Does $f\in L^1$ implies $\log{|f|}\in L^1$?
I think it is not since $|\log|f||$ can be arbitrarily big if $|f|$ is arbitrarily small.
But it seems $\log|f|\in L^1$ is true because of the following inequality,
$$\int_X\log|f|d\mu \leq \log\left(\int_X |f|d\mu \right)$$ if $\mu(X)=1$ due to Jensen's inequality.
I am stuck with it... Thanks in advance for any hint and answers!
What you said proves this holds for finite measure spaces, so instead consider $f(x) = \tfrac{1}{x^2}$ defined on $[1,\infty)$. Then $$\|f\|_1 = \int_1^\infty x^{-2}\,dx = \left. x^{-1}\right|_\infty^1 = 1$$ so $f\in L^1[0,\infty)$ but $$\int_1^\infty |\log f|\,d\mu = \int_1^\infty 2\log x\,d\mu = \infty$$ so $\log f \not\in L^1[0,\infty)$. Zero function works too (see comments) but this is a nontrivial example.