# Convergence of sequence of Riemann-Stieltjes integrals to Riemann-Stieltjes integral

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.

• I'm rusty. When you indicate $f_n \longrightarrow f$, you intend convergence in the sup norm (i.e. uniform convergence)? – Eric Towers Dec 3 '18 at 7:03
• Have you tried the "obvious thing": see if you can show $\int_0^1 \; |f - f_n| \, \mathrm{d}(|F - F_n|) \longrightarrow 0$? – Eric Towers Dec 3 '18 at 7:09
• 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. – Marcos Dec 3 '18 at 7:11
• 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. – Marcos Dec 3 '18 at 7:13

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.

• Thanks so much. As with my linked post, I'll be back as soon as I have worked out the details. – Marcos Dec 3 '18 at 7:52
• 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/… – Marcos Dec 29 '18 at 6:47