Proving that if $f$ is Riemann integrable and $1/f$ is bounded then $1/f$ is Riemann integrable I have to prove the following 
Suppose  $f$ is Riemann integrable on $[a,b]$ and $1/f$ is bounded on $[a,b]$. Prove that  $1/f$  is Riemann integrable on $[a,b]$.
My attempt:
Since $1/f$ is bounded we have that: $p \ge1/f \ge M $ therefore  $p \le f \le M $. Then we know that:
$$|1/f(a) -1/f(b)|=|\frac{f(a)-f(b)}{f(a)f(b)}|\leq|\frac{f(a)-f(b)}{p^{2}}| $$
then we have that 
$$U(P,1/f)-L(P,1/f) \leq 1/p^{2}U(P,f)-L(P,f) \leq \epsilon/p^{2} \leq \epsilon $$
Am I right? and if not, Can you help me to fix the mistakes? I will go to sleep because here in my country is very late, but I promest that tomorrow I'll check the answers and I tell you if I get stuck :), I will appreciate the help  you give thank you :) 

Everything looks good.  Except we could have $p<1$ and so $\epsilon/p^{2}<\epsilon$ is not necessarily true, but irrelevant for the conclusion.

Let $P=\{x_{0},x_{1},\ldots,x_{n}\}$ be a partition of $[a,b]$ such that $|U(P,f)-L(P,f)|<\epsilon$ and put $g:=1/f$.  Since $$M^{g}_{j}=\frac{1}{m^{f}_{j}}$$ and $$m^{g}_{j}=\frac{1}{M^{f}_{j}},$$
and all of these quantities are finite by hypothesis, we have
$$|U(P,g)-L(P,g)|\leq\sum_{j=1}^{n}|M^{g}_{j}-m^{g}_{j}|\Delta x_{j}=\sum_{j=1}^{n}\left|\frac{M^{f}_{j}-m_{j}^{f}}{m^{f}_{j}M^{f}_{j}}\right|\Delta x_{j}\leq (m_{f})^{-2}|U(P,f)-L(P,f)|\leq (m_{f})^{-2}\epsilon.$$
