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$\textbf{Statement of Theorem:}$ 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!

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  • $\begingroup$ You had a period appearing to end a sentence followed by "Where", with a capital initial "W", as if that were the beginning of a new sentence. I changed it. When I've seen this before and asked the person who did it why they wrote it that way, they've said it was unintentional and their editing software did it. But in each of several such instances, the word "Where" began a new line after a line of "displayed" MathJax. This time it was on the same line after "inline" MathJax code. I still don't know what kind of editing software does this. Is that what happened in this case? $\endgroup$ – Michael Hardy May 15 '14 at 15:59
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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.

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).

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 $m_j$.

Yes, that's right.

$$\begin{align} \sum_i ( M_i -m_i) &(g(x_i)-g(x_{i-1}) \le \\ &\sum_i f(y_i)(g(x_i)-g(x_{i-1})-\sum_i f(z_i)(g(x_i)-g(x_{i-1}) + 2 \epsilon (g(b)-g(a)) \end{align}$$

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.

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 $∑_i(M_i−m_i)(g(x_i)−g(x_i−1)<ϵ$. Also, where does the fact that g is monotone ever come in to play?

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.

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Can you show when $g(x) = x$? It uses the exact same argument.

Let us start from the beginning, By definition, $f$ is $R$-$S$ integrable w.r.t. $g$ on $[a,b]$ if for each $\epsilon > 0$ there exists a $\delta >0$ such that whenever $a=x_0<x_1<...<x_n = b$ with $\sup_{1\leq i\leq n} |x_{i}-x_{i-1}| \leq \delta$ and for any tag $x^*_i \in [x_{i},x_{i-1}]$ we have $$\left|\sum_{i=1}^n f(x^*_i)(g(x_i)-g(x_{i-1})) - L\right| \leq \epsilon,$$ we often denote the real number $L =: \int_{[a,b]} f\;dg$.

And here to answer your 3 questions:

  1. Let $\epsilon$ be given (this is not the final $\epsilon$ we are proving in the theorem, you will need to modify it later), we know there exists a partition $P$ such that $$\left|\sum_{i=1}^n f(x^*_i)(g(x_i)-g(x_{i-1})) - L\right| \leq {\epsilon},$$ the above inequality holds for ANY tagged point $x^*_i\in [x_{i},x_{i-1}]$.

  2. Now the idea is pick two sets of tagged points $\{y^*_i\}$ and $\{z^*_i\}$ so that $f(y^*_i)$ is close to $M_i$ and $f(z^*_i)$ is close to $m_i$, say $f(y^*_i) + \epsilon \geq M_i$ and $f(z^*_i) - \epsilon \leq m_i$ (again, the $\epsilon$ is not final). Observe that $$M_i - m_i \leq f(y^*_i) - f(z^*_i) + 2\epsilon. $$

  3. From above inequality, we have $$\sum_{i=1}^n (M_i - m_i) (g(x_i)-g(x_{i-1})) \leq \sum_{i=1}^n (f(y^*_i) - f(z^*_i) + 2\epsilon) (g(x_i)-g(x_{i-1})) $$ rearrange the right hand side, we have $$\sum_{i=1}^n (M_i - m_i) (g(x_i)-g(x_{i-1})) \leq \sum_{i=1}^n f(y^*_i)(g(x_i)-g(x_{i-1})) - \sum_{i=1}^n f(z^*_i) (g(x_i)-g(x_{i-1})) + 2\epsilon (g(b)-g(a)).$$ By triangle inequality, the difference of the two sums on the left hand is small $$\left|\sum_{i=1}^n f(y^*_i)(g(x_i)-g(x_{i-1})) -L + L-\sum_{i=1}^n f(z^*_i) (g(x_i)-g(x_{i-1}))\right| \leq 2\epsilon.$$ We now have $$\sum_{i=1}^n (M_i - m_i) (g(x_i)-g(x_{i-1})) \leq 2\epsilon + 2\epsilon (g(b)-g(a))$$ where we can make the left hand side arbitrarily small.

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