Let $M$ be a smooth manifold in $\mathbb{R}^n$. If Lebesgue measure of $M$ is zero i.e $l(M)=0$, does it mean that volume of manifold is also zero i.e $Vol(M)=0$? Are they the same thing (volume and Lebesgue measure).


No, that doesn't mean $\mathrm{Vol}(M)=0$. The volume of an oriented Riemannian manifold $M$ is an intrinsic quantity defined by $$\mathrm{Vol}(M)=\int_MdV,$$ where $dV$ is the Riemannian volume form of $M$. Hence, for example, if $M\subseteq\Bbb R^3$ is a surface, then $\mathrm{Vol}(M)$ is define as the area of $M$, which is in general non-zero, while its Lebesgue measure in $\Bbb R^3$ is zero. For instance, if $S^2_r\subseteq\Bbb R^3$ is the sphere of radius $r$, then $$\mathrm{Vol}(S^2_r)=4\pi r^2,\quad l_{\Bbb R^3}(S^2_r)=0.$$

However, if $M\subseteq\Bbb R^n$ is an open submanifold of $\Bbb R^n$, then the Riemannian volume form of $M$ is the restriction of that of $\Bbb R^n$ and hence we have $$\mathrm{Vol}(M)=\int_MdV=\int_M dx^1\cdots dx^n=l(M),$$ so the two quantities are the same.

Conclusion: $\mathrm{Vol}(M)$ is an intrinsic quantity, while $l(M)$ is an extrinsic quantity that depends on the particular embedding. Hence, they are in general not equal.

Additional Remark: To prove that if $M\subseteq\Bbb R^n$ is a smooth manifold of dimension $<n$, then $l(M)=0$, we can use Sard's Theorem. The inclusion $\iota:M\hookrightarrow \Bbb R^n$ has critical values everywhere because $\dim M<n$, so $\iota(M)=M$ has measure $0$ in $\Bbb R^n$.

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  • $\begingroup$ Cool :)! Thank you for editing +1 $\endgroup$ – ClassicStyle Jul 18 '15 at 17:40

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