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?
Best Answer
The converse does not hold. A simple counterexample is the characteristic function of $\mathbb{Q}$ with $a$ and $b$ rational.