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 $S=0$.


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 $\frac 1m$.

  • $\begingroup$ 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 ? $\endgroup$ – user657324 Mar 30 '19 at 12:12
  • $\begingroup$ 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. $\endgroup$ – lEm Mar 30 '19 at 12:17

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