3.7. Reduction of Riemann-Stieltjes Integrals to Finite Sums

Function \(\alpha\) in Theorem 3.7 is a special case of step functions.

Definition 3.2 (Step Functions)

A function defined on \([a, b]\) is called a step function if there is a partition

\[a=x_1 < x_2 < \cdots < x_n = b \]

such that \(\alpha\) is constant on \((x_{k-1}, x_k)\) for \(k=2, \ldots, n\).

The number \(\alpha(x_k+) - \alpha(x_k-)\) is defined as the jump at point \(x_k\) for \(k=2, \ldots, n-1\).

At the left endpoint \(x=a\), the jump is defined as \(\alpha(a+) - \alpha(a)\). Similarly, at the right endpoint \(x=b\), the jump is defined as \(\alpha(b) - \alpha(b-)\).

Note

It is possible that \(\alpha(x_k+) = \alpha(x_k -)\). In this case, the jump at \(x_k\) is zero. But this does not mean that \(\alpha\) is constant on \((x_{k-1}, x_{k+1})\) because we might have \(x_k \neq x_k+\).

Step functions provide the link between the Riemann-Stieltjes integrals and the finite sums of functions.

Theorem 3.8

Let \(\alpha\) be a step function on \([a, b]\). Let \(x_1, \ldots, x_n\) be the same as in Definition 3.2 and \(\alpha_k\) be the jump at \(x_k\).

Let \(f\) be a function defined on \([a, b]\) such that

1. at least one of \(f\) and \(\alpha\) is continuous from the left at \(x_k\), and

2. at least one of \(f\) and \(\alpha\) is continuous from the right at \(x_k\) for \(k=2, \ldots, n-1\). And for \(k=1\) and \(k=n\), at least one of \(f\) and \(\alpha\) is continuous from one side at the endpoint.

Then, \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), and we have

(3.13)\[\int_a^b f \dif\alpha = \sum_{k=1}^n f(x_k) \alpha_k\]

Proof. Consider a partition \(P = \{s_0, s_1, \ldots, s_n\}\) on \([a, b]\) where \(s_k\) satisfies that

\[x_k < s_k < s_{k+1}\quad\forall k = 2, \ldots, n-1 \]

Then, we have \(x_k \in (s_{k-1}, s_k) \; \forall k = 2, \ldots, n-1 \). Note that the condition of Theorem 3.7 is satisfied on each subinterval \([s_{k-1}, s_k], \; k=2, \ldots, n-1\). Therefore, \(f \in \mathfrak{R}(\alpha)\) on \([s_{k-1}, s_k]\), and

(3.14)\[\int_{s_{k-1}}^{s_k} f \dif\alpha = f(x_k) [\alpha(x_k +) - \alpha(x_k -)] = f(x_k) \alpha_k \quad\forall k=2, \ldots, n-1\]

By Proposition 3.2, we know that (3.14) also holds for \(k=1\) and \(k=n\). Then by Theorem 3.3, \(f\) is integrable on the entire interval \([a, b]\). We may then conclude this proof by summing up (3.14) over all \(k\).

Substitute \(\alpha(x) = \floor{x}\) in (3.13), we will obtain a formula for representing any finite sum using an integral.

Theorem 3.9

Given a finite sum \(\sum_{k=1}^n a_k\). We have

(3.15)\[\sum_{k=1}^n a_k = \int_0^n a_{\ceil{x}}\dif\floor{x}\]

where \(a_0\) is an arbitrary constant.

Proof. Let \(\alpha(x) = \floor{x}\) on \([a, b]\). Define a function \(f\) on \([0, n]\) by

\[f(x) = a_{\ceil{x}}, \quad x \in[0, n]\]

Note that \(\alpha\) is continuous from the right at \(x=0, 1, \ldots, n-1\), and \(f\) is continuous from the left at \(x=1, 2, \ldots, n\). Therefore, (3.13) is applicable. It yields that

\[\int_0^n f \dif\alpha = \sum_{k=1}^n f(k) \alpha_k = \sum_{k=1}^n a_k \cdot 1 \]

This completes the proof.

Of course, the construction of \(f\) is not unique. The construction is valid as long as \(f(k) = a_k\) and is continuous from the left at \(x=1, \ldots, n\). One may define \(f\) by applying the linear interpolation (or polynomial interpolation or spline interpolation, etc.) on the data \((1, a_1), \ldots, (n, a_n)\), in which case \(f\) is continuous on the entire interval \([0, n]\). But I prefer the one given in the proof since this makes both floor and ceiling functions appear in (3.15), which makes the formula prettier.