I think $L^\infty(\Omega, \mathcal{F}, \mu)\supseteq L^p(\Omega, \mathcal{F}, \mu), \forall p \in (0, \infty)$? My reason is $L^\infty$ is defined as the set of measurable functions that are bounded up to a set of measure zero, and if $f \notin L^\infty$, then there exists a subset of measure nonzero on which $|f|$ is $\infty$, so $f \notin L^p, \forall p \in (0, \infty)$.
So I wonder why "If $\mu$ is finite, then $L^\infty(\Omega, \mathcal{F}, \mu)\subseteq L^p(\Omega, \mathcal{F}, \mu)$ for each $p$"? If we both are right, then If $\mu$ is finite, $L^\infty = L^p$?
This is a spinoff of a reply to my earlier question.
Thanks!