Inverse of a positive $L^{1}$ function on a probability space

Question

Let $(\mathcal{X},\mu)$ be a probability space, and let $f$ be a real-valued, $L^{1}$ function which is also strictly positive $\mu$-a.e. so that we may consider its ($\mu$-a.e.) inverse function $\frac{1}{f}$.

What can we say about the behaviour of $\frac{1}{f}$? Specifically, is it an $L^{p}$ function for some $p\in[1,\infty]$?

Quite naively, since $$\left|\left|f\,\,\cdot\,\,\frac{1}{f}\right|\right|_{1}=1,$$ I would say that Holder's inequality ``is compatible'' with $\frac{1}{f}$ being in $L^{\infty}$. However, I am not able to prove it because I am just starting to (self-)learn measure theory, and I yet don't have the necessary familiarity with it. Therefore, any hint would be greatly appreciated.

Answer 1

Consider the interval $[0,1)$ equipped with Lebesgue measure and $$ f(x) = e^{-1/(1 - x)}. $$ Certainly, $f > 0$ and $f \in L^1$ but $1/f(x) = e^{1/(1 - x)}\not\in L^p$ for any $p\in[1,\infty]$.

Answer 2

On the real line the probability measure $\frac 1 4e^{-|x|/2} dx$ the function $e^{-|x|}$ belongs to $L^{1}$ and is positive but $\frac 1 f$ does not belong to $L^{p}$ (if $p \geq 1$).