Csdrhrt

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.

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,

$$ left| sum_{n=0}^N , f_N(t_n^{(N)}) , [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})]- sum_{n=0}^N , f(t_n^{(N)}) , [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] right|\ leqslant sum_{n=0}^N , |f_N(t_n^{(N)})- f(t_n^{(N)}) | , [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] \ leqslant epsilon sum_{n=0}^N , , [F_N(x_n^{(N)}) - F_N(x_{n-1}^{(N)})] \ = epsilon[F_N(1) - F_N(0)] \ leqslant epsilon[F(1) - F(0) + 2epsilon$$

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.