Problem
If $f(x) > 0$ and Riemann Integrable on $[a,b]$ is $\frac{1}{f(x)}$ Riemann integrable on $[a,b]$?
My attempt
One has to show that $$S(P) - s(P) < \epsilon$$ $$S(P) = \sum_{j=1}^NM_j(x_j - x_{j-1})$$ $$S(P) = \sum_{j=1}^Nm_j(x_j - x_{j-1})$$ Now I consider the difference between these two: $$S(P) - s(p) = \sum_{j=1}^N(M_j-m_{j-1})(x_j - x_{j-1})$$
How do I continue from here?
Here's a way to view the problem. Denote $M_j^f=\sup_{[x_{j-1},x_j]}f(x)$ and $m_j^f=\inf_{[x_{j-1},x_j]}f(x)$. Define $g(x):=\frac{1}{f(x)}$. Then it follows that $M_j^g=\frac{1}{m_j^f}$ and $m_j^g=\frac{1}{M_j^f}$ since $f(x)>0$. That is, assuming there exists $\alpha>0$ such that $m_j^f\geq \alpha >0$ for all $j$. In other words, if $f(x)\geq \alpha >0$ for all $x\in[a,b]$.
Then given $\epsilon>0$, there exists a partition $P$ such that $U(P,f)-L(P,f)<\alpha^2 \epsilon$. For that same partition, we have \begin{align} U(P,g)-L(P,g)&=\sum_{j=1}^n(M_j^g-m_j^g)(x_j-x_{j-1}) \\ &=\sum_{j=1}^n\left(\frac{1}{m_j^f}-\frac{1}{M_j^f}\right)(x_j-x_{j-1}) \\ &=\sum_{j=1}^n\left(\frac{M_j^f-m_j^f}{M_j^f m_j^f}\right)(x_j-x_{j-1}) \\ &\leq\sum_{j=1}^n\left(\frac{M_j^f-m_j^f}{\alpha^2}\right)(x_j-x_{j-1}) \\ &=\frac{1}{\alpha^2}(U(P,f)-L(P,f)) \\ &<\epsilon. \end{align} Unfortunately, we cannot assume $f(x)\geq \alpha>0$ so any such counterexample would have to violate this.