I am beginning to read the text Arithmetic Tales by Olivier Bordellès. Thus-far I am just going through the preliminary material, though I have already stumbled upon a result I would like some advice on. The result is as follows:
Suppose that $a, b\in \mathbb R$ such that $ a < b$, and $f,g: I \to \mathbb R$ are (not necessarily continuous) functions of sets with no common discontinuities for $I := [a,b]$. Furthermore, suppose that $\lambda, \mu \in \mathbb R_{>0}$ are such that $\lambda^{-1} + \mu^{-1} > 1$ and $f \in BV^\lambda(I)$, $g \in BV^\mu(I)$. Then:
$f$ is Riemann–Stieltjes integrable w.r.t. $g$. (i.e. $\int_I f \ \mathrm d \ g$ exists)
for any $\xi \in I$ one has the following inequality:
$\begin{align}\left |\int_I f \ \mathrm d \ g - f(\xi)\left(g(b) - g(a)\right)\right| \leq 2\left(1 + \frac{\zeta}{\lambda + \mu}\right) \cdot Var^\lambda(f)^{-\lambda} \cdot Var^\mu(g)^{-\mu} && (1)\end{align}$
In equation (1), the symbol $\zeta$ is undefined and is presumably a misprint of $\xi$, though I cannot confirm. My question chiefly regards proving part 1. of the theorem. I believe, I can mostly construct a proof, but it contains at least one error. Also, my proof relies on using the Hölder inequality with bounded variation in a way in which I am not used to and I fear I may have made a mistake in my bounding. Additionally I cannot prove the second statement thus far.
Because of this I would like someone therefore to look over my proof, and see if they can fix it Additionally I would like to know about the bound (1) if anyone is willing to provide reference to the bound for part 2., and possibly provide a definition for the missing symbol or a reference to a proof I would be very thankful.
Recap of Definitions
A tagged partition $\mathcal P := \left(\left[x_{i - 1}, x_{i}\right]_{i = 1}^n, \left(t\right)_{i = 1}^n\right)$ is a partition of $I$ by $\left(x_i\right)_{i = 0}^n$ (i.e. $a = x_0$, $x_n = b$, and $x_{i - 1} < x_{i}$ for $i \in [1, n]_{\mathbb N}$) such that for all $i \in [0, n]_{\mathbb N}: x_{i - 1} \leq t_i \leq x_{x_i}$. for a $\delta: I \to \mathbb R_{> 0}$ we say that the tagged partition $\mathcal P$ is $\delta$-fine if for all $i \in [1,n]_{\mathbb N}$ that $x_{i},x_{i - 1} \in B_{\delta(t_i)}(t_i)$. For any $\mathcal P$ we have that $\Sigma(f,g;\mathcal P) = \sum_{i = 1}^nf(t_i)\left(g(x_i) - g(x_{i - 1})\right)$ for $f,g: I \to \mathbb R$.
For $\lambda \in \mathbb R^+$, a function $f: I \to \mathbb R$ is of $\lambda$-bounded variation, $f \in BV^\lambda(I)$, if $\sum_{i = 1}\left|f(x_i)^\lambda - f(x_{i - 1})^\lambda\right|$ is bounded above with a single bound for all partitions $x$ of $I$ and further we let $Var^{\lambda}(f)$ be the corresponding supremum.
Proof of 1.:
We adapt the Lemma H.2 of A Modern Theory of Integration by Robert G.Bartle (and thus we use the terminology of gauges in the proof though it may be omitted). Suppose that $a$, $b$ $f$ and $g$ are as originally stated. Further suppose that $\lambda, \mu \in \mathbb R^+$ satisfy $\lambda^{-1} + \mu^{-1} > 1$ and $f \in BV^\lambda(I)$, $g \in BV^\mu(I)$. As $\lambda^{-1} + \mu^{-1} > 1$ we may find some $\bar{\lambda} > \lambda$ and $\bar{\mu} > \mu$ satisfying $\bar{\lambda}^{-1} + \bar{\mu}^{-1} = 1$. Also pick any $\alpha \in (0,1-\frac{\lambda}{\lambda^\prime})$ and $\beta \in (0, \frac{\lambda}{\bar{\lambda}})$ and note that $\lambda^\prime := \bar \lambda(1 - \alpha) > \lambda$ and likewise $\mu^\prime := \bar\mu(1- \beta) >\mu$.
Note then that $f \in BV^{\lambda^\prime}(I)$ and $g \in BV^{\mu^\prime}(I)$. Suppose that $\varepsilon \in \mathbb R^+$. By continuity argument we find a $\bar{\varepsilon} \in \mathbb R^+$ such that $\bar{\varepsilon}^{\alpha} \cdot Var^{\lambda^\prime}(f)^{\frac{1}{\bar \lambda}} \cdot Var^{\bar{\mu}}(g)^{\frac{1}{\bar{\mu}}} + Var^{\bar \lambda}(f)^{\frac{1}{\bar \lambda}} \cdot Var^{{\mu^\prime}}(g)^{\frac{1}{\bar{\mu}}} \cdot \bar{\varepsilon}^{\beta} \leq \varepsilon$.
Let $\Phi$ be the set-valued predicate given by the formula below: $\begin{equation}\Phi\left(U\right) := \left[ \forall q,q^\prime \in U: f(q) - f(q^\prime) < \bar\epsilon\right] \lor \left[ \forall q,q^\prime \in U: g(q) - g(q^\prime) < \bar\epsilon\right].\end{equation}$
As $f$ and $g$ share no common points of discontinuity so the sentence $\phi := \forall p \in I\ \exists \delta \in \mathbb R^+: p \in B_{\delta}(p) \land\Phi(I \cap B_\delta(p))$ holds. We may then (and with due care without choice) produce a function $\delta: I \to \mathbb >R^+$ such that for all $p \in I$ $p$ and $\delta(p)$ satisfy the above sentence. Any such function may be used as a gauge in the rest of the proof. However, utilising compactness, $\delta$ may be refined so that it is a constant function. In particular let $\mathcal C$ be the set of all open nonempty open subsets $U$ of $I$ such that $\Phi(U)$ holds. By the fact that $\phi$ holds it follows that $C$ is an open cover of $I$. As $I$ is compact so $C$ has a finite subcover $\mathcal D \subset C$. Set then $\bar \delta := \min_{U \in D}\text{ diam} U$. $\bar{\delta} > 0 $ as it is the minimum of a finite set of strictly positive values (an open set has nonzero diameter). Set then $\delta: I \to \mathbb R_{> 0 } \ x \mapsto \bar{\delta} $ Note then for any $p \in I$ and $q,q^\prime \in I$ such that $p - \delta(p) < q^\prime \leq p \leq q < p + \delta(p)$. Then, one has that $p \in U$ for some $U \in D$ and further that $q,q^\prime \in U$ as $p \in U$ and the diameter of $U$ is at least $\delta(p)$ Thus $\left|f(q^\prime) - f(q)\right| < \epsilon$ by $\Phi(U)$.
Now that $\delta$ has been well chosen, suppose that $\mathcal P := \left(\left[x_{i -1}, x_{i}\right]_{i = 1}^n, \left(t\right)_{i = 1}^n\right)$ and $\mathcal Q:= \left(\left[y_{i - 1}, y_{i}\right]_{i = 1}^n, \left(s\right)_{i = 1}^m\right)$ are two $\delta $-fine tagged partitions of $I$ and consider the union tagged partition $\mathcal R = \left(\left[z_i,z_{i - 1}\right]_{i = 1}^l,(r)_{i = 1}^{l}\right)$ Set $\bar{t}: [1,l]_{\mathbb N} \to I$ such that $\bar{t_i} = t_j$ for $j \in [1,n]_{\mathbb N}$ minimal such that $z_i \leq t_j$. It follows then that $\Sigma(f,g,\mathcal P) = \sum_{i = 1}f(\bar{t}_i)\left(g(z_i) - g(z_{i - 1})\right)$. Likewise, set $\bar{s}: [1,l]_{\mathbb N} \to I$ such that $\bar{s_i} = s_j$ for $j \in [1,n]_{\mathbb N}$ minimal such that $z_i \leq s_j$. Then $\Sigma(f,g,\mathcal P) = \sum_{i = 1}f(\bar{t}_i)\left(g(z_i) - g(z_{i - 1})\right)$. It follows by a case analysis that for all $i \in [1,l]_{\mathbb N}$ that either $\left|f(\bar{t}_i) - f(\bar{s}_{i})\right| < \bar{\varepsilon}$ or $\left|g(z_i) - g(z_{i - 1})\right| < \bar \varepsilon$.
Let $\bar I := \left\{k \in [1,l]_{\mathbb N} \ | \ \left|f(\bar{t_i}) - f(\bar{s_i})\right| \leq \bar{\varepsilon}\right\}$ and $ \bar J := [1,l]_{\mathbb N} \setminus \bar I$. Note that $J \subseteq \left\{k \in [1,l]_{\mathbb N} \ | \ \left|g(z_i) - g(z_i)\right| \leq \bar{\varepsilon}\right\}$. Set then $\Delta := \Sigma\left(f,g;\mathcal P\right) - \Sigma\left(f,g;\mathcal Q\right)$. Observe then that: $ \begin{align} \Delta &= \sum_{i = 1}^l \left(f(\bar{t}_i) - f(\bar{s}_{i})\right) \cdot \left(g(z_i) - g(z_{i - 1})\right)\\ &= \sum_{i \in I} \left(f(\bar{t}_i) - f(\bar{s}_{i})\right) \cdot \left(g(z_i) - g(z_{i - 1})\right) + \sum_{j \in J} \left(f(\bar{t}_j) - f(\bar{s}_{j})\right) \cdot \left(g(z_j) - g(z_{j - 1})\right)\\ &= \sum_{i \in I} \left(f(\bar{t}_i) - f(\bar{s}_{i})\right)^{\alpha} \cdot \left(f(\bar{t}_i) - f(\bar{s}_{i})\right)^{1- \alpha} \cdot \left(g(z_i) - g(z_{i - 1})\right) + \sum_{j \in J} \left(f(\bar{t}_j) - f(\bar{s}_{j})\right) \cdot \left(g(z_j) - g(z_{j - 1})\right)^{1 - \beta} \cdot \left(g(z_j) - g(z_{j - 1})\right)^{\beta}. \end{align} $
By the triangle inequality, the Hölder inequality for sums and our choice of gauge observe that: \begin{align} \left|\Delta\right| &\leq \bar{\varepsilon}^{\alpha} \cdot \sum_{i \in \bar I} \left|f(\bar{t}_i) - f(\bar{s}_{i})\right|^{1 - \alpha} \cdot \left|g(z_i) - g(z_{i - 1})\right| + \sum_{j \in \bar J} \left|f(\bar{t}_j) - f(\bar{s}_{j})\right| \cdot \left|g(z_j) - g(z_{j - 1})\right|^{1 - \beta} \cdot \bar{\varepsilon}^{\beta}\\ &\leq \bar{\varepsilon}^{\alpha} \cdot \left[\sum_{i \in \bar I}\left|f(\bar{t}_i) - f(\bar{s}_{i})\right|^{\lambda^\prime}\right]^{\frac 1 {\bar\lambda}} \cdot \left[\sum_{j \in \bar J}\left|g(z_j) - g(z_{j - 1})\right|^{\bar{\mu}}\right]^{\frac 1{\bar \mu}} + \left[\sum_{i \in I}\left|f(\bar{t}_i) - f(\bar{s}_{i})\right|^{\bar \lambda}\right]^{\frac 1 {\bar\lambda}} \cdot \left[\sum_{j \in \bar J}\left|g(z_j) - g(z_{j - 1})\right|^{{\mu^\prime}}\right]^{\frac 1{\bar \mu}}\cdot \bar{\varepsilon}^{\beta}\\ &\leq \bar{\varepsilon}^{\alpha} \cdot Var^{\lambda^\prime}(f)^{\frac{1}{\bar \lambda}} \cdot Var^{\bar{\mu}}(g)^{\frac{1}{\bar{\mu}}} + Var^{\bar \lambda}(f)^{\frac{1}{\bar \lambda}} \cdot Var^{{\mu^\prime}}(g)^{\frac{1}{\bar{\mu}}} \cdot \bar{\varepsilon}^{\beta}\\ &\leq \varepsilon \end{align}
Observe we have found for any $\varepsilon \in \mathbb R^+$ a (constant) gauge $\delta$ that for any two $\delta$-fine partitions have $\epsilon$-close partial RS sums. There-fore, by the Cauchy criterion (every Cauchy net converges) we have that $f$ is indeed $g$-RS-gauge integrable. Furthermore because we may chose for every $\varepsilon$ an appropriate constant gauge then we can conclude straight RS integrability. completing the proof.
Error:
One error in my proof is that the sets of the form $(\bar{t}_i,\bar{s}_i)\cup (\bar{s}_i,\bar{t}_i)$ are not disjoint for $i \in 1,\ldots,l$, so I cannot safely bound $\sum_{i = 1}^l\left|f({\bar t}_i) - f({\bar s}_i)\right|^{\lambda} \leq Var^{\lambda}(f)$. My current thought on rectifying this is appealing to some kind of upper and lower sum notion of integral convergence and then arguing similarly as to before. My first thoughts is that this should work, and most likely even simplify the proof. However I will have to see as I write up an answer, and in any case the proof will be made a little more tricky due to the fact that one is dealing with RS-sums instead of Riemann sums so some alterations would need to be made. Further commentary or answers from the community in the meantime however would always be appreciated, especially if there are other errors that I am not aware of!