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In Rudin's book Real an complex analysis, we have

Let $(X,\mu)$ be a measure space and $1\leq p<q<\infty$, then $\mu(X)<\infty$ if and only if $L^q(X)\subset L^p(X)$.

Now, I want to know that what condition of probability measure $\mu$ (that is $\mu(X)=1$) such that $L^q(X)\equiv L^p(X)$ for given $q>q\geq 1$?

By previous result, we have $L^q\subset L^p$ by Holder's inequality or Jesen's inequality.

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    $\begingroup$ I think this happens iff there is no infinite sequence of disjoint sets in the sigma algebra with positive measure. In the case $L^{p}=L^{q}$ is finite dimensional. $\endgroup$ – Kavi Rama Murthy May 3 at 10:32
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Suppose there exists a sequence $\{A_n\}$ of measurable sets with $\mu(A_n)>0$ for all $n$. Then it is easy to find positive numbers $a_n$ such that $\sum a_n^{p} \mu(A_n) <\infty$ but $\sum a_n^{q} \mu(A_n) =\infty$. This means $f\equiv \sum a_nI_{A_n}$ is in $L^{p}$ but not in $L^{q}$.

Now suppose such a sequence does not exist. Let $f:X \to \mathbb R$ be any measurable function. Consider the disjoint sets $f^{-1}([n,n+1), n \in \mathbb Z$. It follows this set has measure $0$ for $|n|$ sufficiently large which means $f \in L^{\infty} \subset L^{r}$ for every $r >0$. Thus every measurable function belngs to every $L^{p}$, so $L^{p}= L^{q}$.

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