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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.

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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.

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