Consider the function: $$f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if } x \in \mathbb{Q} \\ -1 & \mbox{if } x \notin \mathbb{Q} \end{array} \right.$$

I want to check if this function is Riemann integrable in an interval $[a, b]$. I want to calculate $\overline{S_{N}}$ and $\underline{S_{N}}$ which are the sums of all the areas of the upper and lower rectangles. I know that:

$\overline{S_{N}} = \sum_{i = 1}^{n}M_{i}\cdot \Delta x = \sum_{i = 1}^{n}1\cdot \Delta x = \Delta x \cdot \sum_{i=1}^{n}1 = \Delta x \cdot n = \frac{b-a}{n} \cdot n = b - a$

$\underline{S_{N}} = \sum_{i = 1}^{n}m_{i}\cdot \Delta x = \sum_{i = 1}^{n}(-1)\cdot \Delta x = -\Delta x \cdot \sum_{i=1}^{n}1 = -\Delta x \cdot n = -\frac{b-a}{n} \cdot n = a - b$


$n \in \mathbb{N}$ (is the number division points)
$\Delta x = \frac{b-a}{n}$
$M_{i} = max\{f(x) | x_{i-1} \leq x \leq x_{i}\}$
$m_{i} = min\{f(x) | x_{i-1} \leq x \leq x_{i}\}$

I see that for every $b \neq a$:

$$\lim_{n\to\infty}\overline{S_{N}} \neq lim_{n\to\infty}\underline{S_{N}}$$
What I don't understand is:

1) In the book that I'm reading they say that $\overline{S_{N}} = 1$ and $\underline{S_{N}} = -1$ for every $a, b$. Why?

2) The book also says that the function is not integrable at any point. Buy why? According to my calculations the function is integrable in each interval $[a]$ (I don't know how to phrase this).

  • $\begingroup$ Your book is wrong: your evalutaion of Riemann sums are correct. Moreover, point 2) is meaningless: integrability is NOT a local property, i.e. you can't say "$f(x)$ is integrable at $0$" for example. $\endgroup$ – Crostul Jun 18 '16 at 20:11
  • $\begingroup$ You don't say that a function is integrable at a point; you say that it is integrable on an interval. Some books consider $[a,a]$ to be a degenerate interval and technically this function is integrable on it, the value being zero. But since that is true for any function it isn't a very useful fact. $\endgroup$ – Elliot G Jun 18 '16 at 20:11
  • $\begingroup$ Can you give the title of the book you are reading ? $\endgroup$ – Tony Piccolo Jun 20 '16 at 5:29

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