Given $F(x): [0,1] \rightarrow \mathbb{R}$. $F(x) = 0, x \in \mathbb{Q}$ and $F(x) \ge \frac{1}{2}, x \not \in \mathbb{Q}$. Proof/Disproof $F(x)$ Riemann-Integrable!
My Attempt:
Consider that, $$M_i = \sup\;\{F(x): x_{i-1} \le x \le x_i\}$$ $$U(F;P) = \sum_{i=1}^{n} M_i (x_i - x_{i-1})$$ $$m_i = \inf\;\{F(x): x_{i-1} \le x \le x_i\}$$ $$L(F;P) = \sum_{i=1}^{n} m_i (x_i - x_{i-1})$$
Clearly from the definition that $M_i \ge \frac{1}{2}$ and $m_i = 0$. So we obtain $$U(F;P) \ge \frac{1}{2} \text{ and } L(F;P) = 0.$$
Since $U(F;P) \neq L(F;P)$, $F(x)$ is not Riemann-Integrable.
Is it correct?
Thanks in advance.