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This is problem 1.7 from "Measure Theory and Probability Theory" by K. Athreya and S. Lahiri.

Let $\Omega$ be a nonempty set and let $\mathcal{B} \equiv \{ B_i: 1 \leq i \leq k < \infty\} \subset \mathcal{P}(\Omega)$, not necessarily a partition. Find $ \sigma \langle \mathcal{B} \rangle $.

(Hint: For each $\delta = (\delta_1, \delta_2, ..., \delta_k), \delta_i \in \{0,1\}$ let $B_{\delta} = \bigcap_{i=1}^k B_i{(\delta_i)}$, where $B_i(0) = B_i^c$ and $B_i(1) = B_i$, $i \geq 1$. Show that $\sigma( \mathcal{B} ) = \{E : E = \bigcup_{\delta \in J} B_{\delta}, J \subset \{1,2, ..., k\}\}.$)

Their hint baffles me. If I take $\Omega = \mathbb{N}$ and $\mathcal{B} = \{ B_1, B_2 \}$, $B_1 = \{1,2\}$, $B_2 = \{2,3\}$, then for

$\delta = (0,0) \qquad B_{\delta} = \mathcal{P}(\mathbb{N}) \setminus (B_1 \cup B_2) $;

$\delta = (1,0) \qquad B_{\delta} = \{1\}$;

$\delta = (0,1) \qquad B_{\delta} = \{3\}$;

$\delta = (1,1) \qquad B_{\delta} = \{2\}$;

If I want to show that $\mathbb{N} \in \sigma( \mathcal{B} )$, do I have to take the union of all four $B_{\delta}$ as $E$? But the index set is $J \subset \{1, 2\}$, does it mean I can choose at most two of them? I can't get $\mathbb{N}$ with only two of them or can I?

Thank you for your time.

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You want to write $\Bbb N$ as a set function of $B_1$ and $B_2$, right? We have that $\{2\}=B_1\cap B_2$ and $\{1\}=B_1\setminus \{2\}$. This gives that $\emptyset\in\sigma(\cal B)$. –  Davide Giraudo Nov 9 '12 at 14:43
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1 Answer

  • First, we can assume without loss of generality that $\emptyset$ is one of the $B_i$, otherwise add it and work with $k+1$.
  • Show that the collection $\{B_{\delta},\delta\in\{0,1\}^k\}$ consists of pairwise disjoint sets.
  • Show that if $\{S_1,\dots,S_N\}$ is a partition of $\Omega$, then the $\sigma$-algebra generated by this collection is $\{\bigcup_{j\in J}S_j,J\subset \{1,\dots,N\}\}$.
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Thank you. -Fishy –  Fishy Nov 10 '12 at 11:12
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