Show that the function $f$ in riemann integrable

Question

Let $f:[0,1]\to R$ be a function with $x∈[0,1]$ that is $f(x)=1$ for $x=1/n, n∈N$ and $f(x)= 0 $ for $ x ≠1/n, n∈N$ Show that $f$ is riemann integrable.

Ideas: $\int_{0}^{1}f(x)dx≤1/n$ and work with that?

Answer 1

Let $\varepsilon >0$. For all $p\in\mathbb N$, set $$i_p:=\sup\left\{k\in \{0,...,p\}\mid\frac{k}{p}<\varepsilon \right\},$$ and $$M_i^p:=\sup_{[\frac{i}{p},\frac{i+1}{p}]}f\quad \text{and}\quad m_i^p:=\inf_{[\frac{i}{p},\frac{i+1}{p}]}f.$$

Then, for $p\in\mathbb N$, $$\frac{1}{p}\sum_{i=0}^{p-1}(M_i^p-m_i^p)=\frac{1}{p}\sum_{i=0}^{i_p}(M_i^p-m_i^p)+\frac{1}{p}\sum_{i=i_p}^{p-1}(M_i^p-m_i^p).\tag{1}$$

For the first sum of the RHS of $(1)$, observe that $$\frac{1}{p}\sum_{i=0}^{i_p}\underbrace{(M_i^p-m_i^p)}_{\leq 1}\leq \frac{i_p}{p}<\varepsilon.$$

For the second sum of the RHS of $(1)$, take $N\in\mathbb N$ s.t. $\frac{1}{n}<\varepsilon $ for all $n>N$.

Observe that for all $p$, $$\sum_{i=i_p}^{p-1}(M_i^p-m_i^p)\leq 2N.$$ Therefore, the second sum of the RHS of $(1)$ converges to $0$ whenever $p\to \infty $.

Therefore $$\lim_{p\to \infty }\frac{1}{p}\sum_{i=0}^{p-1}(M_i^p-m_i^p)\leq \varepsilon ,$$ for all $\varepsilon>0 $, and thus, $f$ is Riemann integrable.