Question: Let $\mathbf{f}$ be a function $[a, b] \rightarrow \mathbb{R}^{k}$, and $\alpha:[a, b] \rightarrow \mathbb{R}$ an increasing function.
- Show that $\mathbf{f}$ is integrable with respect to $\alpha$ if and only if for every $\epsilon > 0$ there exists a partition $P = \{x_{0}, \ldots, x_{n}\}$ of $[a, b]$ such that \begin{equation*} \sum_{i=1}^{n} \operatorname{diam}(\mathbf{f}([x_{i-1}, x_{i}])) \Delta \alpha_{i} < \epsilon \end{equation*} where "$\operatorname{diam}$" denotes the diameter of a set.
Assume $\mathbf{f}$ is indeed integrable with respect to $\alpha$, and let $\epsilon$ be a positive real number, and $P = \{x_{0}, \ldots, x_{n}\}$ a particular partition making the above inequality hold.
- Show that for every choice of points $t_{i} \in [x_{i-1}, x_{i}]$ ($i=1, \ldots, n$), we have \begin{equation*} \bigg|\bigg(\sum_{i=1}^{n} \mathbf{f}(t_{i}) \Delta \alpha_{i}\bigg) - \bigg(\int_{a}^{b} \mathbf{f} d \alpha\bigg)\bigg| < \epsilon \end{equation*}
My problem: I cannot work out the 2nd bullet and I know the inequality in the 2nd bullet holds if the partition is finer and finer, but I have no idea to prove it holds for this given partition.
Define $$M=\sum_{i=1}^n \text{diam}({\bf f}([x_{i-1},x_i]))[\alpha(x_i)-\alpha(x_{i-1})]$$ and note that, by hypothesis, $M<\varepsilon$. Therefore $\varepsilon-M>0$.
Fix $\nu>0$ such that for any partition $\{y_0<\ldots<y_q\}$ of $[a,b]$ with $\max_i y_i-y_{i-1}<\nu$ and any $(s_i)_{i=1}^q\in \prod_{i=1}^q [y_{i-1},y_i]$, $$\Bigl|\sum_{i=1}^q {\bf f}(s_i)[\alpha(y_i)-\alpha(y_{i-1})] - \int_a^b {\bf f}d\alpha\Bigr|<\varepsilon-M.$$ Let $Q=\{y_0<\ldots<y_q\}$ be any partition with $\max_i y_i-y_{i-1}<\nu$. Let $R=Q\cup P$. Write $R=\{z_0<\ldots<z_r\}$ and fix $(s_i)_{i=1}^r\in \prod_{i=1}^r[z_{i-1},z_i]$.
There exist $0=m_0<\ldots<m_n=r$ such that $z_{m_i}=x_i$. We estimate \begin{align*}\Bigl|\sum_{i=1}^n {\bf f}(t_i)[\alpha(x_i)-\alpha(x_{i-1})] - \int_a^b {\bf f}d\alpha\Bigr| & \leqslant \Bigl|\sum_{i=1}^n {\bf f}(t_i)[\alpha(x_i)-\alpha(x_{i-1})]-\sum_{i=1}^r {\bf f}(s_i)[\alpha(z_i)-\alpha(z_{i-1})]\Bigr| \\ & +\Bigl|\sum_{i=1}^r {\bf f}(s_i)[\alpha(z_i)-\alpha(z_{i-1})] - \int_a^b {\bf f}d\alpha\Bigr|. \end{align*} We know the second of these two terms is less than $\varepsilon-M$. We will show that the first term is at most $M$, so the sum is strictly less than $M+\varepsilon-M=\varepsilon$.
To handle the first term, we note that, since $$\alpha(x_p)-\alpha(x_{p-1})=\sum_{i=m_{p-1}+1}^{m_p}\alpha(z_i)-\alpha(z_{i-1}) $$ and $$\sum_{i=1}^r {\bf f}(s_i)[\alpha(z_i)-\alpha(z_{i-1})]=\sum_{p=1}^n \sum_{i=m_{p-1}+1}^{m_p} {\bf f}(s_i)[\alpha(z_i)-\alpha(z_{i-1})],$$
\begin{align*}\Bigl|\sum_{i=1}^n {\bf f}(t_i)[\alpha(x_i)-\alpha(x_{i-1})] -\sum_{i=1}^r {\bf f}(s_i)[\alpha(z_i)-\alpha(z_{i-1})]\Bigr| & = \Bigl|\sum_{p=1}^n\sum_{i=m_{p-1}+1}^{m_p} {\bf f}(t_p)[\alpha(z_i)-\alpha(z_{i-1})] \\ & - \sum_{p=1}^n\sum_{i=m_{p-1}+1}^{m_p} {\bf f}(s_i)[\alpha(z_i)-\alpha(z_{i-1})] \\ & \leqslant \sum_{p=1}^n\sum_{i=m_{p-1}+1}^{m_p}|{\bf f}(t_p)-{\bf f}(s_i)|[\alpha(z_i)-\alpha(z_{i-1})] \\ & \leqslant \sum_{p=1}^n \bigl[\text{diam}({\bf f}([x_{p-1},x_p]))\sum_{i=m_{p-1}+1}^{m_p}[\alpha(z_i)-\alpha(z_{i-1})]\bigr] \\ & = \sum_{p=1}^n \text{diam}({\bf f}([x_{p-1},x_p]))[\alpha(x_p)-\alpha(x_{p-1})]\\ &=M. \end{align*}