Show that if $L^p$ is not a subset of $L^q$ for $q \gt p$ then $L^p$ contains indicator functions of sets of arbitrarily small, yet positive measure.
Since $L^p$ is not a subset of $L^q$, there exists $f \in L^p$ such that $f \not \in L^q$.
For each $n_o \in \mathbb{N}$, Let $E_{n_0}=\{x \in X: |f(x)| \gt n_0\}$. Then we have $||f||_p^p=\int |f|^p d\mu \ge n_0^p \mu(E_{n_o})$ which implies that $\mu(E_{n_0}) \le \dfrac{||f||_p^p}{n_o^p}$. Now I claim that $\mu(E_{n_o}) \gt 0$ which will establish the claim.
Suppose that $\mu(E_{n_o})=0$. Now Let $F_1=\{x \in X : 0 \lt |f(x)| \le 1\}, F_2=\{x \in X : 1 \lt |f(x)| \le n_0\}$. Then $E_{n_0}^c=F_1 \cup F_2$. Now $\int_{F_1} |f| ^q \le \int_{F_1}|f|^p \lt \infty$ as $p \lt q$. Thus $\int |f|^q =\int_{E_{n_0}^c}|f|^q=\int_{F_1}|f|^q +\int_{F_2}|f|^q= +\infty$ which implies that $\int_{F_2}|f|^q=\infty$ and therefore $\mu(F_2) =\infty$.
On the other hand $\int |f|^p \ge \int_{F_2}|f|^p \ge \mu(F_2)=\infty $ which is a contradiction. Hence, the claim holds and we are done.
Is this alright?