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,

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

If it does hold true, why would you need to impose conditions on $f$??? — mathworker21, Jan 13 '18 at 16:38
Take the zero function on $[a,b]$ to see the statement doesn’t hold true in general. — Anurag A, Jan 13 '18 at 16:40
I'm sorry my formation of the question is a bit ambiguous, edited. — physicscircus, Jan 13 '18 at 16:42
At least Some boundedness on $f$ And $f \neq 0$ would be required. — Anurag A, Jan 13 '18 at 16:42
See the following:https://math.stackexchange.com/questions/1103545/prove-the-reciprocal-of-a-function-is-integrable-if-it-is-bounded — Anurag A, Jan 13 '18 at 16:43
Most textbooks cover the integrability of reciprocal using the condition that original function is bounded away from $0$ ie there is some constant $k$ such that $|f(x) |>k$ for all $x\in[a, b] $. — Paramanand Singh, Jan 15 '18 at 05:38

Jack D'Aurizio · Accepted Answer · 2018-01-13T17:09:14.727

2

$f$ 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.

edited Jan 13 '18 at 17:09

answered Jan 13 '18 at 16:56

Jack D'Aurizio

353,855

I guess you need the additional condition $inf=d>0$? – physicscircus Jan 13 '18 at 16:58
@Abhikumbale of course, if $f$ is strictly bounded away from zero this issue cannot occur. – Jack D'Aurizio Jan 13 '18 at 16:59
You need the compactness of the interval! In that case, a positive and continous function has always positive minimum, thus respect the condition I gave in the answer below. – Andrea Marino Jan 14 '18 at 01:01

score 0 · Answer 2 · answered Jan 13 '18 at 18:09

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!

Questo commento si autodistruggerà tra poco, ma sappi che qualche divinità potrebbe punirti per aver trascritto Riemann con la lettera minuscola! — Jack D'Aurizio, Jan 13 '18 at 18:23
We don't need to invoke a difficult criterion like Lebesgue's. Usual conditions for integrability like the one based on Darboux sums suffices. If $f$ is bounded away from $0$ and say it's positive then $\sup(1/f)-\inf(1/f)=\dfrac{\sup f - \inf f} {\sup f\inf f} \leq K(\sup f-\inf f)$ for some constant $K$ and we are done with the proof. — Paramanand Singh, Jan 15 '18 at 05:36

Riemann Integrability of a function and its reciprocal

2 Answers2