Problem about Borel-measurability, pushforward measure and $L^1$ spaces.

Question

I am trying to solve this exercise. I get stuck in the last part, but I write all the exercise for context. Also, I am not sure that some explanations I give are enough to justify these results.

(Sorry for my poor English)

Let $\varphi: \mathbb R \to \mathbb R$ be the function given by $$\varphi (x) = \dfrac{2x}{x-1} \chi _{(-\infty, -1)} (x) + \chi _{[-1,1]} (x) + \dfrac{2x}{x+1} \chi _{(1,+\infty)} (x), \quad x \in \mathbb R.$$

1. Prove that $\varphi$ is Borel-measurable over $\mathbb R$.

2. Let $m$ be the Borel-Lebesgue measure in $\mathbb R$. Find the pushforward measure $\varphi (m)$ over the $\sigma$-algebra of the borel sets of $\mathbb R$ and its Lebesgue decomposition with respect to $m$.

3. Let $$f(y) = \sqrt{2-y} \, \arctan (2-y) \, \chi _{(0,2)} (y), \quad y \in \mathbb R.$$ Show that $f\in L^1 (\varphi (m))$.

1. Since $\varphi$ is continuous, it is Borel-measurable.

   (I feel that I have been lucky with this. What would I have to do to prove Borel-measurability if it wasn't continuous?)

2. Drawing $\varphi(x)$ I find that the support of $\varphi(m)$ is $[1,+\infty)$.

   Let $E\subseteq (-\infty, 1)$, then $\varphi(m)(E)=0$.

   Also, $$\varphi(m) (\{ 1 \})= m (\varphi ^{-1} (1)) = m ([-1,1]) =2.$$

   Let $(a,b)\subset (1,+\infty)$, then $$\varphi^{-1}((a,b))=\left( \frac{b}{b-2}, \frac{a}{a-2} \right) \cup \left( \frac{a}{2-a}, \frac{b}{2-b} \right).$$ Then $$m(\varphi ^{-1} ((a,b)))=\left( \frac{a}{a-2}- \frac{b}{b-2} \right) + \left( \frac{b}{2-b}-\frac{a}{2-a} \right)=$$ $$=2\left( \frac{a}{a-2}- \frac{b}{b-2} \right)=-2\left( \frac{b}{b-2}- \frac{a}{a-2} \right)=$$ $$=-2\int _a ^b \! \left( \frac{t}{t-2} \right) ' \, \mathrm d t=-2\int _a ^b \! \frac{-2}{(t-2)^2} \, \mathrm d t=4\int _a ^b \! \frac{1}{(t-2)^2} \, \mathrm d t.$$ Therefore we can write (is this correct?) $$\varphi (m) ((a,b))= \underbrace{2 \delta _{\{1\}}}_{\varphi(m)_s}+\underbrace{4\chi_{[1,+\infty)}\int _a ^b \! \frac{1}{(t-2)^2} \, \mathrm d t}_{\varphi(m)_a},$$ where $\varphi(m)_s \perp m$ and $\varphi(m)_a \ll m$. Thus we have found its Lebesgue decomposition.

3. I don't know how to solve this part. The only thing that crosses my mind is that $f\in L^1 \varphi(m)$ if and only if $$\int \vert f \vert \, \mathrm d \varphi (m)=\int (\vert f \vert \circ \varphi) \mathrm d m \in \mathbb R.$$ But it seems hard to compute that integral and maybe is there something easier that I am not thinking.

Thank you for your help.