Borel Sigma-Algebra of Subset is Trace Sigma-Algebra
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Theorem
Let $\struct {X, \tau}$ be a topological space, and let $A \subseteq X$ be a subset of $X$.
Let $\tau_A$ be the subspace topology on $A$.
Then the following equality of $\sigma$-algebras on $A$ holds:
- $\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
where $\BB$ signifies Borel $\sigma$-algebra, and $\map \BB {X, \tau}_A$ signifies trace $\sigma$-algebra.
Proof
By definition of Borel $\sigma$-algebra, it holds that:
- $\map \BB {X, \tau} = \map \sigma \tau$
and also, by definition of subspace topology:
- $\tau_A = A \cap \tau = \set {A \cap U: U \in \tau}$
Thus, it follows that:
- $\map \BB {A, \tau_A} = \map \sigma {A \cap \tau}$
Thereby, the desired equality:
- $\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$
follows directly from applying Trace Sigma-Algebra of Generated Sigma-Algebra with $\GG = \tau$.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 3$: Problem $10 \ \text{(ii)}$