# A proof about sequence of sets

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