Continuous Mapping is Measurable
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Theorem
Let $\struct {X, \tau}$ and $\struct {X', \tau'}$ be topological spaces.
Denote with $\map \BB {X, \tau}$ and $\map \BB {X', \tau'}$ their respective Borel $\sigma$-algebras.
Let $f: X \to X'$ be a continuous mapping.
Then $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable.
Proof
As $f$ is a continuous mapping, by definition, it holds that:
- $\forall S' \in \tau': \map {f^{-1} } {S'} \in \tau$
Now, by definition, $\map \BB {X, \tau} = \map \sigma \tau$, and so $\tau \subseteq \map \BB {X, \tau}$.
Also, $\map \BB {X', \tau'} = \map \sigma {\tau'}$.
Hence Mapping Measurable iff Measurable on Generator applies.
Therefore, $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $7.3$
- 2005: René L. Schilling: Measures, Integrals and Martingales: $\S 8$: Problem $8$