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This question is inspired by Rudin's Real and Complex Analysis Chapter 3 Problem 5 part (c), which asks for conditions that guarantee that $L^r(\mu)= L^s(\mu)$ for $0<r<s$ on a measure space $(X,\mu)$ which has $\mu(X)=1$. It is not hard to see that if $X$ consists of finitely many elements, the $\sigma$-algebra $\mathcal M$ is the power set, and $\mu$ is finite on $X$, then $L^p(\mu)=L^q(\mu)$ for any $p,q\in(0,\infty]$. We can also consider finite $\sigma$-algebras on arbitrarily large sets $X$, and should obtain the same result.

What I would like to know, as a curiosity, is if this is essentially the only non-trivial situation. My intuition says that it should be, for in the situation of an infinite (hence, uncountable) $\sigma$-algebra on a set $X$ with infinite cardinality, we should be able to construct a measurable function $f$ which lies in $L^q$ but does not lie in $L^r$ for $r>q$.

Arctic Char
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Lentes
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I cannot comment but I redirect you towards uforoboa answer here. In brief, if $X$ does not contains sets of arbitrary small values the reverse inclusion is possible (reverse of the one you proved in the precedent exercise)

Alexandre Sallinen
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  • Great! Then indeed the only possible case in which equality of the spaces can hold is when $X$ can be written as a finite union of atoms (otherwise, since the measure of the space is assumed finite, one must have sets of arbitrarily large positive measure). This is morally what I had in mind. Thanks! I am accepting your answer. – Lentes Oct 22 '20 at 21:33