Is it possible to solve the following exercise without any reference to Lebesgue integral?
Given $\omega_n:=\mu(B_1(0))$ find $\mu(B_r(0))$ and $\mu (\partial B_r(0))$.
First part is very easy: $B_r(0)=T(B_1(0))$, where $T$ is the linear map on $\mathbb R^n$ defined by $T(x_1, \ldots , x_n) := (rx_1, \ldots, rx_n)$: hence, if $E$ is measurable, so is $TE$ and $\mu(TE)=r^n\mu(E)$.
What about $\mu (\partial B_r(0))$? I know the answer is (using the definition of surface integral) $\mu (\partial B_r(0)) = n\omega_n r^{n-1}$ but I don't manage to prove it without using integral theory.
Thanks in advance.