Let $$\mathcal D=\left\{\frac{1}{n}\mid n\in\mathbb N^*\right\}\cup\{0\}.$$ Show that $\boldsymbol 1_{\mathcal D}$ is Riemann integrable over $[0,1]$.
Attempts
Let $f_n(x)=\boldsymbol 1_{\mathcal A_n}$ where $A_n=[0,\frac{1}{2^n}]\cup\bigcup_{k=1}^n\left[\frac{1}{k}-\frac{1}{2^n},\frac{1}{k}+\frac{1}{2^n}\right]\cap[0,1]$. We have that $\bigcup_{n\in\mathbb N^*}A_n=\mathcal D$, and that $(f_n)$ is a sequence of step function that converge pointwise to $\boldsymbol 1_{\mathcal D}$ and $(f_n)$ is decreasing. In particular, $$0\leq \overline{S}(f) \leq \frac{(n+1)}{2^n}\to 0.$$ Since $\underline{S}(f)=0$, the claim follow. I recall that $\overline{S}(f)=\inf_{\sigma }\{\overline{S}_\sigma\mid \sigma \text{ is a subdivision of }[0,1] \} $ and $\underline{S}(f)=\sup_{\sigma }\{\underline{S}_\sigma\mid \sigma \text{ is a subdivision of }[0,1] \} $
and $\overline{S}_\sigma =\sum_{i}M_i(x_{i+1}-x_i)$ and $\underline{S}_\sigma =\sum_{i}m_i(x_{i+1}-x_i)$ where $m_i=\min_{[x_{i},x_{i+1}]}f$ and $M_i=\max_{[x_i,x_{i+1}]}f$.
hint
Let $\epsilon>0$ given.
at $[\epsilon/2,1] $, $f $ is a step function, and it is integrable . thus there exist a subdivision $\sigma $ of $[\epsilon/2,1] $ such that
$U (f,\sigma)-L (f,\sigma)<\frac {\epsilon}{2} $
let $\sigma_1=\sigma \cup \{0\} $. $$U (f,\sigma_1)-L (f,\sigma_1)=$$ $$\leq \frac {\epsilon}{2}+1.\frac {\epsilon}{2} $$
since $f $ is bounded at $[0,\frac {\epsilon}{2}] $.