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I was reading notes on the filtrations and stopping times where I came across this fact: Union of $\sigma$-algebra is not $\sigma$-algebra in general. I am having difficult time comprehending this, so can someone please exemplify this?

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@njguliyev That post addresses a more difficult problem. –  David Mitra Oct 27 '13 at 22:42
    
What are you having difficulty comprehending? –  Don Larynx Oct 27 '13 at 23:07
    
@DonLarynx I think its \sigma-algebra in general, have not yet gotten feel about it. I guess I am in that part of mathematics where I should leave the intuition behind. –  Andrey Carl Oct 28 '13 at 0:50

2 Answers 2

Let $A=\{\,E\subseteq \mathbb R\mid [0,2)\subseteq E\lor E\cap[0,2)=\emptyset\,\}$ and $B=\{\,E\subseteq \mathbb R\mid [1,3)\subseteq E\lor E\cap[1,3)=\emptyset\,\}$. Then $A\cup B$ is not a $\sigma$-algebra because $\{0\}\in B$, $\{2\}\in A$, but $\{0,2\}\notin A\cup B$.

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Consider the universe $\Omega=\{a,b,c\}$ and the sigma-algebras $\mathcal A$ and $\mathcal B$ generated by $\{a\}$ and by $\{b\}$ respectively (can you write down $\mathcal A$ and $\mathcal B$ in extension?). Then $\{a\}$ and $\{b\}$ are both in $\mathcal A\cup\mathcal B$ but not $\{a\}\cup\{b\}$ (can you show this?). Thus, $\mathcal A\cup\mathcal B$ is not a sigma-algebra.

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$\mathscr{A}=\{\Omega, \phi, a, \{b,c\} \}$ $\mathscr{B}=\{\Omega, \phi, b, \{a,c\} \}$. Thus $a, b \in \mathscr{A} \cup \mathscr{B}$ but $\{a,b\} \not\in \mathscr{A} \cup\mathscr{B} $. –  Andrey Carl Oct 28 '13 at 14:57

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