Here is the link to my earlier post here on Math SE on Theorem 6.19 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Theorem 6.19 in Baby Rudin: Do we need the continuity of $\varphi$?
Can we extend this Theorem to vector-valued functions as follows?
Suppose $\phi$ is a strictly increasing continuous function that maps an interval $[A, B]$ onto $[a, b]$. Suppose $\alpha$ is monotonically increasing on $[a, b]$, and suppose $\mathbf{f}$ is a mapping of $[a, b]$ into $\mathbb{R}^k$ such that $\mathbf{f} \in \mathscr{R}(\alpha)$ on $[a, b]$. Define $\beta$ and $\mathbf{g}$ on $[ A , B ]$ by $$ \beta(y) = \alpha \left( \phi(y) \right), \qquad \mathbf{g}(y) = \mathbf{f} \left( \phi(y) \right). $$ Then $\mathbf{g} \in \mathscr{R}(\beta)$ on $[A, B]$, and $$ \int_A^B \mathbf{g} \ \mathrm{d} \beta = \int_a^b \mathbf{f} \ \mathrm{d} \alpha. $$
Proof:
Let $\mathbf{f} \colon= \left( f_1, \ldots, f_k \right)$, where each $f_j$ is a real function defined on $[a, b]$. Then $\mathbf{g} = \left( g_1, \ldots, g_k \right)$, where each $g_j$ is a mapping of $[A, B]$ into $\mathbb{R}^k$ defined by $$ g_j(y) \colon= f_j \left( \phi(y) \right). $$
As $\mathbf{f} \in \mathscr{R}(\alpha)$ on $[a, b]$, so is each $f_j$. Therefore by Theorem 6.19 in Baby Rudin each $g_j \in \mathscr{R}(\beta)$ on $[A, B]$, and $$ \int_A^B g_j \ \mathrm{d} \beta = \int_a^b f_j \ \mathscr{d} \alpha. $$ This implies that $\mathbf{g} \in \mathscr{R}(\beta)$ on $[A, B]$, and $$ \begin{align} \int_A^B \mathbf{g} \ \mathrm{d} \beta &= \left( \int_A^B g_1 \ \mathrm{d} \beta, \ldots, \int_A^B g_k \ \mathrm{d} \beta \right) \qquad \mbox{ [ by Definition 6.23 in Baby Rudin ] } \\ &= \left( \int_a^b f_1 \ \mathrm{d} \alpha, \ldots, \int_a^b f_k \ \mathrm{d} \alpha \right) \\ & \qquad \qquad \mbox{ [ by Theorem 6.19 in Baby Rudin applied to each coordinate ] } \\ &= \int_a^b \mathbf{f} \ \mathrm{d} \alpha, \qquad \mbox{ [ again by Definition 6.23 in Baby Rudin ] } \end{align} $$ as required.
Is there any problem with this theorem?
Is there any problem with this proof thereof?
