# Proving a piecewise function is not Riemann integrable

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

• Use Riemann's criterion. Mar 18 at 6:27
• @geetha290krm Q asks to evaluate both sums first Mar 18 at 6:28
• Upper sum is the same as the one for $g(x)=4x^{3}$. Mar 18 at 6:35
• @geetha290krm is it okay to consider $g(x)$ separately and use the answer for $U(f)$ for this question? Mar 18 at 6:37

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 $$0=x_0\ will be $$\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\ . 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.