Definitions

Suppose $P\colon a=x_0<x_1<\dotsb<x_n=b$ is a partition of $[a,b]$. Let $\Delta x_k=x_k-x_{k-1}$ and $\lVert P\rVert$ denotes $\max_{0<k\le n}\Delta x_k$.

The Cauchy integral of a function $f$ on closed interval $[a,b]$ equals to $I$ if and only if for each $\epsilon>0$, there's some $\delta>0$, for each partition $P$ of $[a,b]$ such that $\lVert P\rVert<\delta$, we have $\left\lvert\sum_{k=1}^nf(x_k)\Delta x_k-I\right\rvert<\epsilon$.

Problem

If $f$ is bounded on $[a,b]$ whose Cauchy integral equals to $I$, then $f$ is Riemann-integrable and $\int_a^bf=I$.

Background

It's an exercise from our calculus(analysis) problemset book, and there's a hint: consider the partitions whose $x_k-x_{k-1}$ is a constant for different $k$'s, and try to estimate the Riemann sum for each of these partitions through the Cauchy integral.

I have no idea about such estimation. After drawing some pictures, I discouraged. I googled on the Internet and found an article. I realized that it's a quite different approach and with some advanced techniques (such as the analysis of a positive measure set -- discontinuities). I hope there will be some simpler approachers, just as the hint says. I need a more detailed hint, or a solution. Can anybody help me? Thanks!

D.C.Gillespie proved the theorem in 1915 (Annals of Mathematics, Vol.17) and what a proof !
To propose the proof as an exercise in a calculus book seems rather strange ...

However see exercise 2.1.19 in Bressoud's A Radical Approach to Lebesgue's Theory of Integration. There is a hint on page 300. Can it help ?

See also theorem 1 in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505.

But the story is the same !

I think you should try to use the definition of the Riemann integral and work with the Darboux sums: here.

I think for a calculus class/book, this might be overcomplicated. Though, before having a proper measure theory course, I was taught that a function is integrable if it has a limited number of discontinuities (or class I and II discontinuities), apart from being bounded.

$$\sum_{k=1}^{n}\left | f(t_{k})\cdot \bigtriangleup x_{k} - I \right | = \sum_{k=1}^{n}\left | f(t_{k})\cdot \bigtriangleup x_{k} - f(x_{k})\cdot \bigtriangleup x_{k} + f(x_{k})\cdot \bigtriangleup x_{k} - I \right | \leq \sum_{k=1}^{n} \left \{ \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} + \left | f(x_{k})\cdot \bigtriangleup x_{k} - I \right | \right \} \leq \varepsilon + \sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} $$

Now, if $f(x)$ is continuous, then it is uniform continuous on $[a,b]$ so that we can pick up $P$ such that $max \left | f(t_{k}) - f(x_{k}) \right | \leq \frac{\varepsilon }{b-a}$ and then: $$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} \leq \frac{\varepsilon }{b-a} \sum_{k=1}^{n} \bigtriangleup x_{k} = \varepsilon$$

If $f(x)$ is bounded and has a limited number of discontinuities, then we can split $[a,b]$ into sub-segments and "make" $f(x)$ continuous on those sub-segments. It should be enough to prove this for the case with one discontinuity, and use induction after that.

However, the necessary and sufficient condition states that the set of points of discontinuity should have measure of zero. If it is so, then: $$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k}$$ can be split into: $$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} \leq M \cdot \mu(D_{n-m}) + \sum_{i=1}^{m} \left | f(t_{k_{i}}) - f(x_{k_{i}}) \right |\cdot \bigtriangleup x_{k_{i}} $$ where summation is on the continuous part of the $[a,b]$, $M=max|f(x)|$, $D_{n-m}$ is the remaining part of $[a,b]$ containing the set of points of discontinuity and $\mu(D_{n-m})$ is its measure which goes to zero. So, this is the biggest question; how, using the fact that Cauchy integral is equal to $I$, to prove that the set of points of discontinuity has the measure of zero? I tend to agree with @Tony Piccolo answer ...