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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!

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  • $\begingroup$ 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? $\endgroup$ – YeZ May 17 '19 at 22:20
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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...

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