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Throughout this post I refer to these as "Darboux integrals/integration" and to these as "Riemann integrals/intnegration". It is a standard result of Analysis that these notions of integrability coincide.


I've tried (and failed) looking for a book that introduces integration in terms of Riemann integrals: Rudin's POMA, Tao's Analysis I, Spivak's Calculus and Abbott's Understanding Analysis all use Darboux integration. Wikipedia explains the reason in its article on Darboux integration:

"The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral. Consequently, introductory textbooks on calculus and real analysis often develop Riemann integration using the Darboux integral, rather than the true Riemann integral."

Itching my curiosity I've shown that some elementary theorems on integration can be proven directly using Riemann integrability (such as that $\int(f+g)=\int f + \int g$, as well as $c\int f = \int cf$). Currently I'm stuck trying to prove Rudin's 6.17 and hence I'm asking for a hand.


Theorem 6.17: Assume $\alpha$ increases monotonically on $[a,b]$ and $\alpha' \in \mathcal{R}$ (i.e. $\alpha'$ is Darboux integrable). Let $f$ be a bounded real function on $[a,b]$. Then $f\in \mathcal{R}(\alpha)$ if and only if $f\alpha'\in\mathcal{R}$. In that case $$\int_a^bf\ d\alpha = \int_a^bf(x)\alpha'(x)\ dx.$$


Can 6.17 be shown by using Riemann integration?

To be clearer, showing the result by directly employing the criterion in the definition of Riemann integrability, as opposed to using other (equivalent) criteria.

There would be no point in using the criterion on the definition of Darboux integrability, nor to use, as Rudin did, the fact that a function is Riemann integrable if and only if for any $\epsilon>0$, there is some partition $P$ such that $$\left|U(f,\alpha,P)-L(f,\alpha,P)\right|<\epsilon.$$

Sam
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  • Poor Stieltjes. – blargoner Oct 24 '22 at 20:18
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    Mathematical Analysis: A Concise Introduction by Bernd Schröder introduces integration in terms of Riemann integrals. Of course, it also discusses the connections to Darboux integrals too. – Abhijeet Vats Oct 24 '22 at 20:22
  • @AbhijeetVats thanks for the reference! The question in my post is (with slight changes in the hypothesis) exercise 5-19 (f) in Schröder's book. – Sam Oct 24 '22 at 20:52
  • (1) In the theorem statement you wrote $\alpha \in \mathcal{R}$. That should be $\alpha' \in \mathcal{R}$, shouldn't it. It is also given that $\alpha$ is monotone increasing. On that basis alone, there could be discontinuities with the derivative $\alpha'$ existing almost everywhere. Is it to be assumed that $\alpha'$ exists everywhere in the interval. – RRL Oct 27 '22 at 19:45
  • @RRL you are right that $\alpha'\in\mathcal{R}$ and that I wasn't fully clear as to what I meant. I have edited the question accordingly. – Sam Oct 27 '22 at 22:13
  • @RRL and I do believe it is assumed that $\alpha'$ exists everywhere on the interval $[a,b]$. – Sam Oct 27 '22 at 22:14
  • To confirm the type of proof you are looking for, lets consider the reverse implication $f\alpha' \in \mathcal{R} \implies f \in \mathcal{R}(\alpha)$. You would want to show that for any $\epsilon > 0$ there exists a partition $P_\epsilon$ such that for any refinement $P$ we have $|S(P,f,\alpha,{t_j}) - \int_a^b f(x) \alpha'(x) , dx| < \epsilon$ for a Riemann-Stieltjes sum $S$ and any choice of intermediate points ${t_j}$. For partition $P=(x_0,x_1,\ldots,x_n)$, the R-S sum is $S(P,f,\alpha,{t_j}) := \sum_{j=1}^n f(t_j)[\alpha(x_j) - \alpha(x_{j-1})]$. Is that what you want? – RRL Oct 27 '22 at 22:56
  • Note that the Riemann integral can be defined either in terms of partition refinement or in terms of the partition norm $|P|$ tending to $0$. The two are not always equivalent for the Riemann-Stieltjes integral. but probably are under the hypotheses given here. – RRL Oct 27 '22 at 22:59
  • @RRL yes, that's what I had in mind. Note it also corresponds with the criterion for Riemann integrability in the link at the beginning of the post. – Sam Oct 28 '22 at 02:15
  • The cited Wikipedia article refers to Apostol's Mathematical Analysis (second edition 1974), which defines the Riemann-Stieltjes integral using Riemann sums, and states three propositions that are similar to but weaker than Rudin's Theorem 6.17. – Calum Gilhooley Oct 30 '22 at 05:57

2 Answers2

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Here is a proof (without refering to Darboux integrals) of the reverse implication of 6.17 that if $f \alpha'$ is Riemann integrable on $[a,b]$ then $f$ is Riemann-Stieltjes integrable with respect to the integrator $\alpha$ on $[a,b]$ and

$$\int_a^b f \, d\alpha = \int_a^b f(x) \alpha'(x)\, dx$$

The objective is to show that for any $\epsilon >0$, there exists a partition $P_\epsilon$ such that for any refinement

$$P = \{x_0,x_1,\ldots,x_n:a = x_0 < x_1 < \ldots <x_n = b\}$$

and any choice of intermediate points $\xi_j \in [x_{j-1},x_j]$, we have

$$\left|S(P,f,\alpha,\{\xi_j\})- \int_a^bf(x) \alpha'(x) \, dx\right|= \left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|< \epsilon$$

By the mean value theorem, there exists $\eta_j \in (x_{j-1},x_j)$ for $j=1,2,\ldots,n$ such that

$$\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]= \sum_{j=1}^nf(\xi_j)\alpha'(\eta_j) (x_j - x_{j-1})$$

Thus,

$$\left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|\\= \left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx + \sum_{j=1}^nf(\xi_j)\alpha'(\eta_j)(x_j-x_{j-1})- \sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})\right|\\ \leqslant \underbrace{\left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx\right|}_{A}+ \underbrace{\sum_{j=1}^n |f(\xi_j)||\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})}_{B}$$ Because $f\alpha'$ is Riemann integrable it follows that there exists a partition $P_A$ such that if $P$ is a refinement, then $$\tag{1}A= \left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx\right|< \frac{\epsilon}{2}$$

Since $f$ is bounded, there exists $M>0$ such that $|f(x)| < M$ for all $x \in [a,b]$ and, hence

$$B = \sum_{j=1}^n |f(\xi_j)|[\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) < M \sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) $$

It will be shown below that because $\alpha'$ is Riemann integrable, there exists a partition $P_B$ such that if $P$ is a refinement, then

$$\tag{2}\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})<\frac{\epsilon}{2M},$$

whence,

$$\tag{3}B < M\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) < \frac{\epsilon}{2}$$

Hence, if P is a refinement of the common refinement $P_\epsilon = P_A \cup P_B$, then by (1) and (3) we have

$$\left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|=A+B < \epsilon,$$

and we have proved that $\int_a^b f \, d\alpha = \int_a^b f(x) \alpha'(x) \, dx$ as was to be shown.

The forward implication can be proved in a similar way.

Proof of (2)

Since $\alpha'$ is Riemann integrable, there exists a partition $P_B$ such that if $P$ is a refinement, then for any choice of intermediate points $t_j \in [x_{j-1},x_j]$, we have

$$\tag{4}\left|\sum_{j=1}^n \alpha'(t_j)(x_j - x_{j-1}) - \int_a^b \alpha'(x) \, dx\right|< \frac{\epsilon}{8M}$$

Denoting $M_j = \sup\{\alpha'(x): x \in [x_{j-1},x_j]\}$ and $m_j = \sup\{\alpha'(x): x \in [x_{j-1},x_j]\}$, by properties of the supremum and infimum there exist points $t_j', t_j''$ such that

$$M_j - \frac{\epsilon}{4M(b-a)} < \alpha'(t_j') \leqslant M_j, \quad m_j \leqslant \alpha'(t_j'') < m_j + \frac{\epsilon}{8(b-a)},$$

and, thus,

$$\sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1}) < \sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1}) -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1}) + \frac{\epsilon}{2M}\\ = \sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1})- \int_a^b\alpha'(x) \, dx + \int_a^b \alpha'(x) \, dx -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1}) + \frac{\epsilon}{4M}$$

Hence, since the LHS is nonnegative,

$$\tag{5} \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1}) \\< \left|\sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1})- \int_a^b\alpha'(x) \, dx \right|+ \left| \int_a^b \alpha'(x) \, dx -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1})\right|+ \frac{\epsilon}{4M}$$

Using (4), it follows from substitution into (5) that

$$ \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1})< \frac{\epsilon}{8M} + \frac{\epsilon}{8M} + \frac{\epsilon}{8M} = \frac{\epsilon}{2M}$$

Noting that $\xi_j, \eta_j \in [x_{j-1},x_j]$ and $|\alpha'(\xi_j)- \alpha'(\eta_j)| \leqslant M_j - m_j$, we obtain the result

$$\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})\leqslant \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1})<\frac{\epsilon}{2M}$$

RRL
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    +1 there, but essentially you proved using Riemann definition that we have partitions with $U(P, \alpha') - L(P, \alpha') <\epsilon/2M$. I think this result is unavoidable. – Paramanand Singh Oct 30 '22 at 02:08
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    Also the same proof works for the direct implication because what you have shown is that the difference between $S(P, f, \alpha) $ and $S(P, f\alpha') $ can be made arbitrarily small so that if one tends to a limit the other also tends to same limit. – Paramanand Singh Oct 30 '22 at 02:16
  • @ParamanandSingh: Thanks. I agree that it is likely unavoidable. Nevertheless I have not invoked the upper and lower Darboux integrals and their properties themselves which is the approach intended by Rudin. Just estimating the sum in (2) in this way is not really going fully into the Darboux framework. It is just a basic property of $\alpha’$ as a Riemann integrable and hence bounded function. – RRL Oct 30 '22 at 02:19
  • Yeah I understand and let's hope asker also gets this. By the way I had begun typing essentially the same proof and yours appeared. :) – Paramanand Singh Oct 30 '22 at 02:24
  • Minor typo: $m_j = \sup{\alpha'(x): x \in [x_{j-1},x_j]}$ should of course be $m_j = \inf{\alpha'(x): x \in [x_{j-1},x_j]}.$ – Calum Gilhooley Nov 02 '22 at 14:48
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This is an alternative approach based on the following criterion of Riemann integrability:

Cauchy's condition for Riemann integrability: Let the function $f:[a, b] \to\mathbb{R} $ be bounded on $[a, b] $. Then $f$ is Riemann integrable on $[a, b] $ if and only if for every $\epsilon>0$ there exists a corresponding $\delta>0 $ (partition $P_{\epsilon} $ of $[a, b] $) such that $$|S(P_1,f)-S(P_2,f)|<\epsilon$$ where $S(P_1,f),S(P_2,f)$ are Riemann sums for $f$ over partitions $P_1,P_2$ and any choice of tags and $||P_1||<\delta,||P_2||<\delta$ ($P_1,P_2$ are finer than $P_{\epsilon} $ ie $P_{\epsilon} \subseteq P_1,P_{\epsilon}\subseteq P_2$).

In what follows we use the norm based definition/criterion (instead of refinement) due to less typing effort involved.

Let us state our hypotheses at the beginning. We are given a bounded function $f$ on $[a, b] $ as well as another function $\alpha$ whose derivative $\alpha'$ is Riemann integrable on $[a, b] $. We don't assume the monotone nature of $\alpha$ and further we don't assume Riemann integrability of $f$.

Let us assume $f\in\mathcal{R} (\alpha) $ with $I=\int_a^b f\, d\alpha$. Let $\epsilon>0$ and by the Riemann-Stieltjes integrability of $f$ with respect to $\alpha$ there exists a $\delta_1>0$ such that $$|S(P, f, \alpha) - I|<\frac{\epsilon} {2}\tag{1}$$ whenever $||P||<\delta_1$.

Let $M$ be a positive upper bound for $|f|$ over $[a, b] $ and applying Cauchy's condition on $\alpha'$ we get a $\delta_2>0$ such that $$|S(P_1,\alpha')-S(P_2,\alpha')|<\frac{\epsilon}{2M}\tag{2}$$ whenever $||P_1||<\delta_2,||P_2||<\delta_2$.

Let $\delta =\min(\delta_1,\delta_2)$ and $P$ be a partition of $[a, b] $ with $||P||<\delta$ and consider an arbitrary Riemann sum $$S(P, f\alpha') =\sum_{k=1}^n f(\xi_k) \alpha'(\xi_k) (x_k-x_{k-1})$$ for $f\alpha'$ over $P$. Next consider a specific Riemann sum $S(P, f, \alpha) $ for $f$ with respect to $\alpha$ over $P$ and tag points $\xi_k$ given by $$S(P, f, \alpha) =\sum_{k=1}^n f(\xi_k) (\alpha(x_k) - \alpha(x_{k-1}))=\sum_{k=1}^n f(\xi_k) \alpha'(\eta_k) (x_k-x_{k-1})$$ (via mean value theorem). We have \begin{align} |S(P, f\alpha') - I| &\leq |S(P, f\alpha') - S(P, f, \alpha) |+|S(P, f, \alpha) - I|\notag\\ &<\left|\sum_{k=1}^n f(\xi_k) (\alpha'(\xi_k) - \alpha'(\eta_k)) (x_k-x_{k-1})\right|+\frac{\epsilon} {2}\text{ (via (1))}\notag\\ &\leq M\sum_{k=1}^n |\alpha'(\xi_k) - \alpha'(\eta_k) |(x_k-x_{k-1})+\frac{\epsilon}{2}\notag\\ &<M\cdot\frac{\epsilon}{2M}+\frac{\epsilon}{2}\text{ (via (2))}\notag\\ &=\epsilon\notag \end{align} The last step which uses Cauchy's criterion needs a bit more explanation. Here we are taking both partitions $P_1,P_2$ equal to $P$ and two sets of tags $t_k, t'_k$ such that $t_k=\xi_k, t'_k=\eta_k$ if $\alpha'(\xi_k) \geq \alpha'(\eta_k) $ and $t_k=\eta_k, t'_k=\xi_k$ if $\alpha'(\xi_k) <\alpha'(\eta_k) $. This ensures that $$|\alpha'(\xi_k) - \alpha'(\eta_k) |=\alpha'(t_k) - \alpha'(t'_k) $$ for all $k$ and we get $\sum |\alpha'(\xi_k) - \alpha'(\eta_k)|(x_k-x_{k-1}) $ as a difference between two Riemann sums for $\alpha'$ over $P$. Note further that this uses the "only if" part of Cauchy's criterion which is an immediate consequence of definition of integral (since two Riemann sums are near the value of integral they are near to each other as well).

This proves that $f\alpha'$ is Riemann integrable with integral $I$.

To prove the reverse implication we assume the Riemann integrability of $f\alpha'$ with integral $I$ and start with a arbitrary Riemann-Stieltjes sum $S(P,f,\alpha)$ and a corresponding Riemann sum $S(P, f\alpha') $ and repeat the above argument starting with $|S(P, f, \alpha) - I|$ and prove that $f\in\mathcal{R} (\alpha) $ with integral $I$.

The above argument can be provided using the definition via refinement of partitions by making appropriate changes. Further note that Riemann integrability (and value of integral) of $f\alpha'$ does not change via change of definitions and thus existence of $\int_a^b f\alpha'$ implies the existence of $\int_a^b f\, d\alpha$ under both definitions with same value of integral (this was mentioned in one of the comments to the question).

  • I see how the Cauchy criterion gives $$\tag{1}\left|\sum_{k=1}^n (\alpha'(\xi_k) - \alpha'(\eta_k)) (x_k-x_{k-1})\right| = |S(P_1,\alpha')-S(P_2,\alpha')|<\frac{\epsilon}{2M}$$ where in this case the partitions $P_1$ and $P_2$ have the same points and different tags. I also see that $$\tag{2}\left|\sum_{k=1}^n (\alpha'(\xi_k) - \alpha'(\eta_k)) (x_k-x_{k-1})\right|\leqslant \sum_{k=1}^n |\alpha'(\xi_k) - \alpha'(\eta_k)| (x_k-x_{k-1})$$ But how do (1) and (2) imply that $\sum_{k=1}^n |\alpha'(\xi_k) - \alpha'(\eta_k)| (x_k-x_{k-1}) < \frac{\epsilon}{2M}$? – RRL Oct 30 '22 at 21:32
  • @RRL: see updated answer. I had same doubt as well but only today I realized that the power to choose arbitrary tags in Riemann sum gives us great leverage. – Paramanand Singh Oct 31 '22 at 00:21
  • @RRL: the expression $\sum(f(\xi_k) - f(\eta_k)) (x_k-x_{k-1})$ is a difference of two Riemann sums. And it takes only a little swapping of tags in each subinterval to realize that non-negative sum $\sum |f(\xi_k) - f(\eta_k) |(x_k-x_{k-1})$ is also a difference of two Riemann sums. Once I realized this I wrote the answer. – Paramanand Singh Oct 31 '22 at 00:30
  • That's good (+1) – RRL Oct 31 '22 at 17:44