# Equivalence of Cauchy integral with Riemann integral

There has already been some discussion on this topic. However my question is about a specific solution to this problem and for the benefit of readers I think it is better to add some context (even though it means repetition of some stuff mentioned in the linked question).

In what follows $f$ is a function of type $f:[a, b]\to\mathbb{R}$ and $f$ is bounded. A partition $P$ of $[a, b]$ is a set of type $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ where $$a = x_{0} < x_{1} < x_{2} < \dots < x_{n} = b$$ The norm $||P||$ of partition $P$ is defined by $||P|| = \max_{i = 1}^{n}(x_{i} - x_{i - 1})$. We define the following sums for $f$ over $P$ \begin{align} C(f, P) &= \sum_{i = 1}^{n}f(x_{i - 1})(x_{i} - x_{i - 1})\notag\\ S(f, P) &= \sum_{i = 1}^{n}f(t_{i})(x_{i} - x_{i - 1})\notag\\ U(f, P) &= \sum_{i = 1}^{n}M_{i}(x_{i} - x_{i - 1})\notag\\ L(f, P) &= \sum_{i = 1}^{n}m_{i}(x_{i} - x_{i - 1})\notag \end{align} where $t_{i}$ are arbitrary points in $[x_{i - 1}, x_{i}]$ and $$M_{i} = \sup\,\{f(x)\mid x\in [x_{i - 1}, x_{i}]\},\,m_{i} = \inf\,\{f(x)\mid x\in [x_{i - 1}, x_{i}]\}$$ The sum $C(f, P)$ is called (left) Cauchy sum for $f$ over $P$. The Riemann sum $S(f, P)$ depends on choice of tags $t_{i}$ but this dependence in not shown in the notation and should be evident from the context. And finally $U(f, P), L(f, P)$ are upper and lower Darboux sums for $f$ over $P$.

Cauchy Integral: The function $f$ is said be said to be Cauchy integrable over $[a, b]$ with Cauchy integral $I$ if for every $\epsilon >0$ there is a number $\delta > 0$ such that $|C(f, P) - I| < \epsilon$ whenever $P$ is a partition of $[a, b]$ with $||P|| < \delta$.

A similar definition is available for Riemann integral if $C(f, P)$ is replaced by $S(f, P)$. Both these notions are equivalent and since every Cauchy sum is also a Riemann sum, the inference from Riemann to Cauchy is trivial. The converse appears to be hard and perhaps not popular enough to be seen in textbooks.

User Tony Piccolo in his answer gives three references for the proof that Cauchy integrability implies Riemann integrability.

It is the second proof from that answer which I want to discuss here (as other two proofs use somewhat complicated ideas and some very non-obvious tricks). This is provided as a hint that

Given any partition $P$ of $[a, b]$ and a number $\epsilon > 0$ there is a partition $Q\supseteq P$ of $[a, b]$ such that $C(f, Q) > U(f, P) - \epsilon$.

Using the counterpart equation $C(f, P) < L(f, P) + \epsilon$ we can easily show that difference $U(f, P) - L(f, P)$ can be made small if sums $C(f, P)$ tend to a finite limit and thus we get Riemann integrability (via Darboux integrability, also this link between Darboux and Riemann integral is popular and available in good textbooks).

Here are my questions:

It is easy to prove that we can choose tags $t_{i}$ such that $S(f, P) > U(f, P) - \epsilon$. We just have to choose tags so that $f(t_{i})$ is sufficiently near $M_{i}$. My hunch is that if we add the tags $t_{i}$ to $P$ we get a partition $Q\supseteq P$ and that is the needed partition which ensures $C(f, Q) > U(f, P) - k\epsilon$ where $k$ is some fixed positive constant. Is this correct? And if so how do we go about proving this?

Another doubt is whether the relation between $C(f, P)$ and $U(f, P)$ is valid in general? Or does it hold only for Cauchy integrable functions? My guess is that it holds only for Cauchy integrable functions. Is this correct?

• I suspect there is a typo in that reference. It would be enough to prove C-integrable implies R-integrable by showing given any $\epsilon > 0$ there exists a partition $Q$ such that $C(f,Q) > U(f,Q) - \epsilon$ and the corresponding result for the lower sum. See arxiv.org/pdf/1409.6770.pdf.
– RRL
May 13 '17 at 7:41
• @RRL: the proof in that paper does not appear to be any simpler than those available in other two referenced given by Tony Piccolo. This shows that the Cauchy integral is theoretically more difficult to work with compare to Riemann or Darboux formulation. May 13 '17 at 8:22
• I'm glad you refer to my answer to a question I consider rather intriguing. As for your previous comment, look at this question about Cousin's lemma .... it seems difficult to treat Cauchy sums. May 14 '17 at 10:37
• @TonyPiccolo I don't think that the failure of Cousin's lemma is essential. I believe that the conclusion of Cousin's lemma could be modified (I did not argue thoroughly) to be: for each $\epsilon>0$, there exists a partition $(a_i)$ such that $\sum_{i\in I}(a_{i+1}-a_i)<\epsilon$ where $I$ is the set of $i$ such that $a_{i+1}-a_i\ge\delta(a_i)$. That is to say, the total length of "bad" intervals could be arbitrarily small. Taking into account that the integrand in question is bounded, this is sufficient. Oct 9 '19 at 11:22

Let $f(x) = x$ on the interval $[0,1]$ and $P = (0,1)$.

Given any refinement $Q = (x_0,x_1, \ldots, x_{n-1},x_n)$ we have, since $f$ is increasing,

$$U(f,P) - C(f,Q) = 1 - \sum_{k=1}^n x_{k-1}(x_k - x_{k-1}) > 1 - \int_0^1x \,dx = 1/2.$$

Take $\epsilon < 1/2$ and we see that the conjectured result cannot be true.

• There could not be a simpler counter-example. +1 May 13 '17 at 8:20
• It's still messy to prove $C(Q,f) > U(Q,f) - \epsilon$. I'm glad you asked this because I saw that hint in the book some years ago and after giving up, never thought about it again.
– RRL
May 13 '17 at 8:25
• But otherwise Bressoud's book is very good and it contains lot of historical information which gives an insight into these difficult topics. May 13 '17 at 8:38
• I worked out an elementary proof independently. Seemingly simple enough. See my answer. Oct 9 '19 at 16:44

I come up with an elementary proof.

Let me first sketch the strategy. Given two real numbers $$a, a function $$f\colon[a,b]\to\mathbb R$$, and an interval $$I\subseteq[a,b]$$, we denote by $$\DeclareMathOperator\osc{osc}\osc(f,I)=\sup f(I)-\inf f(I)$$ the oscillation of $$f$$ on $$I$$.

Theorem (du Bois-Reymond) A function $$f\colon[a,b]\to\mathbb R$$ is Riemann integrable if and only if it is bounded, and for each $$\epsilon,\delta>0$$, there exists a partition $$P=(a=x_0\le x_1\le\dots\le x_n=b)$$ (allowing endpoints equal does not affect anything) of $$[a,b]$$ such that the total length of the intervals on which the oscillation of the function $$\ge\epsilon$$ is less than $$\delta$$, that is to say, $$\sum_{\osc(f,[x_i,x_{i+1}])\le\epsilon}(x_{i+1}-x_i)<\delta$$.

As indicated in OP, we denote by $$C(f,P)$$ the Cauchy sum of a function $$f$$ associated to a partition $$P$$, i.e. $$\sum_{i=0}^{n-1}f(x_i)(x_{i+1}-x_i)$$. We prove the contraposition: suppose that $$f$$ is not Riemann integrable, then by the du Bois-Reymond's characterization, we can pick up $$\epsilon,\delta>0$$ such that for each partition $$P$$ of $$[a,b]$$, the total length of the intervals on which the oscillation of the function $$\ge\epsilon$$ is at least $$\delta$$. Then for each partition $$P$$ of $$[a,b]$$, we construct two refinements $$P',P''$$ of $$P$$ such that the difference of Cauchy sums $$\lvert C(f,P')-C(f,P'')\rvert$$ is at least a constant which only depends on $$a,b,f,\epsilon,\delta$$ but not on the choice of $$P$$, then $$f$$ is not Cauchy-integrable.

# Observations of the simplest case

To do this, we start with the simplest case: consider the trivial partition $$P=(a=x_0. By assumption, $$\osc(f,[a,b])\ge\epsilon$$ and that $$b-a\ge\delta$$. It is natural to pick $$u,v\in[a,b]$$ such that $$f(u)\approx\inf f([a,b])$$ and $$f(v)\approx\sup f([a,b])$$. If $$u, then we consider two refinements $$S=(a\le u\le b)$$ and $$T=(a\le u of $$P$$. Then $$C(f,T)-C(f,S)=(b-v)(f(v)-f(u))$$. Similarly, if $$u>v$$, we consider two refinements $$S=(a\le v\le b)$$ and $$T=(a\le v of $$P$$. Then $$C(f,S)-C(f,T)=(b-u)(f(v)-f(u))$$. In both case, we can choose two refinements so that the difference of Cauchy sums is $$(b-\max(u,v))(f(v)-f(u))\approx(b-\max(u,v))\osc(f,[a,b])\ge(b-\max(u,v))\epsilon$$.

It would be a good choice if $$b-\max(u,v)$$ is large enough. In fact, if we modify the choice a bit, namely taking $$f(u)\approx\inf f([a,(a+b)/2])$$ and $$f(v)\approx\sup f([a,(a+b)/2])$$, then the same argument gives us two refinements so that the difference of Cauchy sums is $(b-\max(u,v))(f(v)-f(u))\ge\frac{b-a}2(f(v)-f(u))\ge\frac\delta2(f(v)-f(u))\approx\frac\delta2\osc\left(f,\left[a,\frac{a+b}2\right]\right)$ If $$\osc(f,[a,(a+b)/2])$$ is large enough, that is to say, $$\ge\epsilon$$, then this difference is approximately no less than $$2^{-1}\delta\epsilon$$, we are done. However, this is of course not guarenteed.

The key idea is to bisect $$[a,b]$$ to obtain a new partition $$(a<(a+b)/2. If $$\osc(f,[a,(a+b)/2])\ge\epsilon$$, then we are done as above. Otherwise, the du Bois-Reymond's characterization forces $$b-(a+b)/2\ge\delta$$. We replace $$[a,b]$$ by $$[(a+b)/2,b]$$ and continue the argument as above. If, for example, $$\osc(f,[(a+b)/2,(a+3b)/4]<\epsilon$$, we again bisect $$[(a+b)/2,b]$$ and restrict to the sub-interval to proceed. This process eventually terminates because the lengths of the intervals we get $$\ge\delta$$, so the process of bisection cannot be interminable.

In the general case, we also repeatedly bisect intervals to refine a partition and then pick out intervals, on the left half of which the oscillation of the function is at least $$\epsilon$$. The total length of the intervals we pick out is at least a constant times $$\delta$$. Then in each of these intervals, we pick out two points approximating infinum and supremum respectively. There is another techical point here: there are two cases, either inf lives to the left of sup, or inf lives to the right of sup. We pick out those intervals such that inf always lives to the left of sup, or always to the right of sup, so that the accumulations of differences on those intervals do not "cancel". By pidgeon-hole principle, either the first choice or the second choice will work. See the following sketch for details.

# Sketch of details of the general proof

For an (closed) interval $$I$$, denote by $$L(I)$$ the left half of $$I$$, i.e. the closed interval from the left endpoint to the midpoint of $$I$$.

Denote by $$\mathcal I(P)$$ the intervals (cut by a partition $$P$$) on which the oscillation of the function $$f$$ is at least $$\epsilon$$, i.e. $$\osc(f,I)\ge\epsilon$$, and denote by $$\mathcal J(P)\subseteq\mathcal I(P)$$ the intervals $$I\in\mathcal I(P)$$ such that $$\osc(f,L(I))\ge\epsilon$$.

For each partition $$P$$, set $$P_0=P$$. We recursively construct a (finite) sequence of $$(P_n)$$: by the du Bois-Reymond characterization and the assumption that $$f$$ is not Riemann-integrable, the total length of intervals in $$\mathcal I(P_n)$$ is no less than $$\delta$$, i.e. $$l_n:=\sum_{I\in\mathcal I(P_n)}m(I)\ge\delta$$, where $$m(I)$$ is the length (i.e. the measure) of $$I$$. If $$g_n:=\sum_{I\in\mathcal J(P_n)}m(I)\ge\delta/2$$, then we terminate. Otherwise, we insert all midpoints of intervals $$I\in\mathcal I(P_n)\setminus\mathcal J(P_n)$$ into $$P_n$$, get a new partition $$P_{n+1}$$. Then we proceed from $$n$$ to $$n+1$$. Note that on one hand, $$l_{n+1}:=\sum_{I\in\mathcal I(P_n)}m(I)\ge\delta$$; on the other hand, by construction, we have bisected each $$I\in\mathcal I(P_n)\setminus\mathcal J(P_n)$$, the oscillation of the left half of which is less than $$\epsilon$$, $$l_n-l_{n+1}=(l_n-g_n)/2$$. Since by assumption, $$l_n\ge\delta$$ and $$g_n<\delta/2$$ (otherwise we terminate), we have $$l_n-l_{n+1}>\delta/4$$. We conclude that this procedure terminates eventually. We denote by $$Q$$ the $$P_n$$ we get when we terminate.

By construction, $$\sum_{I\in\mathcal J(Q)}m(I)\ge\delta/2$$. For each $$I\in\mathcal J(Q)$$, we pick two points $$s_I,t_I\in L(I)$$ such that $$f(s_I)\approx\inf f(L(I))$$ and $$f(t_I)\approx\sup f(L(I))$$. Since $$I\in\mathcal J(Q)$$, we have $$f(t_I)-f(s_I)\approx\sup f(L(I))-\inf f(L(I))\ge\epsilon$$. For each $$I\in\mathcal J(Q)$$, either $$s_I or $$s_I>t_I$$. Denote by $$\mathcal J_1\subset\mathcal J(Q)$$ the subset of intervals $$I\in\mathcal J(Q)$$ such that $$s_I, and by $$\mathcal J_2\subset\mathcal J(Q)$$ the subset of intervals $$I\in\mathcal J(Q)$$ such that $$s_I>t_I$$. We note that $$\sum_{I\in\mathcal J_1}m(I)+\sum_{I\in\mathcal J_2}m(I)=\sum_{I\in\mathcal J(Q)}m(I)\ge\delta/2$$. Hence, the total length of intervals in either $$\mathcal J_j$$ is at least $$\delta/4$$, $$j=1$$ or $$j=2$$. We insert $$\min(s_I,t_I)$$ for each $$I\in\mathcal J_j$$ into $$Q$$, obtaining $$Q'$$ as a refinement of $$Q$$. We insert both $$s_I$$ and $$t_I$$ into $$Q$$, obtaining $$Q''$$ as a refinement of $$Q'$$. Now let's calculate $$\lvert C(Q',f)-C(Q'',f)\rvert$$. As in the simplest case of $$P=(a, we deduce that $\lvert C(Q',f)-C(Q'',f)\rvert\ge\frac{\sum_{I\in\mathcal J_j}m(I)}2\epsilon\ge\frac{\delta\epsilon}8$ where the first inequality is deduced from the fact that $$f(t_I)-f(s_I)$$ is (approximately) no less than $$\epsilon$$ for $$I\in\mathcal J_j\subseteq\mathcal J(Q)$$. Q.E.D.

• Did you read this question ? Oct 10 '19 at 23:22
• @TonyPiccolo Ah, no, thanks for the reference. I believe that this proof should be discovered even before, I mean that there should be some document where this proof is presented, given that the du Bois-Reymond criterion is a powerful tool, it is natural to come up with this proof if one is very familiar with analysis. Personally, I saw an upvote of my ancient post, and now I am teaching a problem session so it is natural for me to rethink, and I succeeded to prove it independently on the metro. Oct 11 '19 at 8:38
• Is your intention to get a proof following Rosentrater's hint ? Oct 11 '19 at 17:46
• @TonyPiccolo No, I did not even hear of it (that is what I meant by independently), until you showed me your question. However, I did read the proof of fundamental theorem of calculus (at Tao's post, Proposition 92) the day before I came up with this proof, and I had the impression that it did stimulate my ideas (I don't know how). Oct 11 '19 at 20:14
• I'm asking if, after reading my post, you want to give a second proof starting from Rosentrater's hint. He refers to the du Bois-Reymond's condition by height-width bounds. It would be interesting to compare the two proofs. Oct 12 '19 at 15:59