Let $T_i:\Bbb R\to\Bbb R$ be defined for $i=1,2,3$ as follows:
$$T_1(x)=x+\beta\\ T_2(x)=-x\\ T_3(x)=|\alpha|\;x\\ \text{where}\;\ \alpha,\beta\in\Bbb R\text{ and}\; \alpha\ne0$$
I want to prove that: $$T_i(\mathcal A_{\Bbb R}^*)=\mathcal A_{\Bbb R}^*,\;\ i=1,2,3$$
I've already proved that $\mathcal A_{\Bbb R}^*\subset T_i(\mathcal A_{\Bbb R}^*),\;\ i=1,2,3$
So It's left to prove that $T_i(\mathcal A_{\Bbb R}^*)\subset \mathcal A_{\Bbb R}^*$, but I don't know how. What I've got is this:
Let $B\in T_i(\mathcal A_{\Bbb R}^*)\Rightarrow\ \exists\ E\in\mathcal A_{\Bbb R}^*$ s.t. $B=T_i(E)$ and also I've already proved that:
$$\lambda^*(T_i(E))=\lambda^*(E)\;\forall E\subset\Bbb R\;\ \text{for}\ i=1,2\\ \text{and}\;\ \lambda^*(T_3(E))=|\alpha|\lambda^*(E)\;\forall E\subset\Bbb R$$
So for i=1,2 I would get that $\lambda^*(B)=\lambda^*(T_i(E))=\lambda^*(E)=\overline{\lambda}(E)$
And that $\lambda^*(B)=\lambda^*(T_3(E))=|\alpha|\lambda^*(E)=|\alpha|\overline{\lambda}(E)$
Does the fact that $\lambda^*(B)=\overline{\lambda}(E)$ implies that B should be Lebesgue measurable and thus $B\in\mathcal A_{\Bbb R}^*$ straightforward?
What do you think or suggest?
All your transformations are invertible. If a set is the preimage of another set, it is also the image of one.