Since every open set in r is an at most countable

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Unformatted text preview: e next theorem shows. Theorem 5.3. The Borel σ -algebra B is generated by intervals of the form (−∞, a] where a ∈ Q is a rational number. Proof. Let O0 denote the collection of all open intervals. Since every open set in R is an at most countable union of open intervals, we must have σ (O0 ) = B . Let D denote the collection of all intervals of the form (−∞, a], a ∈ Q. Let (a, b) ∈ O0 for some b > a with a, b ∈ Q. Let 1 an = a + n so that an ↓ a as n → ∞, and let 1 bn = b − n so that bn ↑ b as n → ∞. Thus, (a, b) = ∞ ￿ n=1 (an , bn ] = ∞ ￿ n=1 {(−∞, bn ] ∩ (−∞, an ]c } which implies that (a, b) ∈ σ (D). That is, O0 ⊆ σ (D) so that σ (O0 ) ⊆ σ (D). However, every element of D is a closed set which implies that σ (D ) ⊆ B . 5–2 This gives the chain of containments B = σ ( O0 ) ⊆ σ ( D ) ⊆ B and so σ (D) = B proving the theorem. The Borel sets of [0, 1] If we now consider the set [0, 1] ⊂ R as the sample space, then B1 ,...
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