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Problem

Let $\mu$ be a measure over $S$, $\sigma$-ring of $X$ . Let $E$ be a $\sigma$-finite set of $S$, and $\mathcal{D}\subset S$ such that $D_1,D_2\in\mathcal{D}$ with $D_1\neq D_2$ implies $D_1\cap D_2=\emptyset$.

Show that $\{D\in\mathcal{D} : \mu(D\cap E)>0\}$ is countable.

Attempt

Following the idea in Is a family of disjoints atoms in $\sigma$-finite neasurable space at most countable? Calling $\mathcal{F}_c=\{D\cap E : D\in\mathcal{D}, \mu(D\cap E)>c\}$, we just need to show that $$ \mathcal{F}_0 =\bigcup_{n=1}^\infty\mathcal{F}_{1/n}\quad \text{is countable}$$ How can I prove $\mathcal{F}_c$ is countable?

Note $\mu$ is not $\sigma$-finite, just $E$.

$E$ is $\sigma$-finite if there exists $\{E_n:n\in\mathbb{N}\}\subset S$ such that $E\subset \cup_n E_n$ and $\mu(E_n)<+\infty$ for all $n\in\mathbb{N}$.

P.D. Would you suggest me an appropriate title for this question?

Psaro
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  • Couldn't we directly prove that $\mathcal{F}_0$ is countable? Skipping the step where we prove $\mathcal{F}_c$ is countable – Psaro Oct 16 '22 at 16:34

1 Answers1

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Let $E_n$ be as in the definition of $\sigma$-finite. Note that any $D$ with $\mu(D\cap E)>0$ must have $\mu(D\cap E_n)>\frac1k$ for some $n$ and $k$. For any $n$ and $k$, the inequality $\mu(D\cap E_n)>\frac1k$ can hold for at most $k\mu(E_n)$ disjoint sets $D$. So $$ \{D\in\mathcal D:\mu(D\cap E)>0\}\subseteq \bigcup_n\bigcup_k \{D\in\mathcal D:\mu(D\cap E_n)>\frac1k\} $$ is a countable union of finite sets, hence countable.

Andreas Blass
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