# Riemann Integral as a limit of sum

One of the standard definitions of Riemann Integral is as follows:

Let $f$ be bounded on $[a, b]$. For any partition $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ of $[a, b]$ and any choice of points $t_{k} \in [x_{k - 1}, x_{k}]$ the sum $$S(P, f) = \sum_{k = 1}^{n}f(t_{k})(x_{k} - x_{k - 1})$$ is called a Riemann sum for $f$ over $P$ and tags $t_{k}$. The norm of $P$ denoted by $||P||$ is defined as $||P|| = \max_{k = 1}^{n}(x_{k} - x_{k - 1})$. A number $I$ is said to be Riemann integral of $f$ over $[a, b]$ if for any arbitrary $\epsilon > 0$ there exists a $\delta > 0$ such that $$|S(P, f) - I| < \epsilon$$ for all Riemann sums $f$ over any partition $P$ with $||P|| < \delta$. When such a number $I$ exists we say that $f$ is Riemann integrable over $[a, b]$ and we write $$I = \int_{a}^{b}f(x)\,dx$$

Note that if $f$ is Riemann integrable over $[a, b]$ then we can choose partition $P$ with points $x_{k} = a + k(b - a)/n$ and $t_{k} = x_{k}$ or $t_{k} = x_{k - 1}$. Thus if $I = \int_{a}^{b}f(x)\,dx$ exists then by definition we have $$\int_{a}^{b}f(x)\,dx = \lim_{n \to \infty}\frac{b-a}{n}\sum_{k = 1}^{n}f\left(a + \frac{k(b - a)}{n}\right)\tag{1}$$

My question is: does the converse hold? If the limit in $(1)$ exists for a certain function $f$ bounded on $[a, b]$ does it mean that $f$ is Riemann integrable over $[a, b]$ according to the definition of Riemann integrability mentioned above?

The reason I ask this question is that many introductory calculus textbooks try to be smart and sort of put $(1)$ as the definition of $\int_{a}^{b}f(x)\,dx$ and hope that they have given a much better definition of integral compared to $F(b) - F(a)$ where $F$ is anti-derivative of $f$.

Update: As can be seen from the answer by user mrf, the converse does not hold and the equation $(1)$ can not be used as a definition of Riemann integral for a bounded. Now I update my question with a pedagogic bent.

Why do many introductory calculus textbook try to define Riemann integral as a limit of sum as mentioned in $(1)$? Does it add any value in terms of pedagogy to teach something which is totally wrong?

• On what level is your book? I've also seen some introductory books that are truly no more than introduction - they clearly focus more on readability than on rigour. However, considering their target readerships is usually, like, high school students, I won't feel very uncomfortable with such "introductory" writing styles. – Vim Aug 31 '16 at 11:53
• @Vim: When I studied calculus in high school at age of 16 years then almost every book at that level in India used the same definition. Maybe things are different in other countries, but I am not sure. – Paramanand Singh Aug 31 '16 at 12:43
• In China there has always been a huge gap between high school calculus and college calculus. In high school calculus teaching there is literally no rigorous formulations of the concepts of limit, differentiation, continuity, etc. Everything is based on intuition. The concept of integration is primarily based on the "equidistant" Riemann sum as in your question, and we only deal with "good" functions in high school. The good news is that more and more provinces are beginning to abandon calculus in high school altogether. – Vim Aug 31 '16 at 12:57
• @vim: let's hope other countries follow suit. – Paramanand Singh Aug 31 '16 at 13:01

The converse does not hold. A simple counterexample is the characteristic function of $\mathbb{Q}$ with $a$ and $b$ rational.
You cannot use $(1)$ as a definition of Riemann integrability. On the other hand, if $f$ is continuous on $[a,b]$ you can prove that the limit on the RHS of $(1)$ exists (using uniform continuity), and you can call it the integral of $f$ over $[a,b]$. But this would be a very restricted concept, since you couldn't even integrate step functions.
Note that all this arguing about Riemann sums has nothing to do with primitives of $f$. It works also for integrals over $n$-dimensional domains.