Under what conditions do $L^p(\mu) = L^q(\mu)$ for $0Q) Let $\mu(X)=1$. Under what conditions do the spaces $L^p(\mu)$ and $L^q(\mu)$ contain the same functions?
I cannot think of any condition on $\mu$ for which this is true except to say that $L^q(\mu)\subset L^p(\mu)$ for $0<p<q$. Any suggestions?
 A: If there exists a sequence $(A_n)_{n \in \mathbb{N}}$ of disjoint measurable sets with $\mu(A_n) >0$, we have $$\mu(\bigcup_{n=1}^\infty A_n) = \sum_{n=1}^\infty \mu(A_n) \leq \mu(X) =1.$$
Thus, we can take a subsequence $(B_n)_n$ of $(A_n)_n$ with $0 < \mu(B_n) \le 2^{-n}$. Next define
$$f(x) := \sum_{n=1}^\infty \frac{1}{\mu(B_n)^{1/q}} 1_{B_n}(x).$$
Then we have
$$\int |f(x)|^p \, d \mu (x) = \sum_{n=1}^\infty \mu(B_n)^{1-p/q} \le \sum_{n=1}^\infty 2^{-n(1-p/q)}<\infty,$$
since $1-p/q>0$, but
$$\int |f(x)|^q \, d \mu (x) = \sum_{n=1}^\infty 1 = \infty.$$
In other words, $f \in L^p$, but $f\notin L^q$.
Thus, we have $L^p(\mu) = L^q(\mu)$ only if there are no such sequences of measurable sets. In this case, there exists a minimal choice $(A_m)_{m=1}^n$ with $\mu(A_m) >0$ and $\sum_{m=1}^n \mu(A_m)=1$ and for any measurable subset $B \subset A_n$ one has $\mu(B) =0$ or $\mu(B) = \mu(A_n)$. (Otherwise we could construct an infinite sequence $(B_n)_n$ of disjoint measurable sets with $\mu(B_n)>0$.)
Now we have the decomposition $$\mu(B) = \sum_{m=1}^n \mu_m(B),$$ where $\mu_1,\ldots,\mu_m$ are purely atomic measures given by $$\mu_m(B) := \mu(B \cap A_m).$$
