3.4. Change of Variables in Riemann-Stieltjes Integrals

Theorem 3.5

Suppose \(f \in \mathfrak{R}(f)\) on \([a, b]\). Let \(g\) be a strictly monotonic and surjective function defined on an interval \(J\) having endpoints \(c\) and \(d\). Suppose also that \(g(c) = a\) and \(g(d) = b\). Let

\[h(x) = f[g(x)]\quad\text{and}\quad\beta(x) = \alpha[g(x)]\quad x \in J \]

Then \(h \in \mathfrak{R}(\beta)\) on \(J\), and \(\int_a^b f \dif \alpha = \int_c^d h \dif \beta\). That is,

\[\int_{a = g(c)}^{b = g(d)} f(t) \dif t = \int_c^d f[g(x)]\dif\alpha[g(x)]\]

Note

Originally in [Apostol, 1974], the condition of function \(g\) is that it is strictly monotonic and continuous. As a matter of fact, these two conditions are equivalent. At the end of the day, we impose these to conditions of \(g\) so that it has an inverse \(g^{-1}\) defined on \([a, b]\). (\(g^{-1}\) is also continuous of course.)

The key of this proof is constructing a one-to-one relation using function \(g\) between a partition \(P\) of \([a, b]\) and a partition \(P'\) of \(J\). In the proof below, we write \(P = g(P^\prime)\) and \(P^\prime = g^{-1}(P)\).

Proof. Without loss of generality, we may assume that \(g\) is strictly increasing. If \(g\) is decreasing, we may easily obtain the same result by applying the linearity of the integrals.

Let \(\varepsilon > 0\) be arbitrary. There exists a partition \(P_\varepsilon\) of \([a, b]\) satisfying the property described in Definition 3.1. As noted previously, we may construct a partition of \([c, d]\), \(P^\prime_\varepsilon = g^{-1}(P_\varepsilon)\). Let \(P^\prime \supseteq P^\prime_\varepsilon\) be any refinement. Similarly, we may construct a partition of \([a, b]\) associated with \(P^\prime\), \(P = g(P^\prime)\).

Note

To exploit the property of \(P_\varepsilon\), we would want to have \(P \supseteq P_\varepsilon\). Luckily, this is indeed true.

Now, we show that \(P \supseteq P_\varepsilon\). For any point \(x \in P_\varepsilon\), we have \(g^{-1}(x) \in P^\prime_\varepsilon \subseteq P^\prime\). Since \(x = g[g^{-1}(x)]\), it then follows that \(x \in g(P^\prime) = P\). This shows \(P \supseteq P_\varepsilon\).

Write \(P^\prime = \{y_0, \ldots, y_n\}\). Suppose \(T^\prime\) is a list of representatives of \(P^\prime\). Let \(T = g(T^\prime)\). Then, of course, \(T\) is a list of representatives of \(P\). This implies that

\[\begin{split}S(P^\prime, T^\prime, h, \beta) & = \sum_{k=1}^n f[g(s_k)]\left\{\alpha[g(y_k)] - \alpha[y_{k-1}]\right\}\\& = \sum_{k=1}^n f(t_k) [\alpha(x_k) - \alpha(x_{k-1}) ]\\& = S(P, T, f, \alpha) \end{split}\]

The remainder of the proof is straightforward and is therefore omitted.