I want to prove a particular case of Sard's Theorem, obviously without using the main Sard's Theorem.
Let $f:M\to N$ be a differentiable ($C^{\infty}$) function. If $m=\dim M<\dim N=n$, then $f(M)$ has measure zero in $N$.
Hint: It is enough to show that if $g:U\subseteq\mathbb{R}^m\to\mathbb{R}^n$ is differentiable, where $U$ is open, then $g(U)$ has measure zero in $\mathbb{R}^n$.
I proved the hint:
Just consider the projection $\pi:U\times\mathbb{R}^{n-m}\to\mathbb{R}^m$ over $U$ and $g\circ\pi:U\times\mathbb{R}^{n-m}\to\mathbb{R}^n$. Because $U\times\mathbb{R}^{n-m}$ and $\mathbb{R}^n$ are in the "same dimention" and $g\circ\pi$ is differentiable, this function sends null sets to null sets. So that $(g\circ\pi)(U\times\{0\}^{n-m})=g(U)$ has measure zero.
But I don't understand why this is enough.
In order to show $f(M)$ has measure zero in $N$, we need to show that if $(V,\phi)$ is a chart in $N$ then $\phi(f(M)\cap V)$ has measure zero in $\mathbb{R}^n$. As the hint suggests, how can I construct some $g:U\subseteq\mathbb{R}^m\to\mathbb{R}^n$ differentiable from here?
Let $(\psi,U)$ be a chart on $M$ and $\varphi$ a local chart on $N$. The overlap map given by;
$$g=\varphi \circ f \circ \psi^{-1}: \psi(U) \subset \mathbb{R}^m \to \mathbb{R}^n$$
is differentiable. Here treat $\psi(U)$ as $U$ since $\psi$ is a local homeomorphism. Therefore, from what you've shown, it follows that $g(\psi(U))=\varphi(f(U))$ has measure zero. So we have a covering of ball $\{B_{\alpha}\}$ of $\varphi(f(U))$ such that $\sum vol(B_{\alpha}) < \delta$. Hence, $\{\varphi^{-1}(B_{\alpha})\}$ is a covering of $f(U)$ which has measure zero.
Now recall that there exists a countable basis $\{U_j\}$ of $M$ and therefore $f(M) \subset \bigcup_j f(U_j)$. However, we know know that each $f(U_j)$ has measure zero; say $vol(f(U_j))< \frac{\delta}{2^j}$ and so,
$$vol(f(M)) \leq \sum_j vol(f(U_j))< \sum_j\frac{\delta}{2^j} = \delta$$