Lebesgue measure has the property of Translation invariance, and my question is whether it is invariant under swaps. In particular, let $A\subseteq \mathbb R^n$ and let $A^{i\leftrightarrow j}$ denote the set obtained by swapping the $i$ and $j$'th entries of every $v\in A$. Does $\lambda (A) =\lambda (A^{i\leftrightarrow j}) $ holds?
Any ideas?
My intuition is that the claim holds, i.e. $\lambda (A) =\lambda (A^{i\leftrightarrow j}) $. Since every rectangle $R=I_1 \times \cdots I_n$ has the same Lebesgue measure as $R^{i\leftrightarrow j}$, which is obtained by swapping $I_i$ and $I_j$, every cover $C$ of $A$ can be easily translated into a cover $C^{i\leftrightarrow j} $ of $A^{i\leftrightarrow j}$, thus the claim should hold. Am I missing something?
(migrated from comments)
In general, if $T$ is a linear transformation on $\mathbb{R}^n$ then $\lambda(TA) = \left|\det(T)\right|\lambda(A)$ holds. Although this general statement needs a bit of work, your claim is less demanding. Indeed, consider the map
$$ T : \mathbb{R}^n \to \mathbb{R}^n, \qquad T(x_1, \cdots, x_n) = (x_{\tau(1)}, \cdots, x_{\tau(n)}) $$
where $\tau : \{1,\cdots,n\} \to \{1, \cdots,n\}$ is the permutation swapping $i$ and $j$. Then $A^{i\leftrightarrow j} = T(A)$ and hence it suffices to prove $\lambda(A) = \lambda(T(A))$ for all measurable $A \subseteq \mathbb{R}^n$.
To this end, you can first establish the claim when $A$ is a rectangle, and then the rest follows from the monotone class argument (or from the Dynkin’s $\pi$−$λ$ theorem).