# Continuous then measurable?

I have this proposition:

Prop. Every continuous functions $$f:\mathbb{R}^n \to \mathbb{R}$$ is $$\mathcal{B^n} - \mathcal{B}$$ measurable.

I assume here $$\mathbb{R}, \mathbb{R}^n$$ count with their standard topology and $$\mathcal{B}$$ is the Borel algebra generated by the open sets respectively.

Question. Does the proposition still hold if we change the topology for any other? Let's say with $$\tau := \{ \emptyset, \mathbb{R}, \mathbb{R} \setminus \{0 \}, \{ 0 \} \}$$, $$\tau' := \{ \emptyset, \mathbb{R}^n, \mathbb{R}^n \setminus \{0 \}, \{ 0 \} \}$$ and $$f:(\mathbb{R}^n, \tau') \to (\mathbb{R}, \tau)$$ continuous. Is still this function measurable? Maybe respect any other $$\sigma$$-algebra?

Thanks!

• What if you consider a continuous mapping $f: X \to Y$, where $X$ and $Y$ are topological spaces? Would $f$ be measurable w.r.t. the Borel $\sigma$-algebras generated by open sets? – YeZ May 17 '19 at 22:20

Assuming you now mean $$\mathcal B$$ to be the Borel algebra generated by $$\tau$$ and similarly for $$\tau'$$, not the canonical Borel algebra: Consider this:
For all $$\mathscr A \in \mathcal B$$, there is a sequence $$\{A_i\}_{i\in\Bbb N} \subseteq \tau$$ with $$\mathscr A = \bigcap_iA_i$$.
A basic property of functions is that for any collection of sets $$\{C_i\}, f^{-1}(\bigcap_iC_i) = \bigcap_if^{-1}(C_i)$$. (Prove this yourself - show that every element of the left is in the right, and every element in the right is in the left.)
So in particular, $$f^{-1}(\mathscr A) = \bigcap_i f^{-1}(A_i)$$, and since the $$A_i$$ are open and $$f$$ is continuous...