TITLE: When $L_p = L_q$?
QUESTION [5 upvotes]: As we know that $L_p \subseteq L_q$ when $0 < p < q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!

REPLY [17 votes]: This is essentially Exercise 6.5 in Folland's Real Analysis.  Let $(X,\mathcal{F},\mu)$ be a measure space, and let $$m = \inf\{\mu(A) : A \in \mathcal{F}, \mu(A) > 0\}$$
$$M = \sup\{\mu(A) : A \in \mathcal{F}, \mu(A) < \infty\}$$
For $0 < p < q < \infty$, it is a fact that $L^p(\mu) \subset L^q(\mu)$ iff $m > 0$, and $L^q(\mu) \subset L^p(\mu)$ iff $M < \infty$ (which in particular holds when $\mu$ is finite).
So for a finite measure $\mu$, it is necessary and sufficient that for some $m >  0$ every set either has measure 0 or has measure at least $m$.  This is going to force your space to be "effectively" finite in some sense.