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 $[a,b]$.
Prove that $1/f$ is Riemann integrable on $[a,b]$
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5Hint, use $|\frac{1}{f(x)}-\frac{1}{f(y)}|\leq \frac{|f(x)-f(y)|}{\rho^2}$ to estimate the Riemann sums. – 2012-12-13
2 Answers
Let $\epsilon>0$. Since $f$ is integrable, by the Riemann Criterion, there exists partition $\mathcal{P}=\left\{ a=x_0
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0nevermind :) sorry – 2016-12-07
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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0Agree with you…we do not actually now which of the equivalent definitions the OP is working with. – 2012-12-13