Continuous Mapping is Measurable

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Theorem

Let $\left({X, \tau}\right)$ and $\left({X', \tau'}\right)$ be topological spaces.

Denote with $\mathcal B \left({X, \tau}\right)$ and $\mathcal B \left({X', \tau'}\right)$ their respective Borel $\sigma$-algebras.

Let $f: X \to X'$ be a continuous mapping.


Then $f$ is $\mathcal B \left({X, \tau}\right) \, / \, \mathcal B \left({X', \tau'}\right)$-measurable.


Proof

As $f$ is a continuous mapping, by definition, it holds that:

$\forall S' \in \tau': f^{-1} \left({S'}\right) \in \tau$

Now, by definition, $\mathcal B \left({X, \tau}\right) = \sigma \left({\tau}\right)$, and so $\tau \subseteq \mathcal B \left({X, \tau}\right)$.

Also, $\mathcal B \left({X', \tau'}\right) = \sigma \left({\tau'}\right)$.


Hence Mapping Measurable iff Measurable on Generator applies.

Therefore, $f$ is $\mathcal B \left({X, \tau}\right) \, / \, \mathcal B \left({X', \tau'}\right)$-measurable.

$\blacksquare$


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