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In connection with my post Convergence to Riemann-Stieltjes integral of sequence of Riemann-Stieltjes-like sums with changing integrand and integrator, an alternative approach to my main objective would be considering the convergence of the sequence of Riemann-Stieltjes (RS) integrals $\int_0^1 \, f_N(x) \, \mathrm{d}F_N(x)$ to the RS integral $\int_0^1 \, f(x) \, \mathrm{d}F(x)$. The properties of the functions involved are as defined in the aforementioned post. I repeat them in the following paragraph for convenience, though.

The functions 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.

Any hints on how to prove the above statement of convergence will be welcome.

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  • $\begingroup$ I'm rusty. When you indicate $f_n \longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)? $\endgroup$ – Eric Towers Dec 3 '18 at 7:03
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    $\begingroup$ Have you tried the "obvious thing": see if you can show $\int_0^1 \; |f - f_n| \, \mathrm{d}(|F - F_n|) \longrightarrow 0$? $\endgroup$ – Eric Towers Dec 3 '18 at 7:09
  • $\begingroup$ Not necessarily uniformly, but if it helps with a proof of convergence (or hints, with the details to be worked out by me), you are welcome to assume it is so. $\endgroup$ – Marcos Dec 3 '18 at 7:11
  • $\begingroup$ No, actually I haven't. This question dawned on me as a rather indirect but actually alternative approach to the linked post. I'll check that. $\endgroup$ – Marcos Dec 3 '18 at 7:13
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Hints:

Note that

$$\left|\int_0^1f_N \, dF_N - \int_0^1 f \, dF \right| \leqslant \left|\int_0^1f_N \, dF_N - \int_0^1 f \, dF_N \right|+ \left|\int_0^1f \, dF_N - \int_0^1 f \, dF \right|$$

(1) We can estimate the first term on the RHS and prove convergence to $0$, if $F_N$ has bounded variation $V_0^1(F_N)$, using

$$\left|\int_0^1f_N \, dF_N - \int_0^1 f \, dF_N \right| \leqslant \int_0^1|f_N - f|\, dV_0^x(F_N), $$ although with the simplification that $F_N$ is montonically increasing we have

$$\left|\int_0^1f_N \, dF_N - \int_0^1 f \, dF_N \right| \leqslant \int_0^1|f_N - f|\, dF_N $$

With uniform convergence of $f_n \to f$ and uniform boundedness of $F_N$ it is easy to progress. Given that $|F_N(x)| \leqslant M$ uniformly in $N$ and $x$ -- which is true if $F_N$ converges uniformly to a bounded function $F$ -- then for all sufficiently large $N$ we have $|f_N(x) - f(x)| < \epsilon/(2M)$ and

$$\left|\int_0^1f_N \, dF_N - \int_0^1 f \, dF_N \right| < \frac{\epsilon}{2M}[F_N(1) - F_N(0)] < \frac{\epsilon}{2M}2M = \epsilon $$

(2) Estimate the second term on the RHS most easily using Rieman sums.

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  • $\begingroup$ Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details. $\endgroup$ – Marcos Dec 3 '18 at 7:52
  • $\begingroup$ Actually, my difficulty is currently with the second term on the RHS. I was thinking of applying the bound $\big|\int_0^1 f d(F_n-F) \big| \leq \sup_{[0,1]} |f| \, V_0^1(F_n-F)$, but I haven't been able to obtainanything useful regarding the limiting total variation of $F_n-F$ on $[0,1]$, whence it came my question math.stackexchange.com/questions/3055573/… $\endgroup$ – Marcos Dec 29 '18 at 6:47

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