Let $\partial C=\{(x,y,z) \in \Bbb{R^3}\enspace\vert \enspace x^2+y^2 = r, \enspace0\leq z\leq h\}.$ I'm trying to show $\partial C$ has Lebesgue-measure zero in $\Bbb{R^3}$.
Let $f: I \to \Bbb{R^2}$, s.t $f=(r\cos x, r\sin x)$, where $I=[0,2\pi]$. Now $f$ is Riemann-integrable and its graph surface $G_f = B(0,r)\subset\Bbb{R^2}.$
Therefore $m_n(G_f)=0$. Then $\partial C=[0,h] \times G_f$. This cartesian product has Lebesgue measure zero, since $[0,h]$ is bounded in $\Bbb{R^2}$. Therefore $m_n(\partial C)=0$. Is this right?
Best Answer
Note that $\partial C = \{(x,y)| x^2+y^2 = r \} \times [0,h]$, so $m_3 (\partial C) = h \cdot m_2 \{(x,y)| x^2+y^2 = r \}$. Hence it suffices to show that the measure of the circle $rS^1$ is zero.
Note that for fixed $x$ the indicator $y \mapsto 1_{r S^1} ((x,y))$ is zero except for at most 2 points. Hence $\int 1_{S^1} = \int_{x \in \mathbb{R}} \int_{y \in \mathbb{R}} 1_{r S^1} ((x,y)) dy dx = \int_{x \in \mathbb{R}} 0y dx = 0$