# Verification on proof of $\sigma(kU)=k^{2n+1}\sigma(U)$?

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 $$\{x\mid \exists x'\in kU:x'=kx\}=U$$:

\begin{align} \sigma(kU) &= \iint_{kU}d(x',y')dx'dy' \\ &= \iint_U d(kx,ky)kdx\ kdy \\ &= \int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} \sqrt{(kx_1-ky_1)^2+\cdots+(kx_n-ky_n)^2}kdx_1\ldots kdx_nkdy_1\ldots kdy_n \\ &= \int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} k\sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}k^{2n}dx_1\ldots dx_ndy_1\ldots dy_n \\ &= k^{2n+1}\int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} \sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}dx_1\ldots dx_ndy_1\ldots dy_n \\ &= k^{2n+1}\iint_Ud(x,y)dxdy \\ \sigma(kU)&= k^{2n+1}\sigma(U) \end{align}

Is this correct? Thanks.

## 1 Answer

One thing : the shape $$U$$ can be pretty complicated so there's no reason it can be written as nicely as $$\int_{x_1^0}^{x_1^1}...\int_{x_n^0}^{x_n^1}\int_{y_0^0}^{y_0^1}...\int_{y_n^0}^{y_n^1}d(x,y)\,dx\,dy.$$

But that's fine because what you only need is that $$d(kx,ky)=kd(x,y)$$ for all $$x,y\in U$$ (which is certainly true) and that $$d(kx)d(ky)=k^{2n}dxdy$$ which is true as well for the reason you mention.