Let K be a solid figure in $\mathbb{R^3}$, $\mu$ the 3-dimensional Lebesgue-measure and suppose we know, that $\int_{\mathbb{R}^n} f(2*x_1,...,2*x_n) d\mu_n=\frac{1}{2^n}\int f(y_1,...,y_n)d\mu_n$ (what we get by substitution of $y_i=2x_i$ and the use of Fubini's theorem). When I now choose $\chi_{2K}$ to be the function that equals 1 for elements of 2K and 0 otherwise (where 2K is the extended figure) it follows: $\mu_3(2K)=vol(2K)=2^3vol(K)=2^3\mu(2K)$. Might someone give me a hint how I might get that $\chi_{2K}(y_1,...,y_3)=\chi_{K}(y_1,...,y_3)$?
Greetings
Rico
What is true is $\chi_{2K}(y_1,y_2,y_3)=\chi_K (x_1,x_2,x_3)$ where $y_i=2x_i$and this is what you need for the proof.