EXTREMELY CONFUSED RIGHT NOW. I explain everything down below.

Let $f(x)$ be a function on the closed interval $[a,b]$. We define the limit of Riemann Sums $S_{P}=\sum_{k=1}^{n} f(c_{k})\Delta x_{k}$ on $[a,b]$ as the mesh $\mu{(P)}\rightarrow 0$ to be the number $I$ if

Given $\epsilon>0$, $\exists\delta$ such that for every partition $P$ of the interval $[a,b]$

$$\mu{(P)}<\delta \Longrightarrow |S_{P}-I|<\epsilon$$

for any $c_{k}$ we choose in the subintervals $[x_{k-1},x_{k}]$.}

\noindent If the definition holds, then the limit of a Riemann Sum would exist and we can write

$$\lim_{\mu{(P)}\to 0} \sum_{k=1}^{n} f(c_{k})\Delta x_{k}=I$$


A bounded function is integrable on $[a,b]$ if and only for every $\epsilon>0$ there exists a partition $P_{\epsilon}$ of $[a,b]$ such that $$U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$$

There are so many criteria for integrability, and when I read different textbooks I find different arguments -- at least that's what I have experienced. From two "definitions" above, which one is correct? Are both correct and Riemann Integrability can be defined in many ways? Is one definition better than the other even though all definitions are correct? How should I interpret this concept?

Thank you!

  • 1
    $\begingroup$ You need to get familiar with a few alternative definitions. The one you cite is based on Riemann sums. There is another based on Darboux sums. The criterion of integrability you mention is not a definition, but rather a theorem which can be proved using any chosen definition. There is another subtle point in definition. Your approach is based on limit as mesh of partitions tends to $0$. There is another approach where we are not dealing with smaller meshes, but finer and finer partitions. Again these forms of limits are equivalent for Riemann integral, but not in general. $\endgroup$
    – Paramanand Singh
    Commented Mar 12, 2021 at 16:06
  • 1
    $\begingroup$ The equivalence between various definitions of Riemann integral are handled nicely in many answers on this site. See for example this answer. $\endgroup$
    – Paramanand Singh
    Commented Mar 12, 2021 at 16:11

2 Answers 2


First, in the general case, there may be several distinct criteria for an object to have an interesting property. The different criteria may each correctly identify some of the objects having that property, but be mute on the other objects having that property. This leads to summary theorems of the form "If any of the following conditions hold: (1) ... (2) ... ... (13) ... Then (object) has (property)." So it is entirely reasonable to have several if-then criteria for an object to have a property.

This is not what is happening here.

Here, you have if-and-only-if criteria. (All definitions are if-and-only-if, even though the mathematical style is to write them as if-thens. For instance, from https://en.wikipedia.org/wiki/Continuous_function , "a function is continuous if arbitrarily small changes in its output can be assured by restricting to sufficiently small changes in its input". Strictly interpreting this sentence, this is "if arbitrarily small changes in a function's, $f$'s, output can be assured by restricting to sufficiently small changes in its input, then $f$ is continuous" and only gives an inference "... $\implies$ $f$ is continuous". In contraposition, it also gives an inference "$f$ not continuous $\implies$ ...". Notice that, strictly, there is no inference given of the form "$f$ is continuous $\implies$ ...". This obvious absurdity is removed when we establish that definitions written as implications are actually biimplications.)

In general, a text takes one definition as "the definition", then proves equivalence between that definition and other characterizations of the property under study.

This could be what is happening here, but it isn't. So what is happening here?

Here, you have the same criterion written in apparently different ways. Can we show this equivalence?

Let $\varepsilon > 0$ and let $\delta$ be chosen so that any partition, $P$, of $[a,b]$ satisfies $$ \mu(P) < \delta \implies |S_P - I| < \varepsilon/2 \text{.} $$ One detail that has been omitted is the choice of $c_k$ in each interval of $P$. In the definition we have been reciting through so far, the choice of the $c_k$ is universally quantified. For all choices of the $c_k$, $S_P$ and $I$ are close in value. This means the inequality holds if we make specific choices. Suppose $c_k$ is chosen to give the minimum (or, if $f$ is so wild it doesn't have a minimum, infimal) value of $f$ in each interval of $P$. Then $S_P = L(f,P)$. Choosing instead the maximal (or supremal) points, $S_P = U(f,P)$. Our inequality promises $$ |L(f,P) - I| < \varepsilon/2 $$ and $$ |U(f,P) - I| < \varepsilon/2 $$ giving $$ |U(f,P) - L(f,P)| = U(f,P) - L(f,P) < \varepsilon \text{.} $$

So the first definition is saying pretty much the same thing as the second: "any choice of the $c_k$" gives small difference between the Riemann sum and $I$" implies "the most extreme difference between choices of $c_k$, the upper sum choice of the $c_k$ and the lower sum choice of the $c_k$, yields small differences in the Riemann sum (because both sums are close to $I$)".

Going in the reverse direction: we have that $L(f,P_\varepsilon) \leq S_{P_\varepsilon} \leq U(f, P_\varepsilon)$ for any choice of the $c_k$, and we apply the squeeze theorem (as $\varepsilon \rightarrow 0$) to show that all three of these (or actually $L(...)$, $U(...)$, and all infinitely many $S_P$ as we range over the possible choices of the $c_k$) converge to the same value, $I$.

One might ask why any particular criterion is "natural" (i.e., is one we might choose as the definition), since once we have all these equivalences, any of them could be the definition. In the cases you cite, the definitions are modeled on two different ways to approach convergence. The first definition is modeled on the definition of convergence of a sequence in the limit, so can be taught very early during rigorous analysis. The second definition is modeled on an alternative definition of convergence of a sequence: that the limit infimum and limit supremum agree. (One nice property of these two limits is that they always exist, so we don't have to say "... both exist and agree".) Why do we have both ideas?

Convergence of a sequence requires a sequence, a collection indexed by a countable set. So, in some sense, the first definition is only useful if you are interested in convergence of a small (ordered) set of values. Limits infimum and supremum can be defined for uncountably large sets, so are useful in more settings that the first definition.

  • $\begingroup$ Much more thorough and informative treatment than my answer. +1 $\endgroup$ Commented Mar 11, 2021 at 14:14
  • $\begingroup$ At some recent point of my career teaching and writing mathematics, I decided to adopt a different mathematical style when writing definitions, in order to avoid the deceptive "if-then" style. For example, in defining integrability I would write: "To say that $f$ is integrable means ...." $\endgroup$
    – Lee Mosher
    Commented Mar 11, 2021 at 14:36

The two definitions are equivalent but the second one involving the upper and lower limits $(U(f,P_{\epsilon}),L(f,P_{\epsilon}))$ was given by Darboux.

We can show that a function is Riemann integrable iff it is Darboux integrable, thus the two are equivalent.

For a proof of this you can check this post-How to prove that a function is Riemann integrable if and only if it is Darboux integrable? .

Because of Riemann's contribution to the development of the theory of defined integrals, in bibliography the dominant term is Riemann Integrability.


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