equivalence for Borel-measurable function

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.

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.
• 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. Commented Jun 11 at 10:36