Say $f(x)=\begin{cases}1&x\in E\\0&\text{elsewhere}\end{cases}$, where $E=\{1/n:n\in\mathbb{N}\}$.
My main roadblock right now is trying to find a suitable partition for $[0,1]$, where a partition of $[0,1]$ is a set $P=\{x_0,x_1,\dots,x_n\}$, where $0=x_0<x_1<\dots<x_n=1$. The most logical choice would be some variation of the set that was given, $E$, but $1/n$ is decreasing, and $1/n\neq0$, so this doesn't work. How should I partition $[0,1]$? Is there a better way of doing it?
On
It does not matter if it is decreasing. Given the set $A=\{1/2,...,1/n\}$, define your partition $P=\{x_0,x_1...,x_{2n+2}\}$ such that for every $k\in\{1,...,n\}$, $$\frac{1}{k}-\frac{\varepsilon}{2n}\in P$$ and $$\frac{1}{k}+\frac{\varepsilon}{2n}\in P,$$ where $0<\varepsilon < 1$. Of course, $0\in P$ and $1\in P$. Giving $x_i$ a proper order, we have $$s(f;P) = 0$$ and $$S(f;P) = \sum\limits_{i=1}^{2n+2}M_i(x_i-x_{i-1}).$$ Note that $M_1(x_1-x_0)<1/n$. And for $i\neq 1$, $M_i = 1$ if $[x_{i-1},x_{i}]$ is of the form $I_k=\left[1/k-\varepsilon/n,1/k+\varepsilon/n\right]$ and $M_i = 0$ otherwise. Since we have $(n-1)$ intervals of the form $I_k$ and their length are $\varepsilon/n$, we conclude that $$\sum\limits_{i=1}^{2n+2}M_i(x_i-x_{i-1})\leqslant \frac{1}{n}+(n-1)\frac{\varepsilon}{n} < \frac{1}{n}+\varepsilon.$$ Thus $$0\leqslant{\int_0^1}f(x)dx = \inf\limits_P \,(S(f;P))<\varepsilon$$ and since $\varepsilon$ can be arbitrarily small, $$\int^1_0f(x)dx =0.$$
If you're thinking Riemannly: What if your $k^\text{th}$ partition is $\left\{\frac{j}{k^k} \mid 0 \leq j < k^k \right\}$?
If you're thinking Lebesguely: What do you think of $[0,1] = \{0\} \cup \bigcup_{i \in \Bbb{N}} \left( (\frac{1}{i+1},\frac{1}{i}) \cup \{\frac{1}{i}\}\right)$?