# Lebesgue measure of a manifold embedded in the euclidean space

• Nov 30th 2011, 07:42 PM
uasac
Lebesgue measure of a manifold embedded in the euclidean space
Let M be a m-dimensional manifold embedded in a n-dimensional euclidean space. If m is strictly less than n, is its Lebesgue measure zero?

Can anyone give me a reference which can prove/disprove this?

Thanks
• Nov 30th 2011, 08:23 PM
Drexel28
Re: Lebesgue measure of a manifold embedded in the euclidean space
Quote:

Originally Posted by uasac
Let M be a m-dimensional manifold embedded in a n-dimensional euclidean space. If m is strictly less than n, is its Lebesgue measure zero?

Can anyone give me a reference which can prove/disprove this?

Thanks

Here's how I would think about doing it. Since $M$ can be embedded into $\mathbb{R}^n$ and subspaces of Lindelof spaces are Lindelof we know that $M$ is Lindelof and so we can find a countable atlas $\{(U_k,\varphi_k)\}_{k\in\mathbb{N}}$ for $M$. Since the countable union of null sets is null it suffices to show that each $U_k$ has measure zero. To do this consider that by considering the map $\mathbb{R}^m\xrightarrow{\iota\circ\varphi_k^{-1}}\mathbb{R}^n$ (where $\iota$ is the embedding $M\hookrigtharrow \mathbb{R}^n$) you get a smooth map from $\mathbb{R}^m$ into $\mathbb{R}^n$ and since $m we know (by lemma 10.3 in Lee) the image of $\iota\circ\varphi_k^{-1}$ has measure zero. And, since measure zero sets are preserved under diffeomorphisms we can pull back along $\iota$ to get that $U_k$ has measure zero.
• Nov 30th 2011, 08:35 PM
uasac
Re: Lebesgue measure of a manifold embedded in the euclidean space
I'm just starting to learn differential geometry, but your arguments seem sound to me. Thank you.