prove $\mathcal{L}^n(A(E))=|\det A|\mathcal{L}^n(E) $

suppose I have a linear map $A:\mathbb R^n \rightarrow \mathbb R^n$I want to prove $\mathcal{L}^n(A(E))=|\det A|\mathcal{L}^n(E) $$\forall E\subset \mathbb R^n$ where $\mathcal{L}^n$ is Lebesgue measure in $ \mathbb R^n$

my professor already proved but there is a point that is not clear to me and I have used bold character

Proof suppose $A$ is invertible than if we define $\mu(E)=\mathcal{L}^n(A(E))$ we have $\mu(E+v)=\mu(E)$and $\mu$ is regular in a topological sense I mean that $\forall E \subset \mathbb R^n, \: \mu(E)= \inf \{\mu(A) : \: A \text{ is open } \}$ so $\mu(E)=c\mathcal{L}^n(E)$

My proof I thought he used the fact that Lebsegue measure is regular in a topologial sense in this way $\mu(E)=\mathcal{L}^n(A(E))= \inf \{ \mathcal{L}^n(A(O)) : \: \text{where } O \supset E \text{ and } O \text{ is open} \}= \inf \mu (O) $

Could it be right ?