Lebesgue $n$-dimensional measure of a hyperplane

Question

Show that every $n-1$-dimensional hyperplane in $\mathbb R^n$ has zero $n$-dimensional Lebesgue measure.

My problem is that I'm stuck on how to cover the hyperplane.

Answer 1

Since Lebesgue measure is invariant under rotations, the problem boils down to showing ($\mathbf x = [\mathit{x}_0,\dots,\mathit{x}_{n-1}]$) $$ S = \{\mathbf x \in \mathbb{R}^n\mid \mathit{x}_0 = 0\} $$ has $0$ Lebesgue measure. For every $j \in \mathbb{n}, \epsilon > 0$, define $$ R_j(\epsilon) = \left[\frac{-\epsilon}{2^{j+n} j^{n-1}}, \frac{\epsilon}{2^{j+n} j^{n-1}} \right] \times \prod_{k=1}^{n-1} [-j,j] \subseteq \mathbb{R}^n $$ Then $R_j(\epsilon)$ has Lebesgue measure (This directly comes from the definition of product measure, since $R_j(\epsilon)$ is a measurable rectangle) $$ \lambda(R_j(\epsilon)) = \frac{\epsilon}{2^{j+n} j^{n-1}}(2j)^{n-1} = \frac{\epsilon}{2^j} $$ Also, $S \subseteq \bigcup_{j=1}^\infty R_j(\epsilon)$ for any $\epsilon > 0$, so by subadditivity, $$ \lambda(S) \leq \sum_{j=1}^\infty \lambda(R_j(\epsilon)) = \sum_{j=1}^\infty \frac{\epsilon}{2^j} = \epsilon $$ Since that holds for all $\epsilon > 0$,$\lambda(S) = 0$.

Note: The value $\epsilon/(2^{j+n} j^{n-1})$ is carefully adjusted so the infinite sum of measures converge.

Answer 2

Case 1: Hyper Plane is of the form $$S = \{\mathbf x \in \mathbb{R}^n\mid \mathit{x}_i = 0\}$$ $i \in \{1,2, ... ,n\}$ is of measure 0 is covered above.

Case 2: Hyper Plane is of the form $a_{1}x_{1} + a_{2}x_{2} + ... + a_{n}x_{n} = c$ such that $\exists i,j$ for which $a_ia_j \neq 0$ W.L.O.G. assume $$x_n = \frac{c - (a_{1}x_{1} + a_{2}x_{2} + ... + a_{n-1}x_{n-1})}{a_n}$$ then $x_n = f(x_1, x_2, ... , x_{n-1})$ is a continious function from $\mathbb R^{n-1} to \mathbb R$, consider the set $$R_j = \prod_{k=1}^{n-1} [-j,j] \subseteq \mathbb{R}^{n-1}$$ $R_j$ is compact, we will show that the graph $G_{j} = \{(x_{n},f(x)) : x \in R_{j}\}$ has lebesgue measure $0$. Since continious function($f$) on compact set($R_{j}$) is uniformly continious we can cover the set $G_j$ by rectangles $\bigcup Rect_{k}\times[a_{k},b_{k}]$ such that $\bigcup Rect_{k} = R_{j}$ and $b_{k} - a_{k} = \frac{\epsilon}{(2j)^{n-1}}$ therefore $\mu(G_{j}) < \frac{\epsilon}{(2j)^{n-1}} \times (2j)^{n-1} = \epsilon$. Hence $\mu(G_{j}) = 0$ and $$HyperPlane \subset \bigcup_{j \in \mathbb{N}} G_{j}$$ Therefore by subadditivity $\mu(HyperPlane) \leq \sum_{j=1}^\infty \mu(R_j) = \sum_{j=1}^\infty 0 = 0$. Since $0 \leq \mu(HyperPlane) \leq 0 \implies \mu(HyperPlane) = 0$