About $L^p$ space

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

Let $$(X,\mu)$$ be a measure space and $$1\leq p, then $$\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.

• 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. – Kavi Rama Murthy May 3 at 10:32

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 $$L^{p}= L^{q}$$.