3.5. Reduction to Riemann Integrals

The next theorem tells us that we may replace the symbol \(\dif \alpha\) with \(\alpha^\prime(x) \dif x\) under some conditions.

Theorem 3.6

Suppose \(f \in \mathfrak{R}(\alpha)\) on \([a, b]\), and \(\alpha\) has a continuous derivative on \([a, b]\). Then \(f \alpha^\prime \in \mathfrak{R}\) on \([a, b]\), and

\[\int_a^b f(x) \dif\alpha(x) = \int_a^b f(x) \alpha^\prime(x) \dif x \]

Proof. First, suppose \(f\) is bounded by \(M > 0\), i.e.,

(3.9)\[\abs{f(x)}\leq M \quad\forall x \in[a, b]\]

Let \(\varepsilon > 0\) be arbitrary.

Because \(\alpha^\prime\) is continuous on \([a, b]\), it is continuous uniformly there. There exists \(\delta > 0\) such that

(3.10)\[\abs{s - t}\implies\abs{\alpha^\prime(s) - \alpha^\prime(t)} < \frac{\varepsilon}{2M(b-a)}\]

Since \(f\) is integrable w.r.t. \(\alpha\) on \([a, b]\), there exists a partition \(P_\varepsilon\) of \([a, b]\) such that for any refinement \(P\) of \(P_\varepsilon\), and any list of representatives \(T\) of \(P\), we have

(3.11)\[\abs{ S(P, T, f, \alpha) - \int_a^b f \dif \alpha} < \varepsilon / 2\]

Then, we can find a finer partition \(P^\prime_\varepsilon \supseteq P_\varepsilon\) such that \(\norm{P^\prime_\varepsilon} < \delta\).

Let \(P \supseteq P^\prime_\varepsilon\) be a refinement such that and \(T\) be a list of representatives of \(P\). Note that \(P\) is of course also a refinement of \(P_\varepsilon\). Applying the mean value theorem, we have

\[\begin{split}S(P,T,f,\alpha) & = \sum_{k=1}^n f(t_k) [\alpha(t_k) - \alpha(t_{k-1}) ]\nonumber\\& = \sum_{k=1}^n f(t_k) \alpha^\prime(s_k) \Delta x_k \end{split}\]

where each \(s_k \in (x_{k-1}, x_k)\).

Taking the difference of \(S(P,T,f \alpha^\prime, x)\) and \(S(P,T,f,\alpha)\), we have

\[\begin{split}\abs{S(P,T,f \alpha^\prime, x) - S(P,T,f,\alpha)}& = \abs{\sum_{k=1}^n f(t_k) [\alpha^\prime(t_k) - \alpha^\prime(s_k)] \Delta x_k}\\& \leq\sum_{k=1}^n \abs{ f(t_k) [\alpha^\prime(t_k) - \alpha^\prime(s_k)] \Delta x_k }\\& = \sum_{k=1}^n \abs{ f(t_k)}\abs{ \alpha^\prime(t_k) - \alpha^\prime(s_k) }\Delta x_k \end{split}\]

Then applying (3.9) and (3.10), the above difference is further bounded by

(3.12)\[\abs{S(P,T,f \alpha^\prime, x) - S(P,T,f,\alpha)} < M \frac{\varepsilon}{2M(b-a)}\sum_{k=1}^n \Delta x_k = \varepsilon / 2\]

Recall \(P \supseteq P_\varepsilon\). Then we may conclude this proof by comparing (3.11) and (3.12).