$\textbf{Statement of Theorem:}$ Assume that $g$ is a monotone function. Then $f$ is integrable with respect to $g$ iff for every $\epsilon$ there exists a sufficiently fine partition such that $\sum_i (M_i -m_i)(g(x_i)-g(x_{i-1}) < \epsilon$, where $M_i=\sup\{f(x) \mid x \in [x_{i-1},x_i]\}$ and $m_i=\inf\{f(x) \mid x \in [x_{i-1},x_i]\}$.
My class is using lecture notes which are quite lacking in rigor and intuition. I am looking for some explanation regarding the parts of the proof I have indicated with stars.
$\textbf{Proof of Theorem:}$
$(\Longrightarrow)$ Suppose $f$ is integrable with respect to $g$. Fix $\epsilon > 0$ an choose a partition fine enough for that $\epsilon$.
$\star$ $\star$ Does there exists many partitions for which $f$ is integrable with respect to $g$? If so, I am assuming we are choosing a particular one of these many partitions.
Fix selections $\{y_j\}$ and $\{z_j\}$ such that $f(y_j)+\epsilon > M_j$ and $f(z_j)-\epsilon < m_j$.
$\star$ $\star$ So for each subinterval of our particular partition we are just choosing $y_j$ and $z_j$ such that $f(z_j)$ an $f(y_j)$ are very close to $M_j$ an $m_j$.
Then
\begin{multline}
\sum_i ( M_i -m_i) (g(x_i)-g(x_{i-1}) \le \\
\sum_i f(y_i)(g(x_i)-g(x_{i-1})-\sum_i f(z_i)(g(x_i)-g(x_{i-1}) + 2 \epsilon (g(b)-g(a)) \tag1
\end{multline}
Since the first two sums converge to the integral $\int_{a}^{b} f \, dg$, we have the result.
$\star$ $\star$ I can't really prove (1) to myself. First of all, I am not sure why the inequality holds and I don't see how this shows that $\sum_i ( M_i -m_i) (g(x_i)-g(x_{i-1}) < \epsilon$. Also, where does the fact that $g$ is monotone ever come in to play??
Anyways, I would really appreciate a more detailed proof of this theorem. Links are welcome!
Best Answer
Yes, if there exists at least one, then any finer partition will also work. The reason we know that such a partition exists is because we are assuming $f$ is integrable with respect to $g$, which you have presumably defined as there existing a partition such that for all finer partitions, the Riemann-Stieltjes sums are at most $\epsilon$ apart. (Or perhaps you've defined it so that for any points chosen within each partition interval, the resulting Riemann-Stieltjes sums differ by at most $\epsilon$ - either way, we get the same thing).
Yes, that's right.
We'll show this step by step. $$\begin{align} \sum_i ( M_i -m_i) (g(x_i)-g(x_{i-1}) &= \sum_i \color{#bb1f1f}M_i (g(x_i) - g(x_{i-1})) - \sum_i \color{#147360}{m_i} (g(x_i) - g(x_{i-1})) \\ &\leq \sum_i \color{#bb1f1f}{(f(y_i) + \epsilon)}(g(x_i)-g(x_{i-1})) - \sum_i \color{#147360}{(f(z_i) - \epsilon)}(g(x_i)-g(x_{i-1})) \\ &= \sum_i f(y_i) (g(x_i) - g(x_{i-1})) - \sum_i f(z_i) (g(x_i) - g(x_{i-1})) + \\ &\quad +\color{#0a4030}{\sum_i 2\epsilon(g(x_i) - g(x_{i-1}))}, \end{align}$$ where the $\color{#0a4030}{\text{last bit}}$ telescopes, as the $g(x_i)$ cancels with the $-g(x_i)$ of the following term, leaving only $2\epsilon (g(b) - g(a))$.
The $\color{#bb1f1f}{\text{red}}$ and $\color{#147360}{\text{teal}}$ inequalities are almost exactly the inequalities described in the choices of $y_j$ and $z_j$ from the previous step, and the reason we know the direction of the sign overall is because we are (apparently) assuming $g$ is monotone increasing, so that $g(x_i) - g(x_{i-1}) \geq 0$. This last bit is important, as, for example, although $2 < 3$, we don't have that $2(1-2) < 3(1 - 2)$. So that is how we use monotonicity of $g$, and hwo the inequalities follow.
You might ask, what if $g$ is monotone decreasing? First of all, you should then realize that your theorem statement is poorly written, and it almost certainly actually wants $\displaystyle \left\lvert\sum_i (M_i -m_i)(g(x_i)-g(x_{i-1})\right\rvert < \epsilon$. Then the easiest thing to do is to do the proof on $-g$ instead.
So almost everything in this point is covered just above, except how this actually shows the desired theorem.
Since $f$ is integrable with respect to $g$, and we chose an $\epsilon$-partition, we know that both $\displaystyle \sum_i f(y_i)(g(x_i)-g(x_{i-1})$ and $\displaystyle \sum_i f(z_i)(g(x_i)-g(x_{i-1})$ are within $\epsilon$ of the value of the integral $I$. So their difference (triangle inequality!) is at most $2 \epsilon$.
Similarly, $g(b) - g(a)$ is just some number, so when multiplied by $2 \epsilon$, we get some constant times $\epsilon$. So the overall size of the final bit is $k \epsilon$ for some constant $k$ independent of $\epsilon$. This can be made arbitrarily small by choosing smaller initial $\epsilon$ values, which is the theorem.