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if $ f\in\mathscr{R(\alpha})$ on $[a,b]$ and if $a<c<b$, then $ f\in\mathscr{R(\alpha})$ $[a,c]$ and $[c,b]$ and $$\int_{a}^{b}fd\alpha = \int_{a}^{c}fd\alpha + \int_{c}^{b}fd\alpha$$


does the implication work backwards too? as in

if $ f\in\mathscr{R(\alpha})$ $[a,c]$ and $[c,b]$ and $a<c<b$ then $ f\in\mathscr{R(\alpha})$ on $[a,b]$ and $$\int_{a}^{b}fd\alpha = \int_{a}^{c}fd\alpha + \int_{c}^{b}fd\alpha$$

i think it does and i've written a proof using upper/lower sums that seems trivial to me. but Rudin didn't write this as an "if and only if" statement so I'm suspicious that there might be a counterexample, perhaps with some weird discontinuity at $c$ for either $f$ or $\alpha$ that makes the backward implication not generally true.

  • I think you should be fine even with a discontinuity at $c$ as you only need to be continuous a.e. but this seems to be very subtle. Can you give the precise definitions of Riemann-integrable Rudin is using? – noctusraid Jan 19 '16 at 16:01
  • (Almost) quote from Wikipedia: "In general, the integral is not well-defined if $f$ and $\alpha$ share any points of discontinuity, but this sufficient condition is not necessary." (https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral). – Martín-Blas Pérez Pinilla Jan 19 '16 at 16:04

3 Answers3

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Counterexample from Counterexamples in Analysis: "functions $f$ and $\alpha$ such that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on both $[a,b]$ and $[b,c]$, but not on $[a,c]$". Let be $$f(x) = 0,\ 0\le x<1,\qquad f(x) = 1,\ 1\le x\le 2.$$ $$\alpha(x) = 0,\ 0\le x\le 1,\qquad \alpha(x) = 1,\ 1< x\le 2.$$ Then, $$\int_0^1 f\,d\alpha = 0,\qquad\int_1^2 f\,d\alpha = 1$$ and $$\int_0^2 f\,d\alpha$$ does not exist.

  • so if $f$ and $\alpha$ do not share a common point of discontinuity, it works? – Baroque Spiderman Jan 19 '16 at 20:51
  • @somedude, I think that if all the integrals exists (is necessary but not sufficient that $f$ and $\alpha$ do not share a common point of discontinuity) then the equality is true. Check your proof with this hypothesis and post a new question if you want. – Martín-Blas Pérez Pinilla Jan 20 '16 at 07:19
  • I just wanted put a comment, but my reputation is low... Anyway, I think above counterexample is wrong, since $f$ is integrable on $[0,2]$.(See exercise 3 of rudin). Also, isn't the theorem works backward? – quicksilver Mar 21 '16 at 07:29
  • @quicksilver, see yourself the book: http://www.kryakin.org/am2/_Olmsted.pdf. $f$ is Riemann integrable, but not Riemann-Stieltjes integrable. – Martín-Blas Pérez Pinilla Mar 21 '16 at 09:40
  • @Martín-Blas Pérez Pinilla, I uploaded an answer. Can you check it out? – quicksilver Apr 25 '16 at 06:52
  • Based on the definitions of Riemann Stieltjes integration, your answer can be correct or can be incorrect. See my answer for more details. – Paramanand Singh Apr 25 '16 at 09:30
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The problem boils down to a proper definition of Riemann-Stieltjes definition. In this case Tom M. Apostol gives much better description (in his Mathematical Analysis) compared to Walter Rudin (in his Principles of Mathematical Analysis).

Definition 1: Let $P = \{x_{0}, x_{1}, x_{2},\ldots, x_{n}\}$ be a partition of interval $[a, b]$ and let $t_{k}$ be a point in the sub-interval $[x_{k - 1}, x_{k}]$. Let $f$ and $\alpha$ be functions defined and bounded on $[a, b]$. A sum of the form $$S(P, f, \alpha) = \sum_{k = 1}^{n}f(t_{k})\{\alpha(x_{k}) - \alpha(x_{k - 1})\}$$ is called a Riemann-Stieltjes sum of $f$ with respect to $\alpha$ on $[a, b]$. We say that $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$, and we write "$f \in R(\alpha)$ on $[a, b]$" if there exists a number $I$ having the following property: For every $\epsilon > 0$ there exists a parition $P_{\epsilon}$ of $[a, b]$ such that for all partitions $P = \{x_{0}, x_{1}, x_{2},\ldots, x_{n}\}$ of $[a, b]$ with $P \supseteq P_{\epsilon}$ and for every choice of $t_{k} \in [x_{k -1}, x_{k}]$ we have $$|S(P, f, \alpha) - I| < \epsilon$$

When $\alpha$ is monotone on $[a, b]$ this definition coincides with the one based on upper and lower integrals given by Rudin.

Apostol mentions another definition of Riemann-Stieltjes integral in the exercises (Problem 7.3, page 174, Mathematical Analysis):

Definition 2: Let $f$ and $\alpha$ be defined and bounded on $[a, b]$. We say that $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$, and we write "$f \in R(\alpha)$ on $[a, b]$" if there exists a number $I$ having the following property: For every $\epsilon > 0$ there exists a number $\delta > 0$ such that for all partitions $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ of $[a, b]$ with norm $||P|| = \max_{k = 1}^{n}(x_{k} - x_{k - 1})< \delta$ and any choice of points $t_{k} \in [x_{k - 1}, x_{k}]$ we have $$|S(P, f, \alpha) - I| < \epsilon$$

Definition 1 is more general than definition 2 in the sense that if $f \in R(\alpha)$ on $[a, b]$ based on definition 2 then $f \in R(\alpha)$ on $[a, b]$ based on definition 1, but there are functions $f, \alpha$ such that $f \in R(\alpha)$ on $[a, b]$ according to definition 1 and $f \notin R(\alpha)$ on $[a, b]$ based on definition 2. An example is given in Problem 7.3(b), Page 174, Apostol' Mathematical Analysis which is integrable according to definition 1, but not according to definition 2 (this example is incidentally the same as the one given by Martin in his answer to the current question).

Moreover if $\alpha(x) = x$ (so that we are just talking about plain Riemann integrability of $f$ on $[a, b]$) then both the definitions are equivalent (proof is not trivial/obvious and is given in Problem 7.4, page 174, Apostol's Mathematical Analysis).

Apostol uses definition 1 in his text and proves the following theorem:

Theorem: Assume that $c \in (a, b)$. If two of the integrals in the equation below exist, then the third also exists and we have $$\int_{a}^{c}f\,d\alpha + \int_{c}^{b}f\,d\alpha = \int_{a}^{b}f\,d\alpha$$

Moreover at the end of this theorem Apostol mentions the following note:

Note: The preceding type of argument cannot be used to prove that the integral $\int_{a}^{c}f\,d\alpha$ exists whenever $\int_{a}^{b}f\,d\alpha$ exists. The conclusion is correct however.

Apostol gives proof of existence of $\int_{a}^{c}f\,d\alpha$ based on existence of $\int_{a}^{b}f\,d\alpha$ in case when $\alpha$ is of bounded variation on $[a, b]$. However the proof for general bounded function $\alpha$ is based on different idea (Cauchy's criterion of integrability of interval functions, see page 22, Functional Analysis by F. Riesz and B. Nagy).

Based on definition 1, the claim by OP is correct. Also the counter-example by Martin in his answer is wrong. But if we use definition 2, then Martin's counter-example is correct and OP's claim is wrong. Thus it is a matter of definitions. In my opinion one should use the more general definition 1 compared to the slightly restrictive (but the one given originally by Riemann) definition 2.

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There are two approaches defining Riemann-Stieltjes integrabilities.

  1. In Rudin's PMA, a function $f$ is r-s integrable w.r.t $\alpha$ when upper r-s integral and lower r-s integral coincides.

    In this case, backward implication also true. Furthermore, exercise 6-3 of rudin tells exactly when $f$ is r-s integrable w.r.t $\alpha$ under the condition $\alpha$ and $f$ have some same discontinuities. You can check Martin's counter-examlple is actually integrable in this definition.

  2. Second approach is limiting 'length of partition' to zero. So in this definition, $f$ is integrable if

    For given $\epsilon>0$, there exist $\delta>0$ and $L$ such that $|R(f,P,Z) - L|< \epsilon $

    if $||P||<\delta$

where $P=\{p_0,p_1, ... , p_n\}$ is a partition,$Z=\{z_0, ... z_{n-1}\}$ is a 'evaluation sequence' ,$R(f,P,Z)= \sum f(z_i)d\alpha _i$ is a riemann-stieltjes sum, $||P||= sup_{i<n}|p_i-p_{i-1}| $. In this case, backward implication is false and Martin's example is not integrable on $[0,2]$ since no matter 'small' $||P||$ is ,we can choose $\{z_i\}$ so that $R(f,P,Z)$ is either 0 or 1.

( Above discussion is due to 세아 from scieng.net)

quicksilver
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