Lebesgue measurable sets are invariant under translations and dilations? From Folland's book, this theorem is proved as following:

Actually I am pretty confused about this proof, is there any other way to prove this theorem? More explicitly, can we use the following theorem to prove this?

I really appreciate your help
 A: I think it might be better to study the proof itself, rather than jump ship and prove it in an unrelated way. To try and make it clearer what is happening in the proof I will step through and then add some details at the end.
The author first "proves" the theorem in terms of Borel sets (by observing behavior of new measures $m_s$ and $m^r$ on finite unions of intervals and then applying Theorem 1.14, which I assume is an extension theorem).
The author also argues that the claim holds for such Lebesgue nullsets.
At this point we know that the equalities hold for both Borel and Lebesgue null sets. Given that any Lebesgue measurable set may be decomposed into a disjoint Borel and Lebesgue null set, we will use the additivity property of the Lebesgue measure to evaluate any Lebesgue measurable set.
Let $E\in\mathcal{L}$, then $E=O \sqcup N$, where $O\in \mathcal{B}_{\mathbb{R}}$ and $N$ is a Lebesgue null set, then:
\begin{align}
m(E+s)
&= m( (O\sqcup N) + s)\\
&= m\left( (O+s) \sqcup (N+s) \right)\\
&= m(O+s) + m(N+s)\\
&\overset{(*)}{=} m(O) + m(N)\\
&= m(O\sqcup N)
= m(E)
\end{align}
$(*)$ is where we apply the results shown on Borel and Lebesgue measurable sets.
There are details that may accidentally be overlooked, such as showing that translating disjoint sets by the same amount leaves the sets disjoint (i.e., after the third equality), but this is an outline of the finishing steps of the proof. It might be worth going through the case of dilation on your own.
