Riemann Integrability of a function and its reciprocal

Question

I was wondering what are the conditions which need to be imposed so that the question given below holds true,

Suppose $f$ is Riemann integrable on $[a,b]$ does this imply that $\frac{1}{f}$ is Riemann integrable?

The condition I can think of is,

1. $f(x)f(y)>0 \forall x,y \in [a,b]$

Answer 1

$f$ can be positive and continuous on $[0,1]$, without $\frac{1}{f}$ being Riemann-integrable over $[0,1]$.
Consider: $$ f(x)=\sum_{n\geq 1} n^2 \exp\left[-n^4\left(x-\frac{1}{n}\right)^2\right] $$ or just $$ f(x) = e^{-1/x^2}\text{ over }(0,1).$$

This function is continuous and positive but it is too close to zero in a right neighbourhood of the origin for $\frac{1}{f}$ to be Riemann-integrable.

Answer 2

A sufficient condition (but not necessary): there exist two positive constants $\epsilon, M$ such that $\epsilon < f(x) < M $ for every $x \in [a, b]$. We use the Lebesgue-integrability criterion:

*A bounded function $f$ on a compact interval is Riemann-integrable iff the set of discontinuity points, $D_f$, has null measure. *

Under the assumptions we made both $f, 1/f$ are bounded on a compact interval. $D_f$ has null measure by Riemann Integrability of $f$. But $D_{1/f}=D_f$ for bounded-from-below amd positive functions, thus $1/f$ is also Riemann integrable.

If $f$ is not bounded from below but still positive, then the notion of riemann integrability of $1/f$ is not properly defined.. I would say that a not bounded $f$ is r.i if $\min \{f, n\} $ is integrable with integral $I_n$, and $I_n$ converges to $I$. If this is the case, then we can reason as follows.

One can afford a "few" values closer and closer to zero, but not that much! Precisely, let's call $A_n := \{ 1/n <= f <= 1/(n-1)\} $, $f_n = \max \{f, 1/n\}$, $g_n=1/f_n$. Now $g_n$ is R. I. by the criterion I exposed. To achieve the convergence of $\int g_n $, we can require the convergence of the series $$ \sum_n n*\mu(A_n)$$ In this case, we have that $|\int g_{j+k} - \int g_k|$ is bounded by the final part of the series (why?), thus convergent to zero for $k \to \infty$.

I wait other people to confirm my rough definition of improper riemann integral!