Lebesgue measure of closed ball and open ball are the same.

Question

Recently, I have been studying the Lebesgue measure, and I have learned some properties of this measure such as regularity,translation-invariance and relation with linear transformations.

Now, I want to verify whether it can be called a volume function.

Let $B(0,1)$ and $\overline{B(0,1)}$ be unit open and closed balls respectively in $\mathbb{R}^n$.

Then how do i prove that $m(B(0,1))=m(\overline{B(0,1)}$?

Moreover, how do I prove that $m(B(0,1))=\pi$ in $2$-dimensional and $\frac{4\pi}{3}$ in $3$-dimensional?(I guess this is a hard one to prove)

Many thanks in advance :)

Answer 1

Since $B(0,1)\cup S=\overline {B(0,1)}$, where $S$ is the unit sphere, the claim of the equality of measures will follow by showing that the measure of $S$ is $0$. This is quite easily shown directly, by covering $S$ by suitable rectangles of arbitrary small total measure (though it can be done more quickly using some rather elementary theory).

As for the particular measures of $B(0,1)$ in dimensions $2$ and $3$, it's not hard at all. Using integrals you can compute these volumes and arrive at the familiar results. In fact, one can, without too much hard work, obtain the values of the measure of the unit balls in all dimensions, see here.

Answer 2

Note that $$ B(0,1)\subset \overline{B}(0,1)\subset B(0,1+\delta)=(1+\delta)B(0,1) $$ for all $\delta>0$, and hence $$ m(B(0,1))\le m(\overline{B}(0,1))\le m(B(0,1+\delta))=(1+\delta)^nm(B(0,1)). $$