Riemann integrable (University of hyderabad 2017 exam)

Question

I know that the function is continuous at $x=0$ by sequential criteria but I can't figure out the Riemann integrability of this function:

Consider $f:[-1,1] \to \mathbb{R}$ defined by $$f(x)=\begin{cases} \sin x, & x \in [-1,1] \cap \mathbb{Q} \\ 0, & x \notin [-1,1] \cap \mathbb{Q} \end{cases}$$ Then at $x=0$, $f$ is

(A) continuous and Riemann integrable in every interval $(a,b)$ containing $0$.

(B) continuous and Riemann integrable in exactly one interval $(a,b)$ containing $0$.

(C) continuous but not Riemann integrable in any interval containing $x$

(D) discontinuos.

Answer 1

Take any partition $P$ of the interval $(a, b) \subset [-1, 1]$ with $0 \in (a, b)$ arbitrary. Clearly, the following holds $a < 0 < b$, therefore there exists $n' \in \mathbb{N}$ with $x_{n'}, x_{n' + 1} \in P$ s.t. $x_{n'} < 0$ and $x_{n' + 1} \geq 0$. $$\begin{align*} U(f, P) - L(f, P) &= \sum_{k = 1}^n(M_k - m_k)(x_k - x_{k - 1})\\ &=\sum_{k = 1}^{n'}(M_k - m_k)(x_k - x_{k - 1}) + \sum_{l=n' + 1}^n(M_l- m_l)(x_l - x_{l - 1})\\ &= \sum_{k = 1}^{n'}(0 - \sin(x_{k - 1}))(x_k - x_{k - 1}) + \sum_{l=n' + 1}^n\sin(x_l)(x_l - x_{l - 1}) \\ &=\sum_{k = 1}^{n'}(-1)\sin(x_{k - 1})(x_k - x_{k - 1}) + \sum_{l=n' + 1}^n\sin(x_l)(x_l - x_{l - 1}) &(1) \\ &\geq \sum_{l=n' + 1}^n\sin(x_l)(x_l - x_{l - 1}) &(2)\\ &\geq \sum_{l=n' + 1}^n \cfrac{2}{\pi}x_l(x_l - x_{l - 1}) &(3)\\ &> 2/\pi\sum_{l=n' + 1}^n 1/2(x_l + x_{l - 1})(x_l - x_{l - 1})\\ &= 1/\pi\sum_{l=n' + 1}^n(x_l^2 - x_{l - 1}^2) = 1/\pi \end{align*}$$ The left sum in the line (1) is greater than $0$, as each term is non-negative, i.e. $x_k - x_{k - 1} > 0$ for any $k \in [n]$, $-\sin(x_{k}) \geq 0$ for any $k \leq n'$. Also the inequality $\sin(x) \geq 2/\pi\cdot x$ for any $0 < x < \pi/2$ was used in the line $(2)$. Finally, note that the inequality $x_l > 1/2(x_l + x_{l - 1})$ for any $l > n' + 1$ was used in line $(3)$. Hence, we conclude that $f$ is not Riemann integrable.

Answer 2

The function $f$ is odd in $[-1,1]$ so it suffices to prove that it is not integradable in $[0,1]$. Clearly $L(f,P)=0$ for every partition $P$. But then if we take the partition $P_n$ of the interval $[0,1]$ with step $\frac{1}{n}$, we would have $\begin{align*} U(f,P_n) &=\frac{1}{n} \sum_{k=1}^{n}\sin(\frac {k}{n} )\\ &= \sin(1/2) \frac{1}{n}\left(\sin(1/2)\cot(1/2n)+\cos(1/2)\right) \end{align*}$ Taking the limit of the above expresion we get $1-\cos 1>0$. That proves the fact that $f$ is not integradable.