I went through the proof of Theorem 3.2.3(1) (the Area Formula) from Federer's GMT book. The statement of the theorem is tha following: Suppose $f:\mathbb{R}^m \to \mathbb{R}^n$ is Lipschitzian with $m\leq n$. If $A$ is an $\mathscr{L}^{m}$-measurable set, then $$ \int_{A}J_{m}f(x) \, d\mathscr{L}^{m}(x) = \int_{\mathbb{R}^{n}}N(f|_{A},y) \, d\mathscr{H}^{m}(y) $$
(Here $J_{m}f(x)=||\wedge_{m} f'(x)||$ where $\wedge_{m} f'(x):\wedge_{m}\mathbb{R}^m \to \wedge_{m}\mathbb{R}^n$ is induced by $f'(x)$ and $\wedge_{m} V$ means the $m$-th exterior power of $V$.)
The second part of the proof deals with the case where $$A\subseteq \{ x; \, \dim \ker f'(x) >0 \} = \{ x ; \, J_{m}f(x)=0 \}.$$ For any $\varepsilon >0$ he considers the decomposition $f=p\circ g$ where $g:\mathbb{R}^m \to \mathbb{R}^n \times \mathbb{R}^m $ and $p:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^n$ are given by $g(x)=(f(x), \varepsilon x)$ and $p(y,z)=y$. After that he concludes: ''$x\in A$ implies $$g'(x)\cdot v=(f'(x)\cdot v, \varepsilon v),$$ $$g'(x) \, \text{is injective},$$ $$||g'(x)||\leq \mathrm{Lip}(f) +\varepsilon,$$ $$J_{m}g(x)\leq\varepsilon (\mathrm{Lip}(f) +\varepsilon)^{m-1}$$ because $|g'(x)\cdot v|=\varepsilon |v|$ for $v\in \ker f'(x)$.'' I was not able to verify the last inequality above. Can anyone give me a hint on how to do it?
If you denote by $Dg(x)$ the Jacobian matrix, there is a formula for $J_mg(x)$, which is $$J_mg(x)=\sqrt{D^tg(x)Dg(x)}$$ now you use the Cauchy-Binet theorem, which tells you that the expression under the square root is given by the sums of the squares of all the $m\times m$ minors. Since you are in the set $A$, all the $m\times m$ minors which use only the derivatives of $f$ are zero, so you are left with the minors where at least one row contains an $\epsilon$. Those you can bound in terms of $\varepsilon^2$ multiplied a constant that depends on $Lip f$ or, in the case of the identity by $\varepsilon^{2m}$. The exact expression is not important. All you care is that $J_mg(x)\le C\varepsilon$, since you are sending $\varepsilon$ to zero.