Existence of $\int_a^bf\,dg$ when $f,g \not\in BV([a,b])$

Question

This question without satisfactory answer asks about necessary and sufficient conditions for existence of the Riemann-Stieltjes integral $\int_a^b fdg$ when both $f$ and $g$ are continuous. Related question is the existence of integral when both $f$ and $g$ have unbounded variation on $[a,b]$.

I have proved that if $g$ is continuous and $g \not\in BV([a,b])$ and $f \in BV([a,b])$ then $\int_a^bfdg$ exists. I use integration by parts argument that $f \in \mathcal{R_g([a,b])} \iff g \in \mathcal{R_f([a,b])}$ and $\int_a^g gdf$ always exists under these conditions.

I tried to find an example of (distinct) continuous functions $f,g \not\in BV([a,b])$ where $\int_a^b fdg$ exists but am unable. Is it possible the integral never exists in this case?

Answer 1

If both integrand and integrator have unbounded variation on an interval, the Riemann-Stieltjes integral can exist.

Consider $$f(x) =\begin{cases}x \cos(\pi/x), &0 < x \leqslant 1\\0 , & x= 0 \end{cases}$$

and $g(x) = f(1-x)$. Both $f$ and $g$ are continuous but of unbounded variation on $[0,1]$.

Since $g$ has bounded variation on $[0,1/2]$ and $f$ has bounded variation on $[1/2,1]$ we have existence of the integrals,

$$\int_0^{1/2}f \, dg, \,\,\,\int_{1/2}^1g \, df $$

Using integration by parts, we also have the existence of

$$\int_{1/2}^1f \,dg = f(1)g(1) - f(1/2)g(1/2) - \int_{1/2}^1 g \, df,$$

and, therefore, existence of

$$\int_{0}^1f \,dg = \int_{0}^{1/2}f \,dg + \int_{1/2}^1f \,dg $$

Answer 2

More a long comment and a complement to the answer of RRL than a full answer: however, after reading these references, which I initially thought only related to the question, I realized that they really contain the solution to core of the posed problem (recalled in the second part of the question), so I decided to write a more complete description.

I tried to find an example of (distinct) continuous functions $f,g \notin BV([a,b])$ where $\int_a^b fdg$ exists but am unable. Is it possible the integral never exists in this case?

No, the integral exists and is well defined provided some conditions are fulfilled, as Renato Caccioppoli and Ivan Petrovskii proved independently in 1934 (see [1] and [5]). The problem they solved arose as an observation made by Henri Lebesgue in the second edition of his "Leçons" ([3], p. 313), pertaining a generalization of the classical version of the fundamental lemma of integral calculus $$ h(x)=\int\limits_a^x f(t)\mathrm{d}t\iff \frac{\mathrm{d}h(x)}{\mathrm{d}x}=f(x), $$ where $f$ is a continuous function. Studying if and when a similar conditions holds for the integral $$ h(x)=\int\limits_a^x f(t)\mathrm{d}g(t) \quad f,g\in C^0,\label{1}\tag{1} $$ Lebesgue was able to prove that \eqref{1} holds for every continuous $f$ and for every continuous function $g$ of bounded variation provided the generalized derivative $$ \frac{\mathrm{d}h}{\mathrm{d}g}=\lim_{\epsilon\to 0}\frac{h(x+\epsilon)-h(x)}{g(x+\epsilon)-g(x)}=f(x)\label{2}\tag{2} $$ exists (and it is obviously continuous, since so is $f$) up to an additive constant: for the case of $g$ being merely continuous, Lebesgue states that the question of whether \eqref{2} is equivalent to \eqref{1} is still an open problem. Noticing Lebesgue's observation, Petrovskii (first in a short note [4] published on the Comptes Rendus and then, with full details, in the paper [5]) and Caccioppoli (inspired by also by [4]) prove the theorem in full generality i.e. proved that, for given $h,g\in C^0$ $$ \frac{\mathrm{d}h}{\mathrm{d}g}=\lim_{\epsilon\to 0}\frac{h(x+\epsilon)-h(x)}{g(x+\epsilon)-g(x)}=0 \implies h\equiv \operatorname{const.}\label{3}\tag{3} $$

Notes

• The proof of Caccioppoli is entirely geometric, very short (five pages!) and fascinating: by analyzing the problem in the $xy$-plane, putting $y=h(t)$ and $x=g(t)$ and analyzing the expression for the direction cosine of the curve $\big\{\big(h(t),g(t)\big)\big\}_{t\in\Bbb R}$, he proves that \eqref{1} implies $$ \frac{\mathrm{d}y}{\mathrm{d}x}=0\implies y=h(x)\equiv\operatorname{const.} $$ for any continuous curve . However, it is not easy to find a copy of the paper (the best place to find it is in his Opere) and also it is not an easy read since his prose is rich and cultured, with very few formulas and calculations. On the other hand, the proof of Petrovskii shows his analytic mastery and, in the last (§4) paragraph, answering a question of D. E. Menshov, he proves that the same result holds even in the case \eqref{2} is not defined on a countable set of points $x\in[a,b]$ (as noted in the commentary by Evgenii Landis from the Selected Works).

• In the later paper [2], Caccioppoli extends the notion of Stieltjes in order to define \eqref{1} for any couple $f,g\in C^0$, and proves \eqref{3} allowing the existence of a "negligible" (in a proper sense) set of singular points for the generalized derivative, in the spirit of the work of Petrovskii.

References
[1] Renato Caccioppoli (1934), "Sul lemma fondamentale del calcolo integrale" (Italian), Atti e Memorie, Accademia di Scienze Lettere ed Arti in Padova, n. Ser. 50, 93-98 (1934). JFM 60.0960.05, Zbl 0009.15801. See also his Opere (1998)[1963], Volume I, pp. 310-314.

[2] Renato Caccioppoli (1955), "L’integrazione e la ricerca delle primitive rispetto ad una funzione continua qualunque" (Italian), Annali di Matematica Pura ed Applicata, IV Serie, 40, 15-34, DOI:10.1007/BF02416519, MR0076851, Zbl 0067.03302.

[3] Henri Lebsgue (1955) [1928], "Leçons sur l'intégration et la recherche des fonctions primitives" (French), deuxième édition, noveau tirage, Collection de monographies sur la théorie des fonctions, Paris: Gauthier-Villars, pp. XIII+342, JFM 54.0257.01, Zbl 0041.02301.

[4] Ivan Georgievich Petrovskii (1929), "Sur les fonctions primitives par rapport à une fonction continue arbitraire", (French), Comptes Rendus Hebdomadaires des Séances de l’Académie des Sciences, Paris, 189, 1242-1244, JFM 55.0146.01.

[5] Ivan Georgievich Petrovskii (1934), "Sur l’unicité de la fonction primitive par rapport à une fonction continue arbitraire", (French), Matematicheskiĭ Sbornik, 41:1, 48–59, JFM 60.0214.03, Zbl 0009.30703. Translated in English as "On the Uniqueness of a Primitive Function Determined by its Generalized Derivatives with Respect to an Arbitrary Continuous Function" in Olga Arsen'evna Oleinik (Ed.) (1996), I.G. Petrowsky Selected Works. Part II. Differential Equations and Probability Theory, Classics of Soviet Mathematics 5, pp. 299-311.

Answer 3

I am not sure if a necessary and sufficient condition for the existence of the integral, even restricted to continuous functions, has been known. But here are some known results that may provide some idea on the existence of Riemann-Stieltjes integral. First, for $f : [a, b] \to \mathbb{R}$, its $p$-variation ($p>0$) is defined as

$$ V^p(f,[a,b])=\sup_{\Pi} \sum_{i=1}^{n} |f(x_i) - f(x_{i-1})|^{p}, $$

where the supremum is taken over all partitions $\Pi=\{a=x_0<\cdots<x_n=b\}$ of $[a, b]$. We easily note the followings:

• $1$-variation is the usual variation discussed in the context of $BV$.
• If $f$ has finite $p$-variation, then it has finite $q$-variation for any $q > p$.
• If $f$ is $\alpha$-Hölder continuous for $\alpha \in (0, 1]$, then $f$ has finite $\frac{1}{\alpha}$-variation. In particular, for each $p > 1$ there exists $f \in C([a, b])$ such that $V^1(f,[a, b]) = +\infty$ but $V^p(f,[a,b])<+\infty$.

Now the following theorem provides a large family of non-trivial examples.

Theorem. (L. C. Young, 1936) Let $p, q > 0$ be such that $\frac{1}{p}+\frac{1}{q} > 1$. Suppose that $f, g : [a, b] \to \mathbb{R}$ have no common discontinuity, $f$ has finite $p$-variation, and $g$ has finite $q$-variation. Then $\int_{a}^{b} f \, \mathrm{d}g$ exists.

This type of integral is called Young integral. There are some extensions to more general types of variations.

On the other hand, the above theorem is sharp in the sense that $1/p+1/q = 1$ case bears counter-examples. Young provided an example of $f$ and $g$ with finite $2$-variations such that $\int_{a}^{b}f\,\mathrm{d}g$ does not exist. Counter-examples for any other values of $p, q$ with $1/p+1/q = 1$ also exist.