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Give example function $f: \mathbb{R} \rightarrow \mathbb{R}$ which $\forall x \in \mathbb{R}$ is not continuous function but is Borel function.

I think that I can take $$f(x) = \begin{cases} 1 & x \in \mathbb{R} \setminus \mathbb{Q} \ \\ -1 & x \in \mathbb{Q} \end{cases} $$

Am I right? How fast prove that it is Borel function?

Michael Hardy
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Thomas
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2 Answers2

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Yes, you're correct. To show that it's a Borel function, begin by showing that if $\mathcal{O} \subseteq \mathbb{R}$ is an open set, then $f^{-1}(\mathcal{O})$ is one of exactly four possible sets: $\emptyset, \mathbb{Q}$, $\mathbb{R} \setminus \mathbb{Q}$, and $\mathbb{R}$.

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Our you can do nothing but write $f=-\chi_{\mathbb{Q}}+\chi_{\mathbb{R}\setminus\mathbb{Q}}$ where $\chi_A$ is the characteristic function of $A$. And use the following :

$\mathbb{Q}$ is countable so it is a borel set and then $\mathbb{R}\setminus{\mathbb{Q}}$ as well (complementary property of $\sigma-$algebra). And the result follows because characteristic function of Borel set are by definition borel function.

But of course you could have just choose $f=\chi_{\mathbb{Q}}$.

user42070
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  • So $\chi_{\mathbb{Q}}$ also i borel function and is not continuous function in every point? – Thomas Dec 14 '13 at 11:09
  • Yes indeed. it is borel because $\mathbb{Q}$ is borel and not continous because of density of both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ in $\mathbb{R}$ (just use sequential continuity to show it). – user42070 Dec 16 '13 at 19:41