Proposition

if $M$ is a $k$-dimensional manifold in $\mathbb{R}^n$ and $k < n$, show that $M$ has measure 0.
If $M$ is a closed $n$-dimensional manifold-with-boundary in $\mathbb{R}^n$, show that the boundary of $M$ is $\partial M$. Give a counterexample if $M$ is not closed.
If $M$ is a compact $n$-dimensional manifold-with-boundary in $\mathbb{R}^n$, show that $M$ is Jordan-measurable.

Solution

1

Let $x \in M$ be given and let $h: U \rightarrow V$ be the diffeomorphism. If $U$ is not bounded, take the intersection of $U$ and an open ball and update $V$ appropriately. Similarly, we can assume $V$ is bounded.

Let $\epsilon > 0$ be given. Let $V_{\epsilon} = V \cap (\mathbb{R}^k \times (-\epsilon, \epsilon) \times \cdots \times (-\epsilon, \epsilon))$. Then $V_{\epsilon}$ is a subset of $V$ and $V_{\epsilon} \cap (\mathbb{R}^k \times \{ 0 \}) = V \cap (\mathbb{R}^k \times \{ 0 \})$.

Let $U_{\epsilon} = h^{-1}(V_{\epsilon})$. We claim that $U_{\epsilon} \cap M = U \cap M$. Since $V_{\epsilon} \subset V$, $U_{\epsilon} \subset U$, so $U_{\epsilon} \cap M \subset U \cap M$. Let $y \in U \cap M$. Then $h(y) \in V \cap (\mathbb{R}^k \times \{ 0 \}) = V_{\epsilon} \cap (\mathbb{R}^k \times \{ 0 \}) \subset V_{\epsilon}$. Then $y \in h^{-1}(V_{\epsilon}) = U_{\epsilon}$. Therefore, $U_{\epsilon} \cap M = U \cap M$.

Hence, $h$ is a diffeomorphism from an open set $U_{\epsilon}$ to an open set $V_{\epsilon}$ such that $h(U_{\epsilon} \cap M) = V_{\epsilon} \cap (\mathbb{R}^k \times \{ 0 \})$. By Theorem 3-13, we can find an $\epsilon > 0$ that makes the area of $U_{\epsilon}$ arbitrarily small while covering $U \cap M$.

The argument above works for any $x \in M$, and for each $x$, we obtain an open set $U$ containing $x$. Since $M \subset \mathbb{R}^n$, there exists a countable open cover of $M$ formed by such $U$’s. Let $x_1, \cdots$ and $U_1, \cdots$ denote them. Let $\delta > 0$ be given. Then using the argument above, find $U_{\epsilon_i}$ such that

$U_{\epsilon_i}$ is open and $U_{\epsilon_i} \cap M = U_i \cap M$.
$h_i\vert_{U_{\epsilon}}$ is a diffeomorphism.
The volume of $U_{\epsilon_i}$ is less than $\delta / 2^i$.

Then $U_{\epsilon_1}, \cdots$ cover $M$ and the total area is less than $\delta$, an arbitrarily chosen positive number. Therefore, $M$ has measure 0.

2

Let $x \in \bd M$. Then $x \in M$ because $M$ is closed. Suppose $x \notin \partial M$. By Theorem 5-2[Spivak], there exists an open set $U$ around $x$ and an open set $W$ in $\mathbb{R}^k$ such that 1-1 differentiable function $f: W \rightarrow U$ such that

$f(W) = M \cap U$,
$f^{-1}: M \cap U \rightarrow W$ is continuous.

This implies that $M \cap U$ is an open subset of $\mathbb{R}^n$. However, there cannot be an open neighborhood of $x$ that is contained in $M$ because $x$ is a boundary point. Therefore, $x \in \partial M$.

Let $x \in \partial M$. We will show that $x \in \bd M$. Suppose $x \notin \bd M$. Then $x$ is an interior point of $M$. Since $x \in \partial M$, there exists an open set $U$ containing $x$, an open set $V \subset \mathbb{R}^n$, and a diffeomorphism $h: U \rightarrow V$ such that $h(U \cap M) = V \cap (\mathbb{H}^k \times \{ 0 \})$ and the $k$th component of $h(x)$ is 0. This implies that $h(x)$ is not an interior point of $V \cap (\mathbb{H}^k \times \{ 0 \})$. This is a contradiction because $h$ is a homeomorphism. Therefore, $x \in \bd M$.

Thus $\bd M = \partial M$.

Let $M = [0, 1)$. Then $M$ is a 1-dimensional manifold-with-boundary and its point-set-topology-boundary is $\{ 0, 1 \}$. However, the boundary of $M$ (as a manifold) is $\{ 0 \}$.

3

Since $M$ is compact, $M$ is closed. Thus $\partial M = \bd M$ by Part 2 above.

As shown before, $\partial M$ is an $(n - 1)$-dimensional manifold.

By Part 1 above, $\partial M$ has measure 0.

Since $M$ is compact, $M$ is bounded. Thus $M$ is a bounded set whose boundary has measure 0. Therefore, $M$ is Jordan-measurable.