Suppose $f$ is Riemann integrable, and $0<m≤|f|≤M$. Then, Prove $1/f$ is Riemann integrable.
My thought is this.
For any $\varepsilon >0$, there exists partition $\Pi = \{a = x_0, ... ,x_ₚ = b\}$ satisfies $U(f, \Pi) - L(f, \Pi) < m^2\varepsilon$.
$U(\pi , f) - L(\pi , f) = \sum_{j=1}^{p}[M_j(f) - m_j(f)]\Delta x_j<m^2\varepsilon$,
when $M_j(f)$ is sup of $f$, $m_j(f)$ is inf of $f$, $\Delta x_j$ is $x_j - x_{j-1}$ on [$x_{j-1}, x_j$].
$M_j(f)-m_j(f) = \underset{x \in [x_{j-1}, x_j]}\sup f(x) - \underset{y \in [x_{j-1}, x_j]}{inf}f(y) = \underset{x, y \in [x_{j-1}, x_j]}\sup |f(x)-f(y)|$.
$M_j(1/f) - m_j(1/f) = \underset{x \in [x_{j-1}, x_j]}{sup}{1/f(x)} - \underset{y \in [x_{j-1}, x_j]}{inf}1/f(y) = \underset{x, y \in [x_{j-1}, x_j]}{sup} |[1/f(x)]-[1/f(y)]| = \underset{x, y \in [x_{j-1}, x_j]}{sup} |\frac{f(x) - f(y)}{f(x)f(y)}| \leq \frac{M_j(f) - m_j(f)}{m^2}$
Then, $U(1/f, \Pi) - L(1/f, \Pi) = \sum_{j=1}^{p}[M_j(1/f) - m_j(1/f)]\Delta x_j \leq \sum_{j=1}^{p}[\frac{M_j(f) - m_j(f)}{m^2}]\Delta x_j <\frac{1} {m^2}\cdot m²ε = ε$.
Since $ 0 < \frac{1}{M} ≤ |\frac{1}{f}| ≤ \frac{1}{m}$, $1/f$ is bounded and holds Riemann's Condition. So, $1/f$ is integrable.
Is it right? Or does it need more condition?