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Let $\mathcal{A}\subseteq\mathcal{P}(X)$ a $\sigma$-algebra, then for all sequence $\{A_n\}\subseteq\mathcal{A}$, exists a disjoint sequence $\{B_n\}\subseteq\mathcal{A}$ such that $$ \begin{align} (1)&\;B_n\subseteq A_n\quad(n\in\mathbb{N});\\ (2)& \bigcup_n A_n=\bigcup_nB_n. \end{align} $$

We consider the following sequence $$ B_1:=A_1,\quad B_k:=A_k\setminus\bigg(\bigcup_{n=1}^{k-1} A_n\bigg)\quad(k=2,\dots). $$ This sequence checks the properties $(1)$ and $(2)$.

proof. (1) The $\{B_n\}$ are disjointed by construction. We work by induction on $n$. Obviously $B_1\subseteq A_1$. Suppose that $(1)$ is true for $n-1$ and we show that it is true for $n$.

Therefore, let $x\in B_n$, then $x\notin B_1,\dots,x\notin B_n$, because $B_1,\dots,B_n$ are disjoined. Then $x\in B_n\setminus(B_1\cup\cdots\cup B_{n-1})$, for hypothesis $x\in B_n$, then $x\in A_n\setminus(A_1\cup\cdots\cup A_{n-1})$. Then $x\in [B_n\cap A_n\setminus(B_1\cup\cdots\cup B_{n-1})\cap(A_1\cup\cdots\cup A_{n-1})]$, by inductive hypothesis $(B_1\cup\cdots\cup B_{n-1})\subseteq (A_1\cup\cdots\cup A_{n-1})$, therefore $x\in [(B_n\cap A_n)\setminus(B_1\cap\cdots\cap B_{n-1})]$.

From the definition of $B_n$ we have $$ x\in A_n\cap A_n\setminus(A_1\cup\cdots\cup A_{n-1})\setminus \underbrace{(B_1\cup\cdots\cup B_{n-1})}_{=(A_1\cup\cdots\cup A_{n-1})}, $$ then $x\in A_n$.

(2) For point $(1)$ $$ \bigcup_n B_n \subseteq \bigcup_n A_n. $$

Now, let $x\in\bigcup_n A_n$, then $\exists\;\overline{n}\in\mathbb{N}$, such that $x\in A_{\overline{n}}$.

It could happen that $x\in A_i$ for same $i<\overline{n}$, therefore we consider $\overline{i}=\min\{i\; :\;x\in A_i,\;i\le\overline{n}\}$, then

$$ x\in B_{\overline{i}}=A_{\overline{i}}\setminus (A_1\cup\cdots\cup A_{\overline{i}-1}). $$

Then $x\in\bigcup_n B_n$.

It's right? Thanks

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