Exercise 1.2.2 (Limit of Riemann integrable functions is not necessarily Riemann integrable)

Let ${(q_i)}_{i\in \mathbb{N}}$ be the sequence of all rational numbers. Define the sequence ${(f_n)}_{n\in \mathbb{N}}$ given by

We now show that

(i)

For all $n\in \mathbb{N}$ the function $f_n$ is Riemann integrable with integral 0.
Pick an arbitrary $f_n$ and fix an $𝜖>0$ and $𝜖<\min <!--\; FUNCTION\; APPLICATION\; -->\{d(q_i,q_j):1\le i,j\le n\}$; we demonstrate that we can always find a majorizing piecewise constant integral of $f_n$ which is less that $𝜖>0$ (which naturally implies that $f_n$ is Riemann integrable with integral of 0). First, build $𝜖∕n$ small disjoint intervals $[-𝜖∕2n+q_1,q_1+𝜖∕2n],\dots ,[-𝜖∕2n+q_n,q_n+𝜖∕2n]$. Define a piecewise constant function $g$ which has value 1 on these intervals, and 0 elsewhere. It is easy to verify that $g$ majorizes $f_n$, and we have $$\begin{array}{llll}\hfill \text{p.c.}{\int \; <!--\; nolimits\; -->}_{[a,b]}g=& 0\cdot m\left({[a,b]\setminus \underset{i=1}{\bigcup }}^n\left[-\frac{𝜖}{2n}+q_i,q_i+\frac{𝜖}{2n}\right]\right)& \hfill & \\ \hfill & +1\cdot m\left(\left[-\frac{𝜖}{2n}+q_1,q_1+\frac{𝜖}{2n}\right]\right)& \hfill & \\ \hfill & +\dots +1\cdot m\left(\left[-\frac{𝜖}{2n}+q_1,q_1+\frac{𝜖}{2n}\right]\right)& \hfill & \\ \hfill =& 𝜖& \hfill & \end{array}$$

This automatically implies that any minorizing piecewise constant integral also has a value of less than $𝜖$; thus, both converge to 0.

(ii)

The function $f:=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}_n$ is not Riemann integrable.
It is easy to verify that the limit equals to

$$f:[a,b]\to \{0,1\}f(x)=\{\begin{array}{c}1\text{ if }x\in \mathbb{Q}\hfill \\ 0\text{ else }\hfill \end{array}$$

This function is not Jordan measurable: any function $h$ which minorizes $f$ has to be equal to 0 (in any interval $[c,d]\in [a,b]$ we can find a rational $c<q<d$ - thus, $h$ cannot be constant on this interval). Similarly, any majorizing $g$ has to be greater than 1. In other words, upper and lower piecewise constant integrals can never converge against each other.