# Integral over balls in $\mathbb{R}^n$ in different norms and measures

I need to calculate the integral

$$\int_{|x| \le r} f(x) \, dx$$ of a function in the $r$ ball in $\mathbb{R}^n$, using the standard Lebesgue measure.

Take $f(x) = 1$, that is naturally the volume of the ball in $\mathbb{R}^n$ if we use the Euclidean norm. But if we use the Chebyshev distance, that ball is an hypercube and the volume, in the Lebesgue measure, is just $r^n$. They are both related by a constant factor tough, namely $\frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$.

So, I wonder what measure could give rise to an integral over the Chebyshev ball that keeps the same value for an arbitrary $f(x)$... is this possible? up to a factor?

I believe/suspect this may have to do with a Radon-Nikodym derivative between measure above and the Lebesgue measure, but again which measure?

Assume that the integral over a cube wrt. measure $\mu_1$ is the same as the integral over a (Euclidean) ball wrt. a measure $\mu_2$ for all functions. Let $Sq=[-1,1]^n$ and $B=\{x: \|x\|_2\leq 1\}$, and $f$ be an integrable function (wrt. both measures). Then we have $$\int_B f\mathrm{d}\mu_1=\int_{Sq} f\mathrm{d}\mu_2=\int_B f\mathrm{d}\mu_2 + \int_{Sq\setminus B} f\mathrm{d}\mu_2$$ which is independent of $f|_{Sq\setminus B}$ due to the left hand side, so $\mu_2$ is zero on $Sq\setminus B$. This holds for any $f$, so you get that $\mu_2(X)=\mu_1(X\cap B)$ for any measurable $X\subseteq Sq$.