Let (X, S , $\mu$) be a $\sigma$-finite measure space and let D $\subseteq S$ a family of disjoint sets. Let $E \in S$ with $\mu(E)>0$ fixed. Prove that D$_E= \{D \in$ D$: \mu(E \cap D) >0 \}$ is at most countable.
The suggestion is to begin with the case $\mu(E) < + \infty$, but I am really lost, I don´t know how to do it.
In measure theory look for countability arguments:
$\mu(E) > 0$ is equivalent to $\exists n: \mu(E) > \frac{1}{n}$.
So for finite $E$, define
$$E_n = \{D \in \mathbf{D}: \mu(E \cap D) > \frac{1}{n}\}\text{,}$$
then
$$\mathbf{D}_E = \cup_n E_n$$
Now show that each $E_n$ is at most finite: $n$ of them many already have union $> 1$ by disjointness (so we can add the measures), so $kn$ many members of $E_n$ have measure $>k$. for any natural number $k$, and when $E$ is finite this must stop at some point (their union is a subset of $E$ so its measure is bounded by $\mu(E)$). So $\mathbf{D}_E$ is a countable union of finite sets, so countable for $E$ with $\mu(E) < +\infty$.
[More details added later:] Suppose $E_n$ consists of the sets $D_1, \ldots D_k$ from $\mathbf{D}$. By assumption the $D_i$ are pairwise disjoint (and so are the $E \cap D_i)$) and $\cup_{i=1}^n (E \cap D_i) \subseteq E$, hence $\mu(E) \ge \sum_{i=1}^k \mu(E \cap D_i)$, and being in $E_n$, means that $\mu(E \cap D_i) > \frac{1}{n}$, so $\mu(E) \ge \sum_{i=1}^k \mu(E_i) > k\cdot \frac{1}{n} = \frac{k}{n}$. So $k \le n\mu(E) < \infty$.
For $E = \cup_n F_n$ where $\mu(F_n) < +\infty$, so $\sigma$-finite, we have countably many countable intersections with the $F_n$ etc.