# Show that $F(M)$ has measure zero in $N$, where $F:M \to N$ is a smooth map with $\dim(M) \lt \dim(N)$

Suppose that $M$ and $N$ are smooth manifolds. Show that $F(M)$ has measure zero in $N$, where $F:M \to N$ is a smooth map with $\dim(M) \lt \dim(N)$

Hint: Define a suitable map from $M \times \mathbb{R^k}\to N$ and use the theorem

Suppose $A \subset \mathbb{R}^n$ has measure zero and $F: A \to \mathbb{R}^n$ is a smooth map. Then $F(A)$ has measure zero.

Let $G: M \times \mathbb{R}^k \to N$ be defined by $G(x,y^1,y^2,\ldots,y^k)=F(x)$ for $x \in M$ and $(y^1,y^2,\ldots,y^k) \in \mathbb{R}^k$. Then $G$ is smooth. Let $(U\times \mathbb{R}^k,\phi \times \text{id})$ and $(V,\lambda)$ be smooth charts for $(x,y^1,y^2,\ldots,y^k) \in M \times \mathbb{R}^k$ and $F(x) \in N$. Then $$\lambda \circ G \circ (\phi \times \text{id})^{-1}: \phi(U) \times \mathbb{R}^k \subset \mathbb{R}^{\dim N}\to \lambda(V) \subset \mathbb{R}^{\dim N}$$ is smooth. It is enough to show that image of $G$ has measure zero.

$$\lambda\left(G\left((U \times \mathbb{R}^k) \cap G^{-1}(V)\right)\right)=\left(\lambda \circ G \circ (\phi \times \text{id})^{-1}\right) \circ (\phi \times \text{id})((U \times \mathbb{R}^k) \cap G^{-1}(V))$$ $$=\left(\lambda \circ G \circ (\phi \times \text{id})^{-1}\right) \circ\left(\phi(U \cap F^{-1}(V)) \times \mathbb{R}^k \right)$$

Clearly $\phi(U \cap F^{-1}(V)) \times \mathbb{R}^k$ is open in $\mathbb{R}^{\dim N}$. All I have to show is that it is an open subset of $\mathbb{R}^{\dim N-1}$. Then it has measure zero and by the above theorem I will be able to show that $G(U \times \mathbb{R}^k)$ has measure zero. This implies that $F(M)$ has measure zero.

How do I conclude that $\phi(U \cap F^{-1}(V)) \times \mathbb{R}^k$ is an open subset of $\mathbb{R}^{\dim N-1}$?

Thanks for the help!!

• If you're applying that theorem, why are you trying to prove it? You only need to argue that $U\subset U\times\Bbb R^k$ is a set of measure zero, no? – Ted Shifrin Mar 18 '17 at 22:31
• @TedShifrin That is true. You are saying that all I have to show is that $\phi(U) \subset \phi(U) \times \mathbb{R}^k$ has measure zero (which is true by the way) and conclude from it? – tattwamasi amrutam Mar 18 '17 at 22:43
• That was my impression. The theorem itself can be proved by using $C^1$ to give bounds on volume. – Ted Shifrin Mar 18 '17 at 22:45
• @TedShifrin That's it then. Right? – tattwamasi amrutam Mar 18 '17 at 23:22