Suppose $f\in R(x)$ on $[a,b]$ and $\frac{1}{f}$ is bounded on $[a,b]$. Prove that $\frac{1}{f} \in R(x) \text{ on } [a,b].$
Where do I begin on this?
Obviously WTS that $f \in R[a,b] \implies \frac{1}{f} \in R[a,b]$. As $f$ is bounded implies that $|\frac{1}{f}|$ is bounded on [a,b]. I am getting stuck past this....
You can use the following
Theorem (Lebesgue’s Criterion for integrablility). Let $f:[a, b]\to\mathbb R$. Then $f$ is Riemann integrable if and only if $f$ is bounded and the set of discontinuities of $f$ has measure $0$.
Clearly, the set of discontinuities of $\frac1{f}$ is the same as for $f$. Since, $f$ is Riemann integrable, this set has measure zero. Lebesgue’s Criterion implies that $\frac1{f}$ is integrable.