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Consider $M\subseteq \mathbb{R}^n$, an (embedded) $C^{k\geq 1}$ manifold.

Is $M$ Borel?

I think yes but I got stuck at proving this. For some context, I was checking that integrals over manifolds with respect to Hausdorff measures make sense.

My try:

  1. Graphs of continuous functions are Borel: let $U$ be open and $f:U \rightarrow \mathbb{R}^q$ be continuous. Let $\phi(x)=(x,f(x))$, a continuous function. We can write $U=\bigcup_n K_n$ for some compact sets $K_n \subseteq U$ and therefore $\text{graf}f=\phi(U)=\bigcup_n\phi(K_n)$, so that the graph of $f$ is the countable union of compact (hence Borel) sets.
  2. A manifold is "countably locally" the graph of a $C^k$ function: by this I mean that $M$ is the countable union of the graphs of some $C^k$ functions. Here I am stuck. I know that $M$ is locally a graph, but I cannot get the countability property. I was thinking of using the density of rationals in the real numbers but I'm not sure if this is the way to go, also beacause I'm not sure that second countability of $M$ is guaranteed with my definition of manifold (see below).

Do you have any hint on how to get through point 2, provided my claim is correct?


$M\subseteq \mathbb{R}^n$ is an embedded $C^{k\geq 1}$ $m-$manifold if for every $p$ in $M$ there exists $U\subseteq \mathbb{R}^m$ open and $\phi:U\rightarrow M$ of class $C^{k}$ such that:

  • $\phi(U)$ is a neighbourhood of $p$, open in $M$
  • $\phi$ is a homeomorphism onto its image
  • $d\phi$ is everywhere injective
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$\bullet$ $\mathbb R^n$ is second countable with a countable basis $B$ given by all balls with rational radii and centers.

$\bullet$ Subspaces of second countable spaces are second countable: If $Y\subseteq X$ and $B$ is a countable basis for $X$, then $\{U\cap Y: U\in B\}$ is a countable basis for $Y$.

$\bullet$ Second countable spaces are Lindelöf: Any open cover admits a countable subcover (see for example here or here).

So an embedded submanifold $M$ of $ \mathbb R^n$ is is "countably locally" the graph of a continous function, which finishes your proof.

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  • $\begingroup$ Then I just need to pick a countable atlas and repeat the argument I used in my first point, right? Other than that, thank you very much! $\endgroup$
    – warm_fish
    Oct 13 '20 at 16:32
  • $\begingroup$ I dont think you need to take a countable atlas to begin with: For each point in your manifold $M$ you find an open (w.r.t $M$) neighborhood which (up to coordinate permutation) is the graph of a continous function. By 1. this gives you a cover of $M$ by Borel sets and since there is a countable subcover, $M$ is a Borel set as a countable union of Borel sets. $\endgroup$ Oct 13 '20 at 16:39
  • $\begingroup$ Okay, I didn't know about the Lindelöf property, thank you again $\endgroup$
    – warm_fish
    Oct 13 '20 at 16:42

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