Let $(X, \mathcal{A})$ be a measure space and $\mu : \mathcal{A} \to [0,\infty]$ a $\sigma-$finite measure. Let $J$ be a non-empty index set and $(A_{j})_{j\in J}$ a disjoint Family in $\mathcal{A}$ where $\mu(A_{j}) > 0$ for all $j \in J$. Show that: $J$ is countable.
I am given the tip $(*)$ to first show that for $B \in \mathcal{A}$ with $\mu(B)< \infty$ that
$J_{n}(B):=\{j \in J: \mu(B \cap A_{j}) > \frac{1}{n} \}$ is finite
My ideas on $(*)$:
As $(X,\mathcal{A})$ is $\sigma-$finite there exists $(C_{n})_{n}\subseteq\mathcal{A}$ each of finite measure so that $X=\bigcup_{n \in \mathbb N}C_{n}$. Since $\mu(B)<\infty$, we can find $(C_{n})_{n=1}^{m}$ so that $B \subseteq\bigcup_{n =1}^{m}C_{n}$ and note that
$\mu((\bigcup_{1=1}^{m}C_{i}) \cap A_{j})\geq \mu(B \cap A_{j})$
and
$\mu((\bigcup_{1=1}^{m}C_{i}) \cap A_{j})=\mu(\bigcup_{i=1}^{m}(C_{i}\cap A_{j}))$
But I do not know how to continue.
Suppose that $J_{n}(B)$ is infinite. Take the union over $J$: $\begin{equation}\mu(\bigcup\limits_{j \in J} B \cap A_{j}) > \sum_{j \in J} \frac{1}{n} = \infty \end{equation}$. Contradiction with the fact that $\mu$ is $\sigma$-finite.