Convergence to Riemann-Stieltjes integral of sequence of Riemann-Stieltjes-like sums with changing integrand and integrator I am considering the limiting behavior of a sequence of Riemann-Stieltjes (RS) (or at least RS-like) sums in the sense of their convergence to a Riemann-Stieltjes integral. The general term has the form 
\begin{equation}
  \sum_{n=0}^N \, f_N(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] \,,  
\end{equation}
where $x_{-1}^{(N)} := 0$ for every value of $N$, every partition $\{ x_n^{(N)} \}_{n=-1}^N$ is of the unit real interval $[0,1]$ with $x_N^{(N)} = 1$, the argument $t_n^{(N)}$ belongs to the subinterval $[x_{n-1}^{(N)}, x_n^{(N)}]$ of $[0,1]$ (actually, in my more specific setting it is $t_n^{(N)} = x_n^{(N)}$), and the partition for $N+1$ is a refinement of that for $N$. I think everything is as usual for an RS sum except for the particularity that there is no fixed function having the role of integrator, but rather a sequence $(F_N)_N$ thereof, and similarly for the integrand, which is defined in terms of the sequence $(f_N)_N$. 
As for the functions, they are all real of a single real variable. Those in the sequence $(f_N)_N$ are continuous and bounded in $[0,1]$, and the sequence converges to a continuous function f bounded in the same interval. In turn, those in $(F_N)_N$ are monotonically increasing step functions bounded in $[0,1]$ and the sequence is uniformly convergent to a function F that is a cumulative distribution function. My overall objective is to prove that the sequence of sums converges to the RS integral $\int_0^1 \, f(x) \, \mathrm{d}F(x)$. 
Any hints on how to proceed will be welcome.
 A: Start by writing
$$\tag{*}\left| \sum_{n=0}^N \, f_N(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})]- \int_0^1 f \, dF \right| \\ \leqslant \left| \sum_{n=0}^N \, f_N(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})]-  \sum_{n=0}^N \, f(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] \right| \\ + \left| \sum_{n=0}^N \, f(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})]-  \sum_{n=0}^N \, f(t_n^{(N)}) \, [F(x_n^{(N)}) - F(x_{n-1}^{(N)})] \right| \\+ \left| \sum_{n=0}^N \, f(t_n^{(N)}) \, [F(x_n^{(N)}) - F(x_{n-1}^{(N)})]- \int_0^1 f \, dF \right|,$$
and try to estimate each term as $N \to \infty.$
Uniform convergence of both $f_N \to f$ and $F_N \to F$ would be helpful.
For the first term on the RHS of (*) , with uniform convergence, we have  for sufficiently large $N$ and all $x \in [0,1]$,
$$|f_N(x) - f(x)| < \epsilon, \,\,\, F_N(1) < F(1) + \epsilon, \,\,\, F_N(0) > F(0) - \epsilon$$ 
Hence,
$$ \left| \sum_{n=0}^N \, f_N(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})]-  \sum_{n=0}^N \, f(t_n^{(N)}) \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] \right|\\ \leqslant  \sum_{n=0}^N \, |f_N(t_n^{(N)})- f(t_n^{(N)}) | \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] \\ \leqslant \epsilon \sum_{n=0}^N \,  \, [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})]  \\ = \epsilon[F_N(1) - F_N(0)] \\ \leqslant \epsilon[F(1) - F(0) + 2\epsilon$$
The convergence of the last term on the RHS of (*) is not always guaranteed when $f$ is R-S integrable with respect to $F$ except under strong conditions. It seems you have that part -- both because the partitions in the sequence are refinements and $f$ is continuous.
