Riemann integrable, bounded

**vaevictis59** · April 15th 2009, 06:06 PM

Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $[a,b]$ and $\frac{1}{f}$ is bounded on $[a,b]$ . Prove that $\frac{1}{f}$ is Riemann integrable on $[a,b]$ .
Relevant Theorems/Definitions
Definition - Riemann integrable - if upper integral of $f(x)dx$ = lower integral of $f(x)dx$ .
Theorem - Riemann integrable iff for each $\epsilon > 0 \exists$ a partition $P$ of $[a,b]$ such that $U(P,f)-L(P,f) \leq \epsilon$ .
Theorem - Riemann integrable iff $\exists A \in \mathbb{R}$ such that $\forall \epsilon >0 \exists$ a partition $P$ of $[a,b]$ such that $\forall$ marked refinements $Q$ of $P$ , $|S(Q, f) - A | \leq \epsilon$ , where $A=\int^a_b f(x)dx$ .
Theorem - Let $f: [a,b] \rightarrow \mathbb{R}$ be a continuous function on $[a,b]$ . Then $f$ is Riemann integrable on $[a,b]$ .
etc.

I don't know how to start this problem. I need a head start on where to go with this one. Thanks in advance.

**Krizalid** · April 15th 2009, 08:08 PM

http://www.mathhelpforum.com/math-he...proof-1-a.html

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