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Thread: Riemann integrable, bounded

  1. #1
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    Riemann integrable, bounded

    Suppose that $\displaystyle f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $\displaystyle [a,b]$ and $\displaystyle \frac{1}{f}$ is bounded on $\displaystyle [a,b]$. Prove that $\displaystyle \frac{1}{f}$ is Riemann integrable on $\displaystyle [a,b]$.
    Relevant Theorems/Definitions
    Definition - Riemann integrable - if upper integral of $\displaystyle f(x)dx$= lower integral of $\displaystyle f(x)dx$.
    Theorem - Riemann integrable iff for each $\displaystyle \epsilon > 0 \exists$ a partition $\displaystyle P$ of $\displaystyle [a,b]$ such that $\displaystyle U(P,f)-L(P,f) \leq \epsilon$.
    Theorem - Riemann integrable iff $\displaystyle \exists A \in \mathbb{R}$ such that $\displaystyle \forall \epsilon >0 \exists$ a partition $\displaystyle P$ of $\displaystyle [a,b]$ such that $\displaystyle \forall$ marked refinements $\displaystyle Q$ of $\displaystyle P$, $\displaystyle |S(Q, f) - A | \leq \epsilon$, where $\displaystyle A=\int^a_b f(x)dx$.
    Theorem - Let $\displaystyle f: [a,b] \rightarrow \mathbb{R}$ be a continuous function on $\displaystyle [a,b]$. Then $\displaystyle f$ is Riemann integrable on $\displaystyle [a,b]$.
    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.
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  2. #2
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