Problem 17 of Chapter 6 of Rudin's Principles of Mathematical Analysis asks us to prove the following:

Suppose $\alpha$ increases monotonically on $[a,b]$, $g$ is continuous, and $g(x)=G'(x)$ for $a \leq x \leq b$. Prove that,


It seems to me that the continuity of $g$ is not necessary for the result above. It is enough to assume that $g$ is Riemann integrable. Am I right in thinking this?

I have thought as follows:

$\int_a^bG\,d\alpha$ exists because $G$ is differentiable and hence continuous.

$\alpha(x)$ is integrable with respect to $x$ since it is monotonic. If $g(x)$ is also integrable with respect to $x$ then $\int_a^b\alpha(x)g(x)\,dx$ also exists.

To prove the given formula, I start from the hint given by Rudin $$\sum_{i=1}^n\alpha(x_i)g(t_i)\Delta x_i=G(b)\alpha(b)-G(a)\alpha(a)-\sum_{i=1}^nG(x_{i-1})\Delta \alpha_i$$ where $g(t_i)\Delta x_i=\Delta G_i$ by the intermediate mean value theorem.

Now the sum on the right-hand side converges to $\int_a^bG\,d\alpha$. The sum on the left-hand side would have converged to $\int_a^b\alpha(x)g(x)\,dx$ if it had been $$\sum_{i=1}^n \alpha(x_i)g(x_i)\Delta x$$ The absolute difference between this and what we have is bounded above by $$\max(|\alpha(a)|,|\alpha(b)|)\sum_{i=1}^n |g(x_i)-g(t_i)|\Delta x$$ and this can be made arbitrarily small because $g(x)$ is integrable with respect to $x$.

  • $\begingroup$ After you write out the discrete sums, you apply the mean value theorem (not the intermediate value theorem). $\endgroup$ Oct 12, 2012 at 14:27
  • $\begingroup$ @chris. thanks. i will make the correction. $\endgroup$ Oct 12, 2012 at 16:07
  • $\begingroup$ Because it is increasing monotonically and is continuous it is the case that it is integrable, so I think that these assumptions are fine -- perhaps not optimal but none the less good enough. $\endgroup$
    – Squirtle
    May 3, 2013 at 5:16

2 Answers 2


Compare with the following theorem,

Theorem: Suppose $f$ and $g$ are bounded functions with no common discontinuities on the interval $[a,b]$, and the Riemann-Stieltjes integral of $f$ with respect to $g$ exists. Then the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and

$$\int_{a}^{b} g(x)df(x) = f(b)g(b)-f(a)g(a)-\int_{a}^{b} f(x)dg(x)\,. $$

  • $\begingroup$ The discontinuity assumption doesn't seem to be necessary (Theorem 7.6, Apostol's Mathematical Analysis). $\endgroup$
    – f10w
    Aug 24, 2022 at 10:50
  • $\begingroup$ More precisely, $f\in R(g)$ already implies that there is no common point of discontinuity. $\endgroup$
    – f10w
    Aug 24, 2022 at 13:17

You are right that continuity is a stronger hypothesis than needed. I haven't checked your proof in detail due to lack of time, but assuming $g$ to be continuous only simplifies the problem. See for example theorem $12.14$ in This book.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .