Proving a piecewise function is not Riemann integrable

Question

$f: [0,1] \to \mathbb{R}$ where

\begin{equation} f(x)= \begin{cases} 4x^3 & \text{if } x \in \mathbb{Q}\\ 0 & \text{if } x \in \mathbb{R}\setminus\mathbb{Q} \end{cases} \end{equation}

Here, it is easy to argue $L(f;p) = 0$ hence $L(f) = 0 $ how do I evaluate the upper Riemann sum, hence the upper Riemann Integral?

Here given any $K^{th}$ subinterval $U(f;p) $ depends on the $p$ partition right? because there exist an irrational number in $K^{th}$ sub interval such that $U(f;p) > 0$ but I can't make the flow of the proof there rigorously.

Answer 1

Method 1: For an interval $[x_{i-1}\ ,x_i]\subseteq [0,1]$, $\sup\limits_{x\in[\ x_{i-1}\ \ \ ,x_i\ ]}f(x)=\sup\{\ 4x^3| \ x\in\mathbb{Q}\cap [x_{i-1},x_i]\}=4x_i^3$, and hence the upper Darboux sum for the partition $0=x_0\ <x_1\ <\cdots\ <x_n=1$ will be $$\sum_{i=1}^n 4x_i^3(x_i-x_{i-1})$$ which is precisely the same Darboux sum of the function $4x^3$ with the partition $0=x_0\ <x_1\ <\cdots\ <x_n=1$. Hence the upper Darboux sum converges to $\int_0^1 4x^3 \mathrm{d}x\ne 0$ as $\max(x_i-x_{i-1})\rightarrow 0$.

Method 2: Use Lebesgue's theorem. $f$ is nowhere continuous in $(0,1]\Rightarrow$ the Riemann integral $\int_0^1 f$ does not exist.