# Continuous Mapping is Measurable

Jump to navigation
Jump to search

## 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 $8$