Let $F$ be a smooth map from $M$ to $N$, which are smooth manifolds with $\dim M < \dim N$. Prove without using Sard's theorem that $F(M)$ has measure $0$ in $N$.
I want to reduce this problem to showing it for open subsets of $\mathbb{R}^m, \mathbb{R}^n$ respectively, where $m=\dim(M), n=\dim(N)$. This is the part that I am unsure about. Please verify if my logic is correct.
My proof:
Let $(U, \varphi)$ be a chart in $N$. We have to show that $\varphi(F(M)\cap U)$ has measure $0$ in $\varphi(U)$. Since $F(M\cap U) =F(F^{-1}(U))$ and $F^{-1}(U) \hookrightarrow M \to N$ is smooth, it is enough to show that the restriction of $F$ to $F^{-1}(U)=:V$ has measure $0$ in $N$. We want to show that $\varphi(F(V))$ has measure $0$ in $\varphi(U)$. Since $\varphi$ is a smooth map, $V\to U \to \varphi(U)$, replacing $\varphi \circ F$ by $F$ and $\varphi(U)$ by $U$, we can assume $U$ is an open subset of $\mathbb{R}^n$. Since $V$ is also a smooth manifold, we can find a countable cover of charts $\{V_i, \psi_i\}$. Since $F(V)=\bigcup F(\psi_i^{-1} ( \psi_i (V_i)))$, it is enough to show that $F(\psi_i^{-1} ( \psi_i (V_i)))$ has measure $0$ in $U$. Since $F\circ \psi_i^{-1}$ is smooth, by replacing we can again assume that $F$ is a smooth map from $V\subset \mathbb{R}^m$ to $U\subset \mathbb{R}^n$.
Now $F$ is the composition of $V\to V\times \mathbb{R}^{n-m} \to U$ where the first map is just identity with $0s$ on the last coordinates and the second map just ignores the lasts coordinates. Both of them are smooth. The image of the first map has measure $0$ (already proven earlier), and since the second map is a smooth map $\mathbb{R}^{n} to \mathbb{R}^{n}$, it takes measure $0$ to measure $0$ sets (also already proven), hence we are finished.