3.6. Step Functions as Integrators

Proposition 3.2

Suppose \(\alpha\) is constant on \([a, b]\) except possibly at point \(x=a\), that is, \(\alpha(x) = \alpha(b)\) for all \(a < x \leq b\). If \(f\) is continuous from the right at \(a\), then \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), and

\[\int_a^b f \dif\alpha = f(a)[\alpha(b) - \alpha(a)]\]

Note

Note that we assume \(\alpha\) possibly has a different value at \(a\). If \(\alpha\) is constant on the entire interval \([a, b]\), the integral clearly exists and is zero.

An analogous result holds when we assume \(\alpha\) is constant on \([a, b]\) except possibly at the right endpoint \(x=b\).

Proof. If \(\alpha(a) = \alpha(b)\), the conclusion is trivial. In the following proof, we assume \(\alpha(a) \neq \alpha(b)\).

Let \(\varepsilon > 0\) be arbitrary. Because \(f\) is continuous at \(x=a\), there exists \(\delta > 0\) such that

\[x - a < \delta\implies\abs{f(x) - f(a)} < \frac{\varepsilon}{\abs{\alpha(b) - \alpha(a)}}\]

Let \(P_\varepsilon = \{x_0, \ldots, x_n\}\) be a partition of \([a, b]\) such that \(x_1 < x_0 + \delta\). For any refinement \(P \supseteq P_\varepsilon\), we have

\[S(P, T, f, \alpha) = \sum_{k=1}^n f(x_k) \Delta\alpha_k = f(t_1) [\alpha(x_1) - \alpha(x_0)] = f(t_1) [\alpha(b) - \alpha(a)]\]

It then follows that

\[\begin{split}\abs{S(P,T,f,\alpha) - f(a)[\alpha(b) - \alpha(a)]}& = \abs{f(t_1) - f(a)}\abs{\alpha(b) - \alpha(a)}\\& < \frac{\varepsilon}{\abs{\alpha(b) - \alpha(a)}}\abs{\alpha(b) - \alpha(a)}\\& =\varepsilon\end{split}\]

This completes the proof.

Theorem 3.7

Given \(a < c <b\). Let \(\alpha\) be constant on \([a, b]\) except at point \(x=c\). That is, let \(\alpha(a)\), \(\alpha(c)\) and \(\alpha(b)\) be arbitrary. Define

\[\begin{split}\alpha(x) := \begin{cases} \alpha(a), & a \leq x < c \\ \alpha(c), & x=c \\ \alpha(b), & c < x \leq b \end{cases}\end{split}\]

If function \(f\) is defined in such a way that

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

2. at least one of \(f\) and \(\alpha\) is continuous from the right at \(c\),

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

\[\int_a^b f \dif\alpha = f(c)[\alpha(c+) - \alpha(c-)]\]

Proof. It follows from Proposition 3.2 and its analogous result that \(f \in \mathfrak{R}(\alpha)\) on \([a, c]\) and \(f \in \mathfrak{R}\) on \([c, b]\). The values of the integrals are

\[\int_a^c f \dif\alpha = f(c)[\alpha(c) - \alpha(a)]\quad\text{and}\quad\int_c^b f \dif\alpha = f(c)[\alpha(b) - \alpha(c)]\]

Then Theorem 3.3 implies that \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), and

\[\int_a^b f \dif\alpha = \int_a^c f \dif\alpha + \int_c^b f \dif\alpha = f(c)[\alpha(b) - \alpha(a)] = f(c)[\alpha(c+) - \alpha(c-)]\]