851lecture05

# Since every open set in r is an at most countable

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 &gt; 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 ,...
View Full Document

Ask a homework question - tutors are online