Let $1 \leq p<\infty$ and $f \in L_\mu^p(X)$. Assume $\mu(X)=\infty$ and $\mu$ is a $\sigma$-finite measure. Prove that:
(a) For any $\varepsilon>0$ there is measurable set $E$ with $\mu(E)<\infty$ such that $|f(x)|<\varepsilon$ on $E$.
(b) If $s$ is a simple function then $s \in L_\mu^p(X)$ iff there is measurable set $E$ with $\mu(E)<\infty$ such that $s(x)=0$ for $x \notin E$.
I don't seem to have much to work with, I would like to let $E=\{x:|f(x)|<\varepsilon\}$, which would be measurable since $f \in L_\mu^p(X)$, but I don't know how to prove its measure is finite.
For a), let $$ X = \bigcup_{n = 1}^\infty X_n. $$ We may assume that $X_i \cap X_j = \emptyset$ if $i \neq j$.
Let $G_n = \{x \in X_n : \vert f(x) \vert < \epsilon\}$. (Hence every $G_n$ is a measurable set.) If $\mu(G_n) = 0$ for all $n$ then $$ \int_X \vert f\vert^p d\mu = \sum_{n=1}^\infty \int_{X_n} \vert f \vert^p d\mu = \sum_{n=1}^\infty \int_{X_n \smallsetminus G_n} \vert f \vert^p d\mu \geq \sum_{n=1}^\infty \int_{X_n \smallsetminus G_n} \epsilon^p d\mu = \epsilon^p \mu(X) = \infty, $$ a contradiction. Let $E = G_n$ for (say the smallest) $n$ such that $\mu(G_n) > 0$. The conclusion is that there exists a measurable set $E$ with $0 < \mu(E) < \infty$ such that $\vert f(x) \vert < \epsilon$ on $E$.