# Prove by the definition that $\ln:[\varepsilon, 1]\to\mathbb{R}$ is Cauchy-integrable and compute its integral

I'll first define the Cauchy integral (which is merely a relaxed version of the Riemann integral).

(Partitions and their norms are defined as usual, just like in the definition of the Riemann integral.)

Definition. Given a function $f:[a, b]\to\mathbb{R}$ ($a < b$) and a partition $P = \{a = x_0 < x_1 < \dots < x_N = b\}$ of $[a, b]$, we define the Cauchy sum for $f$ corresponding to $P$ to be $C(f, P) = \sum_{n = 1}^Nf(x_{n-1})(x_n - x_{n-1})$.

Definition. A function $f:[a, b]\to\mathbb{R}$ ($a < b$) is Cauchy-integrable iff there exists $l\in\mathbb{R}$ such that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that if $P$ is a partition of $[a, b]$ with $|P| < \delta$, then $|C(f, P) - l| < \varepsilon$. We call $l$ the Cauchy integral of $f$. (One can prove $l$ is unique, if it exists.)

For more on the Cauchy integral, see http://emp.byui.edu/BrownD/Mathematics/Calculus-R-R/Calc-R-R-Advanced/Cauchy-integral-intro.pdf

Now here's the exercise I'm having trouble with:

Exercise. Suppose $0 < \varepsilon < 1$ and consider the function $\ln:[\varepsilon, 1]\to\mathbb{R}$. Prove by the definition that such function is Cauchy-integrable and that its Cauchy integral is $-1 + \varepsilon - \varepsilon \ln(\varepsilon)$.

(This exercise comes from a course in Measure and Integration I'm taking right now. Our professor has defined the Cauchy integral and others for historical and pedagogical reasons.)

Of course, I could argue thus: since $\ln:[\varepsilon, 1]\to\mathbb{R}$ is continuous, it must be Riemann-integrable (and, therefore, Cauchy-integrable) and its integral can be computed using the Fundamental Theorem of Calculus. However, the exercise requires me to do it with only the definition of the Cauchy integral. It's not clear to me how to proceed here.

Thoughts?

• Possible duplicate: see the question. Sep 4, 2013 at 22:20
• I don't think this is a duplicate. In the question you mention it is taken for granted that the function $\log$ is Riemann-integrable and it is suggested to use one specific Riemann sum. Sep 4, 2013 at 22:31

## 1 Answer

I would do it as follows (writing "$\log$" instead of "$\rm ln$").

First, note that for any function $f:[a,b]\to\mathbb R$ and any partition $P=(x_0,\dots ,x_N)$, we may write \begin{eqnarray}C(f,P)&=&\sum_{n=1}^N f(x_{n-1})(x_n-x_{n-1})\\ &=&\sum_{n=1}^{N-1} x_n(f(x_{n-1})-f(x_n))+x_Nf(x_{N-1})-x_0f(x_0)\\ &=&\sum_{n=1}^{N-1} x_n(f(x_{n-1})-f(x_n))+bf(x_{N-1})-af(a)\, . \end{eqnarray} When specialized to $f={\log}$ on $[a,b]=[\varepsilon,1]$, this becomes $$C({\log},P)=\sum_{n=1}^{N-1} x_n(\log(x_{n-1})-\log(x_n))+\log(x_{N-1})-\varepsilon\log(\varepsilon)\,.$$ By the mean value theorem, for each $n\in\{ 1,\dots ,N-1\}$ one can find $c_n\in (x_{n-1},x_n)$ such that $\log(x_{n-1})-\log(x_n)=\frac{1}{c_n} (x_{n-1}-x_n)$. This gives $$C(\log, P)=\sum_{n=1}^{N-1} \frac{x_n}{c_n} (x_{n-1}-x_n)+\log(x_{N-1})-\varepsilon\log(\varepsilon)\,.$$

Now, the idea is that if $\vert P\vert$ is small, then all $x_n/c_n$ are close to $1$ (because $c_n\in (x_{n-1},x_n)$) and $x_{N-1}$ is close to $x_N=1$ (so that $\log(x_{N-1})$ is close to $0$). Hence, we get \begin{eqnarray}C(\log, P)&\sim& \sum_{n=1}^{N-1} (x_{n-1}-x_n)-\varepsilon\log(\varepsilon)\\ &=&x_0-x_{N-1}-\varepsilon\log(\varepsilon)\\ &\sim&\varepsilon-1-\varepsilon\log(\varepsilon)\, . \end{eqnarray}

I'm pretty sure that you will be able to write down the missing details.

• Showing me how to rewrite $C(f, P)$ in terms of differences $f(x_{n-1}) - f(x_n)$ and then how to apply the Mean Value Theorem to get a more workable expression was nice. This technique will probably help me solve similar problems in the future. Thank you very much for taking the time help me, Etienne! (As a side note, I must say this is not the first time I'm unable to solve an exercise and later I find out that a solution involves the MVT. I definitely have to increase MVT's priority in my "techniques-to-try" queue.) Sep 6, 2013 at 14:02
• The MVT is indeed sometimes helpful! Sep 6, 2013 at 17:16