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Let $\mathcal F_{i,j}$, $1\leq i\leq n$ and $1\leq j\leq m_i$ be independents $\sigma -$algebra. Then $\mathcal G_i=\sigma (\bigcup_{j}\mathcal F_{i,j})$ are independents.

Proof : Set $$\mathcal A_i=\left\{\bigcap_{j\in J}A_{i,j}\mid J\subset \{1,...,m_i\}, A_{i,j}\in \mathcal F_{i,j}\right\}.$$ Then, $\mathcal A_i$ are $\pi-$system that contain $\bigcup_{j=1}^{m_i}\mathcal F_{i,j}$ and $\Omega $. Since $\mathcal A_i$ are independents, then so are the $\mathcal G_i=\sigma (\bigcup_{j=1}^{m_i}\mathcal F_{i,j})$'s by a theorem.

Question : I don't understand why the fact that $\Omega $ are contained in the $\mathcal A_i$ is important ? Any idea ?

user659895
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1 Answers1

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First, I guess that you work on $\Omega $ (it's mentioned no where). Then, I think that $\mathcal A_i$ should be $$\mathcal A_i=\left\{\bigcap_{j=1}^{m_i}A_{ij}\mid A_{ij}\in \mathcal F_{ij}\right\},$$ instead of what you wrote. Otherwise it wouldn't be a $\pi-$system. Using my definition of $\mathcal A_i$, if the $\mathcal A_i$'s don't contained $\Omega $, then $\mathcal A_i$ wouldn't contained $\bigcup_{j}\mathcal F_{ij}$, and thus, $\sigma (\mathcal A_i)=\sigma \left(\bigcup_{j}\mathcal F_{ij}\right)$ wouldn't be true (we would have only $\sigma (\mathcal A_i)\subset \sigma \left(\bigcup_{j}\mathcal F_{ij}\right)$.)

Surb
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