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Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ be the function $f\left( x \right)=\left| x \right|$ and $\mathcal{A}= f^{-1}\left( \mathcal{B_1} \right)$ denote the preimage $\sigma$- algebra of $f$.

show that:

a Borel-measurable function $g:\mathbb{R}\longrightarrow \mathbb{R}$ is $\mathcal{A}$-measurable $\iff$ $g$ is an even function

To show this measurability, it suffices to prove that $f^{-1}\left( E' \right)$ $\in$ $\mathcal{B_1} $ for all $E' \in \mathcal{E'}$ for a generator $\mathcal{E'}$ of $\mathcal{B_1}$.

I know roughly how to deal with measurable maps, but I have problems dealing with this two given maps. I guess we have to work with the composition $g\circ f$ at some point.

I would be grateful for help :)

$\underline{\text{my try of the other direction : }}$

Let $g$ be $\mathcal{A}$-measurable. So for any Borel set $B\subseteq \mathbb{R}$, $g^{-1}\left( B \right) \in \mathcal{A}$ .

Let $B=\left[ a,b \right]$ since $g$ is $\mathcal{A}$-measurable $g^{-1}\left( \left[ a,b \right] \right) \in \mathcal{A}$.

Like you already said the sets in $\mathcal{A}$ would look like this $\left[ a,b \right]\cup \left[ -b,a \right]$ in my case

$\Rightarrow$ g must be symmetric

$\Rightarrow$ for $\lambda \in \mathbb{R}$ holds $\lambda \in g^{-1}\left( \left[ a,b \right] \right)$ implies -$\lambda \in g^{-1}\left( \left[ a,b \right] \right)$.

so $g\left( \lambda \right) \in \left[ a,b \right] \Rightarrow g\left( -\lambda \right) \in \left[ a,b \right]$ $\Rightarrow$ g must be an even function.

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In order for $g$ to be $\mathcal{A}$-measurable, the preimage of every $\mathcal{A}$-measurable set has to be in $\mathcal{A}$. The preimage $f^{-1}$ of any interval $[a,b]$ with $b<0$ is empty. If the interval is contained in the positive numbers, then: $$ f^{-1}([a,b])=[-b,-a]\cup[a,b]. $$ If $0$ is in the interval, then we have: $$ f^{-1}([a,b])=[-c,c], $$ where $c=\max \{a,-b\}$. We have calculated the generators $\mathcal{E}_i$ of $\mathcal{A}$. An even function $g$ satisfies that $g^{-1}(\mathcal{E}_i) \in \mathcal{A}$. For the converse, consider intervals of the form $[-b,-a]\cup[a,b]$, what can we say about $g$? In particular, what if the interval is just a point?

Looking at the set $\{ a \}$, the preimage $g^{-1}(a)$ has to be a countable union of points (why?). In particular, that union is symmetrical with respect to zero, because $\{a \}= [0,a] \cap [a,a+1]$(assuming that $a>0$, you can check the other case). There are some details to be filled, because you cannot assume that the preimage of an interval is another interval. That preimage will be a countable union of symmetrical intervals and points, but it is not difficult to see that it does not matter. Thus, for those points $c_i$, we have: $g(-c_i)=a=g(c_i)$. If the preimage is empty, there is nothing to prove, so $g$ is even.

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  • $\begingroup$ thank you very much for your help. I have added my attempt above. is that' enough? it feels a bit like i've skipped a step $\endgroup$
    – joel07 _
    Commented Jun 10 at 21:07
  • $\begingroup$ No, because it is not sufficient to show that if $g^{-1}(-\lambda)$ is in $\mathcal{A}$. You want it to be exactly $g^{-1}(\lambda)$. I will add a spoiler with the solution to my answer. $\endgroup$
    – Gowexx
    Commented Jun 11 at 10:36
  • $\begingroup$ thank you very much for your patience and time. I got it :) $\endgroup$
    – joel07 _
    Commented Jun 11 at 18:06
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    $\begingroup$ Great! Keep it up! $\endgroup$
    – Gowexx
    Commented Jun 12 at 10:50

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