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Show that $$f(x)=\begin{cases}\frac{1}{b}& x=\frac{a}{b}\in [0,1],a,b\in\mathbb Z\\ 0&x\in \mathbb R\backslash \mathbb Q \cap [0,1]\end{cases}$$ is integrable on $[0,1]$.

Let $$S_\sigma =\sum_{i=1}^n m_i(x_{i+1}-x_i)\quad \text{and}\quad S^{\sigma }=\sum_{i=1}^n M_i(x_{i+1}-x_i),$$ where $$M_i=\max_{ [x_{i+1},x_i]}f,\quad \text{and}\quad m_i=\min_{[x_{i+1},x_i]}f.$$

I have to show that $\overline{S}=\underline{S}$ where $$\overline{S}=\sup_{\sigma }S_\sigma \quad \text{and}\quad \underline{S}=\inf_\sigma S^\sigma .$$

Obviously $S_\sigma =0$ for all partition $\sigma $, and thus $\overline{S}=0$. But for $\underline{S}$ I have some problem. I just can't find $M_i$. But maybe something as :

I consider $ \sigma _n : 0<\frac{1}{n}<...<\frac{n-1}{n}<1$. Then I would say that $M_i\leq \frac{1}{i}$, and thus

$$S^{\sigma _n}\leq \frac{1}{n}\sum_{k=1}^n\frac{1}{k}\underset{n\to \infty }{\longrightarrow }0,$$ and thus, the claim follow. It work ?

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    $\begingroup$ Just to clarify: I suppose you mean Riemann integrable? $\endgroup$
    – Wojowu
    Dec 11, 2017 at 20:44
  • $\begingroup$ Yes !!! @Wojowu $\endgroup$
    – user380364
    Dec 11, 2017 at 20:47
  • $\begingroup$ The simplest way is to show that it is bounded and continuous except on a set of measure zero. Do you know that theorem? $\endgroup$
    – Zach Boyd
    Dec 12, 2017 at 2:54
  • $\begingroup$ @ZachBoyd: If it would be enough, then $\boldsymbol 1_{\mathbb Q\cap [0,1]}$ would be Riemann integrable, what is wrong, no ? $\endgroup$
    – user380364
    Dec 12, 2017 at 9:14
  • $\begingroup$ The function you cited is everywhere discontinuous, whereas the one from the problem is continuous except at the rationals, which is the difference. $\endgroup$
    – Zach Boyd
    Dec 12, 2017 at 10:15

1 Answer 1

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$f$ is integrable iff $\exists P$ for every $\epsilon >0$ such that $S^{\sigma}-S_{\sigma}<\epsilon$.

Choose $P=\{x_0, ....,x_n\}$ such that $x_i - x_{i-1} = \frac{b-a}{n}$.

$S_{\sigma} = 0$ on $P$ because for every interval $[x_i, x_{i+1}]$, $m_i=0$.

Also, since $f(x)<1$ for $x<1$, we also have $M_i < 1$ and so $$\sum_{i=1}^n M_i < n$$

Put $\sum_{i=1}^n M_i = M <n$

Therefore $S^{\sigma}-S_{\sigma} = \frac{b-a}{n}M - 0<\epsilon$ on $P$ for large enough $n$.

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  • $\begingroup$ In what your proof wouldn't work if $f$ would be $\boldsymbol 1_{\mathbb Q\cap [0,1]}$ ? Because as you made it, $\boldsymbol 1_{\mathbb Q\cap [0,1]}$ looks Riemann integrable... $\endgroup$
    – user380364
    Dec 12, 2017 at 9:13
  • $\begingroup$ I don't understand what you mean. Can you better explain? $\endgroup$
    – Yousif Mansour
    Dec 12, 2017 at 14:20
  • $\begingroup$ The fact that $M<n$ don't implies that $\frac{b-a}{n}M\to 0$ when $n\to \infty $. $\endgroup$
    – Surb
    Dec 12, 2017 at 16:03
  • $\begingroup$ @Surb you are right. I see now. But we can easily show that M < 2. The proof would hold then. $\endgroup$
    – Yousif Mansour
    Dec 12, 2017 at 16:34
  • $\begingroup$ It's even not true ! The best you can do is that $M=\mathcal O(\sqrt{n})$ but not better. $\endgroup$
    – Surb
    Dec 12, 2017 at 16:36

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