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Prove that if $f$ is Riemann integrable on $[a,b]$ and $0<\rho \le f(x)$ for all $x \in [a,b]$ then $1/f$ is Riemann integrable on $[a,b]$.

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Hint, use $|\frac{1}{f(x)}-\frac{1}{f(y)}|\leq \frac{|f(x)-f(y)|}{\rho^2}$ to estimate the Riemann sums. –  ougao Dec 13 '12 at 14:22

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Let $\epsilon>0$. Since $f$ is integrable, by the Riemann Criterion, there exists partition $$\mathcal{P}=\left\{ a=x_0<x_1<...<x_{n-1}<x_n=b \right\}$$ such as that \begin{equation}U_{f,\mathcal{P}}-L_{f,\mathcal{P}}<\epsilon\rho^2\end{equation} Therefore, for arbitrary $x,y\in [x_{i-1},x_i]$, \begin{equation}\left|\frac{1}{f(x)}-\frac{1}{f(y)}\right|=\frac{\left|f(x)-f(y)\right|}{\left|f(x)f(y)\right|}\le\frac{\left|f(x)-f(y)\right|}{\rho^2}\le \frac{M_i(f)-m_i(f)}{\rho^2}\end{equation} Taking the supremum and infimum as $x,y\in [x_{i-1},x_i]$ yields, \begin{gather} M_i\left(\frac{1}{f}\right)-m_i\left(\frac{1}{f}\right)\le \frac{M_i(f)-m_i(f)}{\rho^2}\Rightarrow\\ U_{\frac{1}{f},\mathcal{P}}-L_{\frac{1}{f},\mathcal{P}}\le \frac{1}{\rho^2}(U_{f,\mathcal{P}}-L_{f,\mathcal{P}})<\frac{1}{\rho^2}\rho^2\epsilon=\epsilon\end{gather} Therefore $\frac{1}{f}$ is integrable by the Riemann Criterion

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As usual you can get a quick proof using the (Riemann-)Lebesgue criterion: a function $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable iff it is bounded and its set of discontinuities has measure zero. Here:

1) Since $0 < \rho \leq f(x)$ for all $x$, $0 \leq \frac{1}{f(x)} \leq \frac{1}{\rho}$ for all $x$: $\frac{1}{f}$ is bounded.

2) Since $f$ is nowhere zero, $\frac{1}{f}$ is continuous at exactly the same points as $x$. Since $f$ is assumed Riemann integrable, this common set of discontinuities has measure zero.

I should say that I like the other answer better -- it is more elementary and arguably more instructive. But I think it is nice to see this answer as an alternative.

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Agree with you…we do not actually now which of the equivalent definitions the OP is working with. –  user22705 Dec 13 '12 at 15:09

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