Almost equivalent definitions of the Riemann–Stieltjes integral Below, I will present two definitions of the Riemann–Stieltjes integral, the second of which is more general. My question concerns the relationship between these two definitions. 
Definition 1: Let $f,g:[a,b] \to \mathbb{R}$. For a partition $P=\{x_0, x_1,x_2 \cdots x_{n-1},x_n\}$ of $[a,b]$, consider the sum $$S(P,f,g)  \stackrel{\rm def}{=} \sum_{i=0}^{n-1} f(c_i) \left[g(x_{i+1}) - g(x_{i})\right]$$ where we have "sample points" $c_i \in [x_i, x_{i+1}]$.
$f$ is then said to be Riemann–Stieltjes integrable with respect to $g$ if there is a real number $L$ with the following property: for all $\epsilon>0$ there is a $\delta>0$ such that for any partition $P$ with $\text{sup}_{{0\leq i \leq n-1}}(x_{i+1} - x_i) < \delta$ and any sequence of points $\{c_i\}_{{0\leq i \leq n-1}, c_i \in [x_i, x_{i+1}]}$ we have 
$$\left|S(P,f,g) -
 L\right| < \epsilon$$
Definition 2: We modify the above definition so that it is like this instead: for all $\epsilon>0$ there is partition $P_{\epsilon}$ such that any refinement $P' \supset P_{\epsilon}$ satisfies $$\left|S(P',f,g) -
 L\right| < \epsilon$$ independent of the sequence of points $\{c_i\}_{{0\leq i \leq n-1}, c_i \in [x_i, x_{i+1}]}$ we choose. 
Remark: The first definition implies the second. Simply let $P_{\epsilon}$ be any partition with $\text{sup}_{{0\leq i \leq n-1}}(x_{i+1} - x_i) < \delta$. However, interestingly, the second definition does not imply the first. Take 
$$g(x) =
\begin{cases}
0  & x \in [0, \frac 12) \\
1, & x \in [\frac 12, 1]
\end{cases}$$
$$f(x) =
\begin{cases}
0  & x \in [0, \frac 12] \\
1, & x \in (\frac 12, 1]
\end{cases}$$
as a counterexample. For this example, the integral exists and is equal to $0$ in the sense of the second definition by ensuring our chosen partition $P{_\epsilon}$ is such that $\frac 12 \in P_{\epsilon}$. This ensures $g(x_{i+1}) - g(x_i) = 0$ except in the interval $[x_k, \frac 12]$; however, this interval does not affect the sum since $f \equiv 0$ in $[x_k, \frac 12]$. 
Conversely, for the first definition we needn't have $\frac 12 \in P$. $\frac 12$ may be in the interior of some subinterval $[x_i, x_{i+1}]$ (ie., $x_i < \frac 12 < x_{i+1}$). This would mean that $g(x_{i+1}) - g(x_i) = 1$, and depending on the "sample point" $c_i$ we choose in this subinterval, the sum may be $1$ or $0$. This can happen regardless of how fine the partition is, and hence the integral does not exist.   
Problem:

Are there any regularity conditions we can impose on $g$ to ensure the equivalence of the above definitions? Strict monotonicity is a natural example. If that doesn't work, consider stronger conditions (e.g., $g$ is homeomorphism onto its image, or a $C^{1}$ diffeomorphism). 

 A: Depending on how the Riemann-Stieltjes integral is defined (at least two ways that you mention) there are a variety of joint conditions on the integrand $f$ and integrator $g$ that guarantee existence.  There is  some but not complete overlap.
The Riemann-Stieltjes integral is less flexible than the Riemann integral. One impediment is that the continuity of $g$ comes into play. A function that is continuous is Riemann integrable and so too is one that is discontinuous only on a set of measure zero.  Taking definition (2), if the integrator is increasing, then if $f$ is continuous, the Riemann-Stieltjes integral exists, but may not exist if $f$ is only continuous almost everywhere.  That is because it is necessary that the integrand and integrator have no common points at which they are discontinuous.
Some basic relationships (of which I am aware) are:
Definition (1) holds if and only if definition (2) holds for Riemann integrals where $g(x) = x$.
Definition (1) implies definition (2) for Riemann-Stieltjes integrals when $f$ is bounded and  $g$ is increasing. 
Definition (2) implies definition (1) for Riemann-Stieltjes integrals when  $g$ is increasing, and either $f$ or $g$ is continuous. You found a counterexample if the continuity requirement is relaxed.
To prove the third implication, first consider that $f$ is continuous and R-S integrable with respect to $g$ under definition (2). Then for any partition $P = (x_0,x_1, \ldots,x_n)$ and choice of tags we have
$$\left|S(P,f,g) - \int_a^bf \, dg\right| = \left|\sum_{j=1}^n f(\xi_j)[g(x_j) - g(x_{j-1})]  - \sum_{j=1}^n \int_{x_{j-1}}^{x_j}f \, dg\right|$$
Since $f$ is continuous we can apply the integral mean value theorem to find points $\eta_j$ such that 
$$\left|S(P,f,g) - \int_a^bf \, dg\right| = \left|\sum_{j=1}^n [f(\xi_j)-f(\eta_j)]\,[g(x_j) - g(x_{j-1})] \right| \\ \leqslant \sum_{j=1}^n |f(\xi_j)-f(\eta_j)|\,[g(x_j) - g(x_{j-1})]. $$
By uniform continuity of $f$, for any $\epsilon >0$ there is a $\delta > 0$ such that if $\|P\| < \delta$ then $|f(\xi_j)-f(\eta_j)| < \epsilon/(g(b) - g(a))$ and 
$$\left|S(P,f,g) - \int_a^bf \, dg\right| < \epsilon.$$
Proof of the implication assuming that the integrator $g$ is continuous, rather than $f$, is lengthier.  In brief, we choose a partition $P' =(x_0,x_1,\ldots,x_n)$ such that the upper sum $U(P',f,g)$ and lower sum $L(P',f,g)$ are within $\epsilon/2$ of the integral. Using the uniform continuity of $g$, we find $\delta >0$ such that $|g(x) - g(y)| < \epsilon/(2nM)$ when $|x-y| < \delta$, where $M$ bounds $f$. Then a partition $P$ with $\|P\| < \delta$ is constructed through a tedious process such that
$$\int_a^b f \, dg - \epsilon < L(P',f,g) < L(P,f,g) \leqslant S(P,f,g) \leqslant U(P,f,g) < U(P',f,g) < \int_a^b f \, dg + \epsilon. $$ 
A: Asking that $f$ be continuous is sufficient. In fact, all you need is that $f$ be left continuous when $g$ is right-discontinuous, and vice versa. This is a rather easy exercise that can be done by considering the difference between the maximum and the minimum values attainable by choosing sample points on a given partition. As a corollary of this, $f$ must be continuous where $g$ is both left- and right-discontinuous. 
Where $g$ is continuous, it can be shown fairly easily that $f$ is allowed to have finitely many discontinuities. In fact, as long as the set of discontinuities of $f$ is measure zero (in the sense of Lebesgue measure), the integral ought converge. This proof is identical to that on the Riemann integral.
