# Proof of $\boldsymbol 1_{\{1/n\mid n\in \mathbb N^*\}}$ Riemann integrable on $[0,1]$.

A friend has to prove that $$\boldsymbol 1_{\{\frac{1}{n}\mid n\in \mathbb N^*\}}$$ is Riemann integrable on $$[0,1]$$ as Homework. I made the proof for him, but at the end the teacher put a grade of $$5/15$$. I made as follow :

I denote $$s_\sigma [0,1]=\sum_{i=0}^{n-1}m_i(x_{i+1}-x_i)\quad \text{and}\quad S_\sigma [0,1]=\sum_{i=0}^{n-1}M_i(x_{i+1}-x_i),$$ where $$\sigma =\{x_0,...,x_n\}$$ is a subdivision of $$[0,1]$$, $$m_i=\min_{[x_{i},x_{i+1}]}f\quad \text{and}\quad M_i=\max_{[x_{i},x_{i+1}]}f.$$ Then I denote $$s=\sup_\sigma s_\sigma [0,1]\quad \text{and}\quad S=\inf_\sigma S_\sigma [0,1].$$ The fact that $$s=0$$ is clear. I try to show that $$S=0$$.

Proof

Let $$m\in\mathbb N^*$$. Let a sequence of subdivision $$(\sigma _n)$$ of $$[\frac{1}{m},1]$$ s.t. $$|\sigma _n|\to 0$$. They have a theorem that says that $$\boldsymbol 1_{\{a_1,...,a_n\}}$$ is Riemann integrable over any interval of the form $$[a,b]$$. Let $$\tau_n=\{0\}\cup \sigma _n=\{y_0,y_1=x_0,y_2=x_1,...,y_{n+1}=x_n\}$$. It's a subdivision of $$[0,1]$$. Now we have that $$\sum_{i=0}^{n}M_i(y_{i+1}-y_i)\leq \frac{1}{m}+\sum_{i=1}^{n}M_i(y_{i+1}-y_i)=\frac{1}{m}+\sum_{i=0}^{n-1} (x_{i+1}-x_i)M_i.$$

Therefore, $$\lim_{n\to \infty }\sum_{i=0}^{n}M_i(y_{i+1}-y_i)\leq \frac{1}{m}+\lim_{n\to \infty }\sum_{i=0}^nM_i(x_{i+1}-x_i)=\frac{1}{m}.$$ Therefore, if we let $$m\to \infty$$, we get $$\lim_{n\to \infty }\sum_{i=0}^n M_i(y_{i+1}-y_i)=0,$$ and thus $$S=0$$.

Question : What's wrong here ? The comment of the teacher is : In your proof you implicitly consider $$\tau_{m,\infty }$$ as a partition of $$[0,1]$$ whereas it's not. But modify a bit your proof, you get that $$S\leq \frac{1}{m}$$ for all $$m$$, and thus $$S=0$$.

I agree that $$\tau_n$$ in my proof should have been $$\tau_{m,n}$$, but I don't understand his point when he says that I consider $$\tau_{m,\infty }$$ as a subdivion of $$[0,1]$$. Is my proof really wrong ?

The teacher's comment is correct, though I would say that $$5/15$$ is a bit harsh as your proof is almost correct, as pointed out by the teacher.
First of all, you haven't actually shown that the Riemann upper sum $$S$$ is $$0$$, as the infimum should be taken over all partitions of $$[0,1]$$. It seems that you intended to prove that there exists a sequence of partitions $$\tau_n$$ so that the sequence of upper sums (Darboux upper sum) associated the partitions tend to $$0$$. However, in your proof, $$\tau_n$$ depends on the choice of $$m$$, and for different $$m$$, your $$\tau_n$$ should be different. And the way that you phrased it seems to suggest that: $$S=\lim_{m\to\infty}\lim_{n\to\infty}\sum_{i=0}^n M_i(y_{i+1}-y_i)=0$$ Which is not correct because $$\tau_{m,n}$$ does not form a sequence of decreasing partitions. Instead, what you should have said was $$0=s\leq S\leq \frac 1m$$ for all $$m\in\mathbb N^*$$, each $$(\tau_{m,n})_{n=1}^\infty$$, for fixed $$m$$, is sequence of partitions so that the upper sum tends to $$\frac 1m$$.
• I see, but $$S\leq \lim_{m\to \infty }\lim_{n\to \infty }\sum_{i=0}^n M_i(y_{i+1}-y_i)=0$$ is correct, right ? – user657324 Mar 30 '19 at 12:12
• Yes, it is correct. I think "If we let $m\to \infty$, we get $\lim_{n\to \infty }\sum_{i=0}^n M_i(y_{i+1}-y_i)=0$, and thus $S=0$." might have led the teacher think that you are saying the limit on the left is $S$. That's also why I said your proof was almost correct, as it can be justified with slightly better wordings. – lEm Mar 30 '19 at 12:17