Let $(X,μ)$ be a measure space. Find a necessary and sufficient condition on $(X,μ)$ that $L_q(E) ⊂ L_p(E)$ for all $1 ≤ p < q ≤ ∞.$
I want to say that the condition is that $E$ is finite. This is clearly sufficient. Also it is clearly necessary for normal lebesgue measure. The counter example for arbitrary measure I want to use is $f(x) = \sum_{n = 1}^\infty \frac{1}{n^x}1_{E_n}(x)\mu(E_i)^{-1}$ where $xp < 1$ but $xq > 1$. Then it is p-test from calculus and done.
But this depends on writing $E = \sum_{n = 1}^\infty E_i$ with the $E_i$ finite and disjoint. Can I assume this? Thanks. If not, is there another approach.
Actually I don't think what I wrote made much sense since then $f$ is not measurable.
The wanted necessary and sufficient condition is that $$C:=\sup\{\mu(A), A\in\mathcal F, \mu(A)<+\infty\}<+\infty.$$
This can be found in Rudin's book Real and complex analysis.
Indeed, assume that $C$ is finite, and fix $f\in L_q(E)$. Define $E_k:=\{1/(k+1)^2\leqslant |f|\lt 1/k^2\}$. Then $$\int |f|^p\mathrm d\mu\leqslant \int_{\{|f|\lt 1\}}|f|^p\mathrm d\mu+\int_{\{|f|>1\}}|f|^q\mathrm d\mu\leqslant \sum_{k=1}^{+\infty}k^{-2p}\mu(E_k)+\int |f|^q\mathrm d\mu.$$ Since the sets $E_k$ have a finite measure, we have $\mu(E_k)\leqslant C$ for each $k$ hence $$\int |f|^p\mathrm d\mu\leqslant C\sum_{k=1}^{+\infty}k^{-2p}+\int |f|^q\mathrm d\mu<\infty$$ and we conclude that $L_q\subset L_p$.
Conversely, assume that $L_q\subset L_p$ for each $1\leqslant p\lt q\leqslant +\infty$, in particular $L_2\subset L^1$. Then the inclusion is continuous (this can be seen from the closed graph theorem): this means that there exists a constant $K$ such that for each $f\in L_2$, $\lVert f\rVert_1\leqslant K\lVert f\rVert_2$. Using this inequality when $f$ is the characteristic function of a measurable subset of finite measure, we obtain the finiteness of $C$.