# If the Riemann-Stieltjes integral $\int_a^b{f\,dg}$ exists for every $f\in\mathcal{R}[a,b]$, then $g$ is absolutely continuous

In Remark 18.13 of Real Analysis: Foundations and Functions of One Variable, the author states that

The class of Riemann integrable functions and the class of absolutely continuous functions are also dual: a function $$f$$ is Riemann integrable if and only if $$\int_a^b{f\,dg}$$ exists for every absolutely continuous function $$g$$, and a function $$g$$ is absolutely continuous if and only if $$\int_a^b{f\,dg}$$ exists for every Riemann integrable function $$f$$. The proof of this theorem, however, uses concepts from measure theory that we do not deal with in this book.

I am interested in finding a proof for these two statements. I have proved the following lemma:

Lemma. If $$f$$ is Riemann integrable on $$[a,b]$$ and $$g$$ is absolutely continuous on $$[a,b]$$, then the Riemann-Stieltjes integral $$\int_a^b{f\,dg}$$ exists with $$(\text{R-S})\int_a^b{f\,dg}=(\text{L})\int_a^b{f(x)g'(x)\,dx}.$$

The proof is not hard, but mostly relies on the fundamental theorem of calculus in Lebesgue integrals, so an elementary proof seems impossible. For future reader's reference, I added a sketch of the proof here.

Sketch of Proof. Since $$g$$ is absolutely continuous, it is differentiable a.e. on $$[a,b]$$. It is easy to see that $$fg'$$ is Lebesgue integrable on $$[a,b]$$. Let $$(P,\xi)$$ be a tagged partition of $$[a,b]$$. Consider the estimate \begin{align*} &\left|\sum_{i=1}^{n}{f(\xi_i)[g(x_i)-g(x_{i-1})]}-(\text{L})\int_a^b{f(t)g'(t)\,dt}\right| \\ \le~&\sum_{i=1}^{n}{(\text{L})\int_{x_{i-1}}^{x_i}{|f(\xi_i)-f(t)||g'(t)|\,dt}} \\ \le~&\omega\sum_{i=1}^{n}{(\text{L})\int_{x_{i-1}}^{x_i}{|g'(t)|\,dt}}=\omega\cdot(\text{L})\int_a^b{|g'(t)|\,dt}\le\omega V_a^b(g). \end{align*} Here $$V_a^b(g)$$ is the total variation of $$g$$ on $$[a,b]$$, while $$\omega$$ is the largest oscillations of $$f$$ on $$[x_{i-1},x_i]$$ for each $$i=1,\ldots,n$$. Since $$f$$ is Riemann integrable on $$[a,b]$$, we shall have $$\omega\to 0$$ as $$P$$ is refined. The desired assertion thus follows. $$\square$$

From this, the first statement is proved:

Proposition 1. $$f$$ is Riemann integrable on $$[a,b]$$ if and only if $$\int_a^b{f\,dg}$$ exists for every $$g\in\operatorname{AC}[a,b]$$.

Proof. ($$\Longrightarrow$$). This is precisely the lemma presented above.

($$\Longleftarrow$$). Note that the identity function $$x\mapsto x$$ is absolutely continuous, and the Riemann integral is a special case of the Riemann-Stieltjes integral with $$g(x)=x$$. $$\square$$

Proposition 2. $$g$$ is absolutely continuous on $$[a,b]$$ if and only if $$\int_a^b{f\,dg}$$ exists for every $$f\in\mathcal{R}[a,b]$$.

The direction ($$\Longrightarrow$$) is also trivial, as it is again the preceding lemma. However, I am stuck in the proof of the direction ($$\Longleftarrow$$), i.e., proving that $$g\in\operatorname{AC}[a,b]$$ whenever $$\int_a^b{f\,dg}$$ exists for every $$f\in\mathcal{R}[a,b]$$, as indicated in the title.

It is not hard to show that $$g$$ is (uniformly) continuous on $$[a,b]$$: For every $$c\in[a,b]$$, let $$f:=\chi_{\{c\}}$$. Then $$f$$ is clearly Riemann integrable on $$[a,b]$$ and discontinuous at $$c$$. By assumption, $$\int_a^b{f\,dg}$$ exists, so $$g$$ must be continuous at $$c$$ (see this post for the reason why).

It would be nice if I could hear from you guys about the proof for absolute continuity.

Update. Inspired from Oliver Díaz's comment, I realized that we had the following nice result:

Lemma 2. $$g$$ is of bounded variation on $$[a,b]$$ if $$\int_a^b{f\,dg}$$ exists for every continuous function $$f$$ on $$[a,b]$$.

Since $$C[a,b]\subseteq\mathcal{R}[a,b]$$, the function $$g$$ must be of bounded variation, i.e., $$g\in\operatorname{BV}[a,b]$$. The proof of Lemma 2 can be found from this post.

Update 2. Now that $$g$$ is continuous and has bounded variation on $$[a,b]$$. By Banach-Zaretsky Theorem, it suffices to show that $$g$$ has the following Luzin property (N), namely if $$N\subseteq[a,b]$$ with Lebesgue measure $$\lambda(N)=0$$, then $$\lambda(g(N))=0$$ as well. However, I still do not get how to prove this directly. Any help will be appreciated.

• @OliverDíaz Not necessarily. The original text did not make such assumption there. Nov 24, 2022 at 16:24
• @OliverDíaz But thanks for your reminder. Here $g$ is indeed of bounded variation. I have edited my post there. Nov 24, 2022 at 16:44
• Your lemma 2 is available on MSE :math.stackexchange.com/a/2428728/72031 Nov 24, 2022 at 23:47
• @ParamanandSingh Thanks, I shall replace my parts with this link (as my OP is too long). Nov 25, 2022 at 0:47

In this posting we present a solution to OP; however, we do make use of some results from Lebesgue-integration, namely the Radon-Nikodym theorem, Lebesgue's decomposition theorem, and the regularity (inner and outer) of Borel measures. One less common result that we will use this is the following:

Lemma C (Horst): Suppose $$\alpha$$ in monotone nondecreasing on $$[a,b]$$. For any bounded function $$f$$ on $$[a,b]$$, if $$f\in\mathcal{R}(\alpha)$$, then $$f\in L_1(\mu_\alpha)$$ where $$\mu_\alpha$$ is the unique regular measure such that $$\mu_\alpha((c,d])=\alpha(d+)-\alpha(c+)$$, and $$RS-\int^b_af\,d\alpha=L-\int^b_a f\,d\mu_\alpha$$. Furthermore, $$f\in\mathcal{R}(\alpha)$$ iff $$f$$ is $$\mu_\alpha$$-almost surely continuous on $$Z_\alpha=\{x\in[a,b]: \mu_\alpha(\{x\})=0\}$$, and there are no $$x \in[a,b]$$ for which $$f$$ and $$\alpha$$ are simultaneously discontinuous from the left or from the right.

Lemma C is a generalization of Lebesgue integrability criteria for Riemann integrals to the setting of Riemann-Stieltjes integration. A full presentation of these types of results can be found in the paper by Horst, H. J, T., Riemann-Stieltjes and Lebesgue-Stieltjes Integrability, The American Mathematical Monthly Vol. 91, No. 9 (Nov., 1984), pp. 551-559.

Throughout this posting, $$m$$ denotes the Lebesgue measure on the real line, $$\mathcal{R}$$ denotes the space of all Riemann integrable functions on $$[a,b]$$, and $$\mathcal{R}(\alpha)$$ denotes the space of all $$\alpha$$-Riemann-Stieltjes integrable functions. The assumption of the problem in the OP is $$\mathcal{R}\subset\mathcal{R}(\alpha)$$. In particular, $$C([a,b])\subset\mathcal{R}(\alpha)$$.

The following facts are already discussed in the OP:

• By Banach space methods, in particular the Banach-Steinhouse theorem a.k.a uniform boundedness theorem, $$\alpha$$ is of total bounded variation over $$[a,b]$$ (see for example this posting).

• Since a function $$f$$ is $$\alpha$$-Riemann-Stieltjes integrable implies that $$f$$ and $$g$$ don't share the same points of discontinuity from the same side, and $$\mathcal{R}\subset\mathcal{R}(\alpha)$$, it follows that $$g$$ must be continuous. In the setting of Lemma C, $$Z_\alpha=[a,b]$$

Now, since $$f\in\mathcal{R}(\alpha)$$ implies that $$f\in\mathcal{R}(V_{\alpha})$$, where $$V_\alpha$$ is the variation function of $$\alpha$$, $$V_\alpha\pm\alpha\geq0$$ are monotone nondecreasing, and $$\alpha=\frac12((V_\alpha+\alpha)-(V_\alpha-\alpha))$$, it is enough to consider the case where $$\alpha$$ is monotone nondecreasing.

Extend $$\alpha(x)=\alpha(a)$$ and $$\alpha(x)=\alpha(b)$$ for $$a and $$x>b$$ respectively. The Lebesgue-Stleiltjes theorem yields a unique regular measure supported in $$\mu$$ on $$[a,b]$$ such that $$\mu((c,d])=\alpha(d)-\alpha(c)$$ for all $$-\infty. Notice that $$\mu_\alpha(\mathbb{R})=\mu_\alpha(a,b])=\alpha(b)-\alpha(a)<\infty$$; hence, by Lebesgue's decomposition theorem there is $$f\in L^+_1(m)$$ and a measure $$\mu_s$$ that is singular with respect to $$m$$ such that $$\mu_\alpha(A)=\int_A f dm + \mu_s(A),\qquad A\in\mathscr{B}([a,b])$$ We claim that $$\mu_s=0$$. Otherwise, there is $$E\subset[a,b]$$ such that $$m(E)=0=\mu_s(E^c)$$ and $$\mu_s(E)>0$$. Since all finite Borel measures are regular, there is a compact set $$K\subset E$$ such that $$\mu_s(K)>\mu_s(E)/2>0$$. By Lemma C, the function $$g:=\mathbb{1}_{K\setminus\mathbb{Q}}$$ is not in $$\mathcal{R}(\alpha)$$ since it is discontinuous on $$K$$ which was positive $$\mu_s$$ measure. However, as $$m(K)\leq m(E)=0$$ and $$g$$ is only discontinuous on $$K$$, $$g\in\mathcal{R}\subset\mathcal{R}(\alpha)$$ which is a contradiction. Hence, $$\mu_s=0$$ and so, $$\mu_\alpha\ll m$$. This in particular means that $$\alpha$$ is absolutely continuous and $$\alpha'=f$$ $$m$$-a.s.