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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.

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    1) By noting that $\partial B_r(0)\subset B_t(0)\setminus B_s(0)$ for every $s\lt r\lt t$ and using the argument in your post. 2) The way to give to $\partial B_r(0)$ the measure we know is **NOT** through $(n-1)$-dim Lebesgue. // I did not downvote but if I did, it would be because the lack of exactness of its formulation makes it difficult to answer. You could do the question a big favor by reformulating it with the most possible rigor and precision.2012-11-03

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I'm not sure what definition of surface area you are using, or how your proof by "integral theory" goes. The simplest proof should be the following, which uses some elementary properties about the surface of an $n$-ball. You should check your definition of surface area to make sure that you can derive these properties.

Let $f(r)$ be the volume of the $n$-ball with radius $r$, and let $S_r$ be the $(n-1)$-dimensional surface area of the same ball. If $h>0$ then $f(r+h)-f(r)>h\cdot S_r$ This follows from the fact that $S_r$ is increasing with respect to $r$, and the left hand side is the volume of a spherical shell with thickness $h$ and inner surface area $S_r$.

In the same way, $f(r)-f(r-h) Put these inequalities together and divide by $h$. ${f(r)-f(r-h)\over h} Let $h\to 0$, the squeeze theorem then gives $S_r = f'(r)$ Since you have shown that $f(r)=r^n\cdot \omega_n$, where $\omega_n$ is the volume of the unit $n$-ball, it follows that $S_r=nr^{n-1}\omega_n$