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Suppose $f\in R(x)$ and $\frac{1}{f}$ is bounded on $[a,b]$. Prove that $\frac{1}{f}\in R(x)$ on $[a,b]$.

We need to show $U(P,f)-L(P,f)\leq\epsilon$ to prove Riemann Integrability.

To prove this function is Riemann integrable we take for granted that the function is bounded. So my approach is to let $\epsilon>0$ be given and $P=\{x_0,x_1,...,x_n\}$ be an arbitrary partition of $[a,b]$. By definition, $U(P,f)=\sum\limits_{i=1}^nM_i(f)(x_i-x_{i-1})$ and $L(P,f)=\sum\limits_{i=1}^nm_i(f)(x_i-x_{i-1})$. However, this is all arbitrary, and I'm not sure how to construct $M_i(f), m_i(f), U \text{ or }, L$ Should I be going about this another way? Maybe say $g=\frac{1}{f}$?

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Since $1/f$ is bounded (above), there is $Μ>0$ such that for all $x\in [a,b]$ we have $f(x)\geqslant M.$ Let $\varepsilon>0$. Since $f$ is integrable, there is a partition $P=\left\{ a=x_0<x_1<...<x_{n-1}<x_n=b \right\}$ of $[a,b]$ such that: $$U(f,P)-L(f,P)<\varepsilon M^2.$$ Since for $k=0,1,\ldots,n-1$: $$Μ_k\left(\frac{1}{f}\right)= \frac{1}{m_k(f)},~~~m_k\left(\frac{1}{f}\right)= \frac{1}{M_k(f)},$$ we get: $$Μ_k\left(\frac{1}{f}\right)-m_k\left(\frac{1}{f}\right)=\frac{M_k(f)-m_k(f)}{M_k(f)m_k(f)}\leqslant\frac{M_k(f)-m_k(f)}{M^2},$$since $M_k(f)\geqslant m_k(f)\geqslant M$. So: $$U\left(\frac{1}{f},P\right)-L\left(\frac{1}{f},P\right)\leqslant\frac{U(f,P)-L(f,P)}{M^2}<\varepsilon$$ and $1/f$ is integrable.

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    $\begingroup$ There's no specific motivation for $M^2$ is there? You could have written $3M$ instead, so long as it's not $M, \text{ or } 2M$ $\endgroup$ – help Nov 26 '19 at 8:17
  • $\begingroup$ $M^2$ appears during the evaluations, since $M_k(f)\geqslant m_k(f)\geqslant M$. So we would like to to cancel it out. Thats why we choose $P$ such that $U(f,P)-L(f,P)<\varepsilon M^2.$ If $3M$ appeared we would choose $P$ such that $U(f,P)-L(f,P)<3\varepsilon M.$ $\endgroup$ – Nikolaos Skout Nov 26 '19 at 8:25

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