Let $T\colon\mathbb{R}\to\mathbb{R}$ a linear trasformation of $\mathbb{R}$ defined as $Tx:=ax+b$, where $a,b\in\mathbb{R}$, $a\ne 0.$
We denote with $\mathcal{L}(\mathbb{R})$ the Lebesgue $\sigma-$ algebra and we denote with $\lambda^*$ the Lebesgue outer measure. Let $E\in2^{\mathbb{R}}$, we must prove that
Proposition. $T(E)\in\mathbb{\mathcal{L}(\mathbb{R})}$ $\iff$ $E\in\mathcal{L}(\mathbb{R})$
I just proved that $$\lambda^*(T(E))=|a|\lambda^*(E)\tag1.$$
Regarding the implication $(\Leftarrow)$ there are not problems.
$(\Rightarrow)$ Proof. Suppose that $T(E)\in\mathcal{L}(\mathbb{R})$, then $\lambda^*(Z)=\lambda^*(Z\cap T(E))+\lambda^*(Z\cap\complement T(E))$ for all $Z\in 2^{\mathbb{R}}$.
Therefore we have that $$\lambda^*(T(Z))=\lambda^*(T(Z)\cap T(E))+\lambda^*(T(Z)\cap \complement T(E)),$$ so $$\lambda^*(T(Z))=\lambda^*(T(Z\cap E))+\lambda^*(T(Z\cap\complement E)).$$
Multiplying the members for $|a|^{-1}$ we have $$|a|^{-1}\lambda^*(T(Z))=|a|^{-1}\lambda^*(T(Z\cap E))+|a|^{-1}\lambda^*(T(Z\cap\complement E)),$$ for $(1)$ we obtain $$\lambda^*(Z)=\lambda^*(Z\cap E)+\lambda^*(Z\cap \complement E),$$ then $E\in\mathcal{L}(\mathbb{R}).$
Proof 2. Observe that $$\lambda^*(E)=\lambda^*(T[T^{-1}(E)])=|a|\lambda^*(T^{-1}(E)),$$ then $$\lambda^*(T^{-1}(E))=\frac{1}{|a|}\lambda^*(E)\tag2.$$
Let $T(E)\in\mathcal{L}(\mathbb{R})$ then $\lambda^*(Z)=\lambda^*(Z\cap T(E))+\lambda^*(Z\cap\complement T(E))$, therefore $$\lambda^*(T(Z))=\lambda^*(T(Z\cap E))+\lambda^*(T(Z\cap\complement E)).$$ Now, $$\frac{1}{|a|}\lambda^*(T(Z))=\frac{1}{|a|}\lambda^*(T(Z\cap E))+\frac{1}{|a|}\lambda^*(T(Z\cap\complement E)),$$ then for $(2)$ we have $$\lambda^*(T^{-1}[T(Z)])=\lambda^*(T^{-1}[T(Z\cap E)])+\lambda^*(T^{-1}[T(Z\cap\complement E)])$$ and so $$\lambda^*(Z)=\lambda^*(Z\cap E)+\lambda^*(Z\cap \complement E),$$ then $E\in\mathcal{L}(\mathbb{R}).$
Thanks!
Question Are the proofs correct?