Given a probability space $(\Omega,\mathcal{F}, P).$ Suppose $\mathcal{A}$is a subclass of $\mathcal{F}$. Let $\mathcal{A}_{x,\epsilon}$ be the class of $\mathcal{A}-$sets satisfying $x\in A^{\circ}\subset A\subset B(x,\epsilon)$ and let $\partial \mathcal{A}_{x,\epsilon}$ be the class of their boundaries.
Claim: If $\partial\mathcal{A}_{x,\epsilon}$ contains uncountably many disjoint sets, then at least one of them mush have $P-$ measure zero.
This is in section 2 from the book " convergence of probability measures(Billingsley)".
I think to proof this claim, we may need Borel-Cantelli lemma, but Borel-Cantelli lemma is for countably many sets.
Denote the metric on $\Omega$ by $\rho(x,y)$. Since for any $A\in\mathcal{A}_{x,\epsilon}$, $x\in A^{\circ}\subset A\subset B(x,\epsilon)$, then $\partial A\subset \partial B(x,\epsilon)\subset \{y:\rho(x,y)=\epsilon\}$.
Therefore $\cup_{A\in\partial\mathcal{A}_{x,\epsilon}} A\subset \{y:\rho(x,y)=\epsilon\}$. Thus $P(\cup_{A\in\partial\mathcal{A}_{x,\epsilon}} A) \leq P(\{y:\rho(x,y)=\epsilon\})<\infty$ But then how can I get the conclusion?
Any help will be appreciated. Thanks
You don't need anything fancy like Borel-Cantelli. Just the axioms of measure theory. In fact, a more general result is true, in your same probability space:
Proof: For any $n\in \mathbb N$, let $\mathcal A_n=\{A\in \mathcal A\mid P(A)>\frac1n\}$. Note that $\mathcal A_n$ has fewer than $n$ elements: if there were $n$ sets $E_1,E_2,\dots,E_n\in \mathcal A_n$, then you would have $$ P(E_1\cup E_2\cup\dots\cup E_n)=P(E_1)+\dots+P(E_n)>\frac1n+\dots+\frac1n=1, $$ contradicting that $P$ is a probability measure. Now, let $\mathcal A_0=\{A\in \mathcal A\mid P(A)>0\}$. Note that $\mathcal A_0=\bigcup_{n\ge 1}\mathcal A_n$, so $\mathcal A_0$ is a countable union of finite sets. This implies $\mathcal A_0$ is countable, as claimed.