So I have the following exercise and I want to check my solution:
For $A \subset \mathbb{R}^n$ and $t \in \mathbb{R}$ let $tA :=\left \{y \in \mathbb{R}^n|y=tx, x \in A \right \}$.
Let $t>0$. I have to prove that $\lambda_{n}(tA)=t^n \lambda_{n}(A)$ for $\lambda_{n}$ being the lebesgue measure.
What I thought is: $\lambda_{n}(tA)=(tb_1-ta_1)\dot ... \dot (tb_n-ta_n)=\prod_{i=1}^{n}(tb_i-ta_i)=\prod_{i=1}^{n}t(b_i-a_i)=t^n\prod_{i=1}^{n}(b_i-a_i)=t^n\lambda_{n}(A)$
Is everything right?
Note: $t>0$ is necessary first of all - if $t \leq 0$ then the statement is obviously false. Next, $A$ must be measurable for us to talk about its measure to begin with. I am assuming this though you haven't stated it explicitly.
The following approach is used a lot in measure theory - you prove the result on the "building blocks" and you compound on that to generalize to the entire sigma algebra, sorta like induction.
In this case, you can start with intervals or open balls, both being a basis, where it's easy to show that $\lambda(t\Pi_i[a_i, b_i]) = t^n \lambda(\Pi_i[a_i, b_i])$ using something similar to what you wrote.
Then we generalize to open sets in general:
Any open set can be written as the disjoint union of countably many intervals (standard result) so given an open $A$ we can find disjoint intervals $U_i$ such that $A = \cup U_i$. Then it should be easy to see that $tA = \cup tU_i$ and from there find that $\lambda (tA) = t^n \lambda(A)$.
Finally, any lebesgue measurable (and for any regular measure for that matter) set's measure is witnessed by open sets, meaning for any measurable set $A$
$\lambda (A) = \inf \{\lambda(U) | U \supseteq A, U$ is open $\}$
Finally again you should be able to conclude that:
$\lambda(tA) = inf \{\lambda(tU)| U \supseteq A, U$ is open $\}$
by noticing that $tA \subseteq U \implies A \subseteq \frac{1}{t}U$ and so on. From here, we get the final result for every measurable set.