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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http://math.stackexchange.com/questions/26888/the-union-of-a-strictly-increasing-sequence-of-sigma-algebras-is-not-a-sigm – njguliyev Oct 27 '13 at 22:37
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@njguliyev That post addresses a more difficult problem. – David Mitra Oct 27 '13 at 22:42
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What are you having difficulty comprehending? – Don Larynx Oct 27 '13 at 23:07
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@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 00:50
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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 $\{0,2\}\notin A\cup B$.
Hagen von Eitzen
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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