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So, I have to show that the function $f\colon \mathbb{R} \to \mathbb{R}$ given by:

$f(x) = \begin{cases} -x & \text{if $x < 0$,} \\ 2 & \text{if $x \geq 0$,} \end{cases}$

is $\mathcal{B}/\mathcal{B}$-measureable. And I'm not quite sure how to do it. Going by the definition, I'd have to show that

$f^{-1}(A) \in \mathcal{B}, \quad \forall A \in \mathcal{B}$.

I've already shown, that

$\{f \geq a\} = \begin{cases} \mathbb{R} & \text{if $a\leq 0$,} \\ (-\infty , -a] \cup [0,\infty) & \text{if $0<a\leq 2$,} \\ (-\infty , -a] & \text{if $a>2$,} \end{cases}$

and I'm thinking that I can use this to show measurability, but I'm not sure? Since all of the above are Borel sets, does it follow that $f^{-1}$ of any of the above are also Borel sets?

Any help would be much appreciated!

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You can write $f$ as $$ f(x) = 2\chi_{[0,\infty)}(x)-x\cdot\chi_{(-\infty,0)}(x)$$ (where $\chi$ denotes indicator function) which is measureable as sum/product of measurable functions.

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  • $\begingroup$ Of course, thank you very much! :-) $\endgroup$ – Phillip Bredahl Sep 28 '15 at 7:10
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    $\begingroup$ @phibre, you are welcome. You can probably find in any measure theory textbook details needed for this to work, i.e., measurability of sum, product and composition of measurable functions, and of course measurability of indicator functions. $\endgroup$ – Ennar Sep 28 '15 at 7:15

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