da_i < e₀  (all i). The function f is monotone increasing on [a,b], so M_i = f(x_i) and m_i = f(x_i-1). Hence

Defn.  A Riemann-Stieltjes sum for f with respect to a for a partition P of an interval [a,b] is defined by

where the x_j, satisfying x_j-1 £ x_j £ x_j  (1 £ j £ n), are arbitrary.

Corollary.  Suppose that f is Riemann-Stieltjes integrable on [a,b], then there is a unique number g ( = ò_a ^b f  da) such that for every e > 0 there exists a partition P of [a,b] such that if P £ P₁,P₂, then

where R(P₁,x) is any Riemann-Stieltjes sum of f with respect to a for the partition P₁. In this case, we can interpret the integral as

although a careful proof is somewhat involved.

Pf. Since L(P₂;f,a) £ g £ U(P₁;f,a) for all partitions, we see that parts i.) and ii.) follow from the definition of the integral. To see part iii.), we observe that m_j £ f(x_j) £ M_j and hence that

But we also know that both

and condition (*) hold, from which part iii.) follows.   ^[¯]