When does $L^r(\mu)=L^s(\mu)$?

Question

Full question:

Assume that $\mu(X)=1$. Prove that $L^r(\mu)\supset L^s(\mu)$ if $0<r<s$. Under what conditions do these two spaces contain the same functions?

I have already shown the first part is true. Which follows from the inequality $$||f||_r \leq ||f||_s \quad (0<r<s\leq \infty)$$

The second part is tricky. It is clear that $L^r(\mu)=L^s(\mu)$ when the only measurable functions are simple, but it isn't clear if there are any additional conclusions.

Answer 1

An exercise in Folland's "Real Analysis" states:

Suppose $0<p<q<\infty$. Then $L^p\not\subseteq L^q$ iff $X$ contains sets of arbitrarily small positive measure, and $L^q\not\subseteq L^p$ iff $X$ contains sets of arbitrarily large finite measure.

From this it would appear that you need to not have sets of arbitrarily small positive measure, like for the counting measure on $\mathbb{N}$ to form the $\ell^p$ spaces.