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Let $ \Omega $ a set and $ \mathcal{A}\subseteq \mathcal{P}(\Omega) $ a finite $ \sigma $-Algebra over $ \Omega $. Show there exists a partition $ \Omega=A_1\cup...\cup A_n $ such that $$ \mathcal{A}=\{A_{k_1}\cup...\cup A_{k_r}:r\in \mathbb{N},k_i\in \{1,...,n\}\}. $$

My idea: By condition we have $ \mathcal{A}=\{A_1,...,A_m\} $ with $ A_i\subseteq \Omega $ for all $ i=1,...,m $. Define new set's as follows: $$ \begin{align}A_1'&:=A_1\\[20pt]A_k'&:=A_k\setminus{\left (\bigcup_{l=1}^{k-1} A_l\right )},\quad k=2,...,m\\[20pt]A_{m+1}'&:= \Omega\setminus{\left (\bigcup_{l=1}^m A_l\right )}. \end{align}$$

By construction the set's $ A_1',...,A_m',A_{m+1}' $ are disjoint and we have $$ \begin{align}\bigcup_{w=1}^{m+1} A_w'&=A_1'\cup \left (\bigcup_{w=2}^{m} A_w'\right)\cup A_{m+1}'\\[20pt]&= A_1'\cup \left (\bigcup_{w=2}^{m} \left(A_k\setminus{\left (\bigcup_{l=1}^{k-1} A_l\right )}\right)\right)\cup A_{m+1}'\\[20pt]&= A_1'\cup \left (\bigcup_{w=2}^{m} \underbrace{\left(\bigcap_{l=1}^{k-1} (A_w\setminus{A_l})\right )}_{=A_w}\right)\cup A_{m+1}'\\[20pt]&=A_1 \cup\left ( \bigcup_{w=2}^m A_w\right)\cup \left ( \Omega\setminus{\left (\bigcup_{l=1}^m A_l\right )}\right)\\[20pt]&=\left ( \bigcup_{w=1}^m A_w\right)\cup \left ( \Omega\setminus{\left (\bigcup_{l=1}^m A_l\right )}\right)=\Omega\end{align}$$

I want to show that $$ \mathcal{A}=\{A_{k_1}'\cup...\cup A_{k_r}':r\in \mathbb{N},k_i\in \{1,...,m,m+1\}\}=:\mathcal{M} $$

The inclusion "$\supseteq $" follows directly by the condition that $ \mathcal{A} $ is a $\sigma$-Algebra but I don't know how to show the other other inclusion "$\subseteq $".

hallo007
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    Does this answer your question? Every finite $\sigma$-algebra is of the form...? – Anne Bauval Nov 03 '22 at 15:26
  • You can't prove this, because it's not true. Since $\Omega\in\mathcal A$, it could be that $A_1=\Omega$, but then $A_1'=\Omega$ and $A_k'=\emptyset$ for all $k>1$. These do partition $\Omega$, but they don't generate $\mathcal A$ in the way specified. Instead, try this: $\mathcal A$ is a partially ordered set wrt the inclusion relation. Let's call a set $A\in\mathcal A$ minimal if $B\subseteq A$ implies $B\in{A,\emptyset}$. Show that the set of all minimal sets is a partition of $\Omega$ generating $\mathcal A$ in the specified way. – Vercassivelaunos Nov 03 '22 at 15:31
  • @ Anna Bauval This linked answer is not really convincing me because I don't know how to use these facts to prove my claim. – hallo007 Nov 03 '22 at 15:44
  • @hallo97 You can't use them to prove your claim because your claim is not true. See my comment above. You have to use another avenue. – Vercassivelaunos Nov 03 '22 at 17:28

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