Is $1/f$ integrable when $f$ is finite on compact sets, absolutely continuous and $f \geq 1$?

Question

Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is an absolutely continuous function, such that $f \geq 1$ and such that it is finite on any compact set.

Can I conclude that $1/f$ is integrable on $\mathbb{R}^n$? If not, what condition could I add to make this true?

Answer 1

Seems like you need more then, like some conditions specifying how $f$ behaves at infinity. For example as it is $f(x)=1$ satisfies your conditions, but is not integrable. If you have a condition like $|f(x)| > C \|x\|^\alpha$ for all $x$ with $\|x\|>M$, $M,C$ positive constants, and $\alpha>1$, then you can get a bound on the tail of $1/f$.

Answer 2

You can take $f$ constant equal to 1. The function $1/f$ is not integrable on ${\bf R}^n$.