# Riemann integrable, bounded

• Apr 15th 2009, 07:06 PM
vaevictis59
Riemann integrable, bounded
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.
• Apr 15th 2009, 09:08 PM
Krizalid