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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.

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

1 Answer 1

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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 stated below.

Notation: If $f$ is a real-valued function on $[a,b]$, and $f$ is intebrable in some sense $\mathcal{S}$, where $\mathcal{S}$ is Riemann (\mathcal{R}), Lebesgue (\mathcal{L}), Riemann-Stieltjes ($\mathcal{RS}$, or Stieltjes-Lebesgue ($\mathcal{SL}$), we use $\mathcal{S}-\int^b_af$ to denote the value of the integral of $f$ over $[a,b]$.

WE have the following result:

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 $a<a$ 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 $-\infty<c<d<\infty$. 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.


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