Verification on proof of $sigma(kU)=k^{2n+1}sigma(U)$?
1
$begingroup$
Let $U$ be a bounded open subset of an $n$ -dimensional euclidean space endowed with the usual topology and the usual metric $d$ . Now, let $$sigma(U)=iint_U d(x,y)dxdy$$ Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape. I think I have a proof of $sigma(kU)=k^{2n+1}sigma(U)$ , but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof: Each of its points of $U$ will have coordinates $x=(x_1,ldots,x_n)$ . Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations $$x_i'=kx_i$$ $$dx_i'=kdx_i$$ Then, because $x_i=x_i'/k$ , the shape $kU$ gets transformed to ${xmid exists x'in kU:x'=kx}=U$ : begin{align} sigma(kU) &= iint_{kU}d(x...
