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I have the following definition for independence of events (from lecture notes):

Let $(\Omega, \mathcal{A}, P)$ be a probability space, let $I$ be a set, and let $A_i \in \mathcal{A}, i \in I$ be a family of elements of $\mathcal{A}$ with the property that for every finite nonempty subset $J \subseteq I$ of $I$ it holds that $$ P\big( \cap_{i \in J} A_i \big) = \prod_{i \in J} P(A_i).$$ Then the family $A_i, i \in I$, is called independent (is called P-independent).

With this definition $A$ and $A^c$ from $\mathcal{F_1} = \{ \emptyset, \Omega, A_1, A_1^c \}$ where $A_1$ and $A_1^c$ both occur with probability larger than zero are not independent (which makes sense to me since they are mutually exclusive).

I further have the following definition for independence of $\sigma$-algebras:

Let $(\Omega, \mathcal{A}, P)$ be a probability space, let $I$ be a set, and let $\mathcal{A}_i \subseteq \mathcal{A}, i \in I$ be a family of sigma-algebras with the property that for every family $A_i \in \mathcal{A}, i \in I$, of events with $\forall i \in I: A_i \in \mathcal{A}_i$ it holds that $A_i, i \in I$, is P-independent. then the family $\mathcal{A}_i, i \in I$, is called independent (is called P-independent).

Now, this definition confuses me because for me it reads like the $\sigma$-algebra $\mathcal{F_1}$ from my example can never be belong to a family of independent $\sigma$-algebras, say $\mathcal{F}_1, \mathcal{F}_2$ because it contains a family of events $A_1, A_1^c$ that is not independent. But then no $\sigma$-algebra at all could ever belong to an independent family of $\sigma$-algebras because by definition every $\sigma$-algebra contains complementary sets like $A_1$ and $A_1^c$. Clearly, this is nonsense..

What am I missing here?

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    $\begingroup$ What you overlooked is that, in the definition of independence of $\sigma$-algebras, you take one set $A_i$ from each $\sigma$-algebra $\mathcal A_i$ (and require that any such selection of sets be independent). You don't take two sets (like $A_1$ and $A_1^c$) from the same $\sigma$-algebra. $\endgroup$ – Andreas Blass Oct 1 '15 at 8:41
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    $\begingroup$ Thanks, that made things clear! You can post it as answer if you want and I will mark it as the correct one. $\endgroup$ – mwater Oct 1 '15 at 9:12
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Promoting my comment to an answer, as suggested by the OP:

What you overlooked is that, in the definition of independence of $\sigma$-algebras, you take one set $A_i$ from each $\sigma$-algebra $\mathcal A_i$ (and require that any such selection of sets be independent). You don't take two sets (like $A_1$ and $A_1^c$) from the same $\sigma$-algebra.

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