I'm going to include two precursor sections here to introduce all of the content referenced in my actual question (if you're familiar with both continued fractions and the Riemann-Stieltjes integral, scroll down to the Question section).
Precursors
1. Continued Fractions
A continued fraction is a nested fraction of the form $$ \alpha = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \ddots}}} $$ For conciseness, we typically write simply $\alpha = [a_0;a_1,a_2,\ldots]$ (provided the continued fraction corresponding to $\alpha$ is infinite, which is only the case if $\alpha$ is irrational), and, for the sake of the measure-theoretic concerns associated with continued fractions, we also typically confine $\alpha$ to $(0,1)$, so that $a_0 = 0$. Moreover, we will, in the case of this question, at least, restrict the terms $a_1$, $a_2$, and so on, to $\mathbb{N}$ (as an aside, these terms are known as the partial quotients of $\alpha$).
2. The Riemann-Stieltjes Integral
Given an interval $[a, b]\subset\mathbb{R}$, we call $\mathcal{P}$ a partition of $[a,b]$ into $n\in\mathbb{N}$ intervals if $\mathcal{P} = \lbrace x_0, x_1, \ldots, x_n\rbrace$ is such that $x_0 = a$, $x_n = b$, and $x_{i - 1}<x_i$, $\forall i\in\mathbb{N}$ satisfying $i\leq n$. We denote the set of all partitions of $[a,b]$ using $\mathcal{P}([a, b])$. Now, given a function $f:[a, b]\to\mathbb{R}$, we define the Riemann integral of $f$ over $[a, b]\subset\mathbb{R}$ as $$ \int_a^bf(x)~dx = \inf_{\mathcal{P}\in\mathcal{P}([a, b])}\mathcal{U}(f, \mathcal{P}) = \sup_{\mathcal{P}\in\mathcal{P}([a, b])}\mathcal{L}(f, \mathcal{P}) $$ where $$ \mathcal{U}(f, \mathcal{P}) = \sum_{i = 1}^n(x_i - x_{i - 1})\sup_{x\in[x_{i - 1},x_i]}f(x)~~~\mathrm{and}~~~\mathcal{L}(f, \mathcal{P}) = \sum_{i = 1}^n(x_i - x_{i - 1})\inf_{x\in[x_{i - 1},x_i]}f(x) $$ provided these supremums/infimums exist (which is not always the case, since we can construct non-integrable functions). On the other hand, the Riemann-Stieltjes function is not defined in terms of sums of the form $$ \sum_{i = 1}^n(x_{i} - x_{i - 1})f(c_i) $$ for some $c_i\in[x_{i - 1},x_i]$, but rather Riemann-Stieltjes integrals, for a function $g:\mathbb{R}\to\mathbb{R}$ called an integrator, are defined in terms of sums of the form $$ \sum_{i = 1}^n(g(x_i) - g(x_{i - 1}))f(c_i) $$ for some $c_i\in[x_{i - 1}, x_i]$.
Question
We have the following theoremSuppose that, for some $f:\mathbb{N}\to\lbrace x\in\mathbb{R}:x>0\rbrace$, $\exists C,\delta>0$ such that $f(r)<Cr^{1/2 + \delta}$, $\forall r\in\mathbb{N}$. Then, for almost all $\alpha = [0;a_1, a_2,\ldots]\in(0, 1)$ $$ \lim_{N\to\infty}\frac{1}{N}\sum_{k = 1}^Nf(a_k) = \sum_{r = 1}^{\infty}\frac{f(r)}{\ln 2}\ln\bigg{(}1 + \frac{1}{r(r + 2)}\bigg{)} $$
Now, the question, finally, is: is there any connection between this result and the notion of the Riemann-Stieltjes integral? The closest I could come to any reasonable interpretation of the above result in terms of Riemann-Stieltjes integrals is that you can rewrite the logarithmic terms on the RHS of the above equation in the form $$ \frac{1}{\ln 2}\ln\bigg{(}\frac{r(r + 2) + 1}{r(r + 2)}\bigg{)} = \frac{\ln(r(r + 2) + 1) - \ln(r(r + 2))}{\ln 2} = \frac{\ln(r^2 + 2r + 1)}{\ln 2} - \frac{\ln(r^2 + 2r)}{\ln 2} $$ which would lead me to believe that, on the RHS, at least, we have an obvious integrator function, namely $$ g(x) = \frac{\ln x}{\ln 2} $$ Now, given the form of the inputs to the integrator function, we know also that we are integrating over the region $$ I = \bigcup_{r = 1}^{\infty}[r^2 + 2r, r^2 + 2r + 1] $$ meaning this is an improper integral. Finally, we require the argument of $f$ to be confined to $[r^2 + 2r, r^2 + 2r + 1]$, $\forall r\in\mathbb{N}$. In the original statement of the equation in question, this is clearly not the case, but we can simply suppose that $f$ is composed with a function, say $\varphi:\mathbb{R}\to\mathbb{N}$, which satisfies $\varphi:c_r\mapsto r$, $\forall c_r\in[r^2 + 2r, r^2 + 2r + 1]$, $\forall r\in\mathbb{N}$. With all of this in mind, we might therefore be tempted to believe that the RHS of the given equation is in some way related to the Riemann-Stieltjes integral $$ \int_{I}(f\circ\varphi)(x)~d\frac{\ln x}{\ln 2} = \sum_{r = 1}^{\infty}\int_{r^2 + 2r}^{r^2 + 2r + 1}(f\circ\varphi)(x)~d\frac{\ln x}{\ln 2} $$ Although, given my very limited experience with the Riemann-Stieltjes integral, I really have no clue whether I'm on the right track here. Any pointers in a more productive direction here would be greatly appreciated.
Let $\lfloor z \rfloor$ be the integral and $\{z\}$ be the fractional part of $z$, and suppose that for $\alpha = [0;a_1,a_2,\ldots]$ and nonnegative integers $j$ you define $\alpha_{(j)}:=[0;a_{j+1},a_{j+2},\ldots]$. Then $\alpha_{(j)}$ is gotten from $\alpha$ by $j$ applications of the map $\phi: z \mapsto \{z^{-1}\}$, and $g_0(z):=(\log \, (z+1))/\log 2$ is an integrator function which is invariant for $\phi$ in the sense that $$\int_0^1 h(z)\, dg_0(z)=\int_0^1 h(\phi(z)) \, dg_0(z)$$ for all continuous functions $h$ on $[0,1]$; also, $\int_0^1 dg_0(z) = 1$. Because of this, it's possible to use ergodic theory to prove (On the ergodic theorems (II) (Ergodic theory of continued fractions), C. Ryll-Nardzewski, Studia Mathematica 12 (1951), 74-79, Theorem 3) that, provided that $h$ is a function on $[0,1]$ such that $\int_0^1 h(z)\, dg_0(z)$ exists, then for almost all $\alpha$ in $[0,1]$, \begin{eqnarray*} \lim_{N\to\infty}\frac{1}{N}\sum_{0\le k\le N-1} h(\phi(\phi(\cdots(\alpha)\cdots))) &=&\int_0^1 h(z) \,dg_0(z), \\ \qquad \qquad \hbox{where there are $k$ applications of $\phi$;} \end{eqnarray*} the left-hand side of this can be rewritten as $$ \lim_{N\to\infty}\frac{1}{N}\sum_{0\le k\le N-1} h(\alpha_{(k)}). $$ In fact, if $h$ is nonnegative, it's enough to assume that the integral exists (or is infinite) as a Riemann-Stieltjes integral which is improper at 0, 1, or both.
The result you want now follows if you set $h(z):=f(\lfloor z^{-1} \rfloor)$, since $\lfloor \alpha_{(k)}^{-1} \rfloor = a_{k+1}$ and, allowing the integral to be improper at 0, \begin{eqnarray*} \int_0^1 f(\lfloor z^{-1} \rfloor) \,dg_0(z)&=& \sum_{n\ge 1} \int_{1/(n+1)}^{1/n} f(n) \,dg_0(z)\\ &=& \sum_{n\ge 1} f(n)\left(g_0(\frac 1 n)-g_0(\frac 1 {n+1})\right)\\ &=& \sum_{n\ge 1} \frac{f(n)}{\log 2}\left(\log(1 + \frac 1 n)-\log(1+\frac 1 {n+1})\right)\\ &=& \sum_{n\ge 1} \frac{f(n)}{\log 2}\log \frac {(n+1)^2}{n(n+2)}\\ &=& \sum_{n\ge 1} \frac{f(n)}{\log 2}\log \left(1 + \frac {1}{n(n+2)}\right). \end{eqnarray*} If $f$ is nonnegative it's not necessary to have a growth condition on $f$. If you like, you can write this as an improper Riemann-Stieltjes integral of $f$ by using the integrator function $$g_1(z):=1-\frac1 {\log 2} \log\left(1+\frac 1 {\lfloor z \rfloor + 1}\right).$$