I have been working on this problem for a while, but I cannot seem to get it right each time I try to attempt it. Can anyone help?
This is the problem:
Suppose $f:[a,b] \to R$ is Riemann integrable, and suppose $g = \frac{1}{f}$ is well defined and bounded on $[a,b]$. Prove that $g$ is Riemann integrable on $[a,b]$.
This is what I have worked on so far:
Let $\phi(x) = \frac{1}{x}$. Note that $\phi$ is continuous for all $x \neq 0$. Since $f \in R[a,b]$, and $f$ is bounded, then $\phi \circ f$. Let $g= \phi \circ f$. Then there exists $M > 0$ such that $|f(x)| \le M$ for all $x \in [a,b] \implies f(x) \in [-M,M]$. $g$ is bounded so there exists $m > 0$ such that $|g(x)| \le m$ for all $x \in [a,b]$ which results into $|\frac{1}{f}| \le m \implies |f(x)| \ge \frac{1}{m}$ for all $x \in [a,b]$. Let $S$ be the union of two compact intervals as $S=\{y \in R \mid \frac{1}{m} \le |y| \le M \}$. $\phi$ is continuous because $0 \not\in S$. So $\phi: S \to R$ is continuous with $S$ compact. Using this Theorem which states "Suppose $f:[a,b] \to S$, $f \in R(x)$ on $[a,b]$, and $\phi: S \to R$ is continuous with $S$ compact. Then $\phi \circ f \in R(x)$ on $[a,b]$", we know that the composition of a continuous function with a compact domain and an integrable function with be integrable.
From here on out, I am struggling to find/prove a lemma to conclude this proof with $\phi \circ f \in R[a,b]$. Any help will be appreciated.