TheoremBase

The Borel σ-algebra of ℝ is the σ-algebra generated by the open sets, hence a σ-algebra; an integrable function is by definition an element of the set of measurable maps into ℝ, whose elements are exactly the measurable maps X → ℝ.

Proof

Proof of (i). By Borel Sigma-Algebra of a Metric Space (ii) and Borel Sigma-Algebra of a Metric Space (i),

B(R)=σ(TdR),\mathcal{B}(\mathbb{R})=\sigma(\mathcal{T}_{d_{\mathbb{R}}}),

the σ\sigma-algebra generated by the set TdR\mathcal{T}_{d_{\mathbb{R}}} of open subsets of R\mathbb{R}, which is a set of subsets of R\mathbb{R}. By Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii), taking the set to be R\mathbb{R}, it is a σ\sigma-algebra on R\mathbb{R}, so (R,B(R))(\mathbb{R},\mathcal{B}(\mathbb{R})) is a measurable space.

Proof of (ii) and (iii). By Integrable Functions and Their Lebesgue Integral (ii),

f∈Meas(X,F;R,B(R))f\in\mathrm{Meas}(X,\mathcal{F};\mathbb{R},\mathcal{B}(\mathbb{R}))

(formed for the measurable space (X,F)(X,\mathcal{F}) of Measure, Measure Space, and Probability Measure (ii) and that of (i)). By Measurable Map, and the Set of Measurable Maps (ii) its elements are exactly the F/B(R)\mathcal{F}/\mathcal{B}(\mathbb{R})-measurable maps X→RX\to\mathbb{R}, which gives (ii) and, by Measurable Map, and the Set of Measurable Maps (iii), (iii). ■\blacksquare

Citations

Loading…

Dependencies

Uses0

Loading…

Comments

Log in to comment.

Loading…