Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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,

$$\int_a^b\alpha(x)g(x)\,dx=G(b)\alpha(b)-G(a)\alpha(a)-\int_a^bG\,d\alpha.$$

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 $x$.

share|improve this question
    
After you write out the discrete sums, you apply the mean value theorem (not the intermediate value theorem). –  Chris Janjigian Oct 12 '12 at 14:27
    
@chris. thanks. i will make the correction. –  Jyotirmoy Bhattacharya Oct 12 '12 at 16:07
    
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. –  Squirtle May 3 '13 at 5:16

2 Answers 2

up vote 1 down vote accepted

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

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.