# Proving $\frac{1}{f}$ is Riemann Integrable

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 $$g=\frac{1}{f}$$?

## 1 Answer

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 $$P=\left\{ a=x_0 of $$[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.

• There's no specific motivation for $M^2$ is there? You could have written $3M$ instead, so long as it's not $M, \text{ or } 2M$ – help Nov 26 '19 at 8:17
• $M^2$ 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 $U(f,P)-L(f,P)<3\varepsilon M.$ – Nikolaos Skout Nov 26 '19 at 8:25