Problem: I am interested in the following question in order to be to apply Jensen's inequality to prove that under the assumptions below, we have $$\int_\Omega\log(|f|)\,d\mu\leq\log(\|f\|_p)\quad\text{where we assume }0<p\leq q<\infty\text{ with }f\in L^q(\Omega).$$ Let $(\Omega,\mathcal F,\mu)$ be a measure space and let $\mu(\Omega)=1$. If $f\in L^1(\Omega)$, do we have that $\log(|f|)\in L^1(\Omega)$?
My Troubles: After reading the following post: https://math.stackexchange.com/a/819742/595519, I tried using that criteria for integrability. But the set $\{\log(|f|)\leq-n\}$, which is the same as the set $\{|f|\leq e^{-n}\}$ gives me trouble since the fact that $f$ is integrable implies that this set may well have a positive measure for all $n\in\mathbb N$ which would make the series diverge. The same happens when I try to use the bound $1-1/x\leq\log(x)$ for $x>0$.
Therefore, I would like to ask if anyone knows of a correct way of showing that indeed $\log(|f|)\in L^1(\Omega)$ or whether we simply cannot conclude the result under the mere hypothesis that $f\in L^1(\Omega)$?
Thank you very much for your time and appreciate any help.
Here is a counterexample to your statement:
Consider $f(x)=e^{-\tfrac1x}$ for $0<x\leq1$ and $f(0)-0$.
This function is continuous and this integrable over $[0,1]$ (Lebesgue measure), yet the function $\log\circ f(x)=\frac{1}{x}$ for $0<x\leq1$ is not integrable over $[0,1]$.
It is true however, that $\int^1_0\log\circ f\leq \int^1_0 f$.
AS for the use of Jensen's inequality, for any $f\in L_1(\Omega,\mu)$, also long as $-\log\circ f\in L_1$ or $-\int_\Omega \log\circ f\,d\mu =\infty$, it holds that $$-\log\Big(\int_\Omega f\,d\mu\Big)\leq -\int_\Omega \log\circ f\,d\mu$$