Here is my proof using Riemann sum.
We know that by Archimedian property, for any $\epsilon>0$, there exists $n\in\mathbb{N}$ such that $\frac{1}{n}<\frac{\epsilon}{2}$. Let $A_n = \{x|g(x)\geq\frac{1}{n}\}.$ If $x\in A_n$, then $x\in\frac{j}{k}$ where $j\leq k\leq n$ and $gcd(j,k)=1.$ Hence, $A_n$ is finite. Let $B=\{i|A_n\cap[x_{i-1},x_i]=\emptyset\}.$ If $i\in B$, then $g(x_i^*)<\frac{1}{n}<\frac{\epsilon}{2}.$ If $i\notin B$, then $g(x_i^*)\leq 1.$
Then, for any $\epsilon>0$, choose $0<\delta = \frac{\epsilon}{2|A_n|}\leq 1$ such that for any partition satisfying $||P||<\delta$, and for any Riemann sum $R(g,P)$ of $g$ relative to $P$, we have
\begin{align}|R(g,P)-0|&=|\sum_{i=1}^{n}g(x_i^*)(x_i-x_{i-1})|\\& = |\sum_{i\in B}g(x_i^*)(x_i-x_{i-1})+\sum_{i\notin B}g(x_i^*)(x_i-x_{i-1})|\\&\leq |\sum_{i\in B}g(x_i^*)(x_i-x_{i-1})|+|\sum_{i\notin B}g(x_i^*)(x_i-x_{i-1})|\\&< |\sum_{i\in B}\frac{\epsilon}{2}(x_i-x_{i-1})|+|\sum_{i\notin B}|x_i-x_{i-1}|\\&\leq\frac{\epsilon}{2}\sum_{i\in B}|x_i-x_{i-1}|+\sum_{i\notin B}|x_i-x_{i-1}|\\&<\frac{\epsilon}{2}+|A_n|\frac{\epsilon}{2|A_n|}=\epsilon \end{align} as required.
Is my proof correct?