We have the following theorem:
Let $f: A \subset \mathbb{R}^n \to \mathbb{R}^m$ be Lipschitz, with Lipschitz constant Lip $f$, $1 \leq n \leq m$. Then we have
$$ \int_{\mathbb{R}^m} H^0(f^{-1}(y) \cap A) dH^n(y) = \int_A J_f(x) dL^n $$ Where $H^0$ is the counting measure, and $f^{-1}(y) = \{ x : f(x) = y \}$, $J_f(x)$ is the Jacobian of $f$ at $x$.
The proof given in Maggi's book, consists of three steps:
We prove that $ H^0(f^{-1}(y) \cap A)$ is an $H^m$ measurable function of $y$ on $\mathbb{R}^m$
We prove that for any Lebesgue measurable set $A \subset \mathbb{R}^n$, we have
$$\int_{\mathbb{R}^m} H^0(f^{-1}(y) \cap A) dH^n(y) \leq (Lip f)^n L^n(A) $$
Then Maggi has a comment: "As a consequence, we may freely modify $A$ on subsets of measure zero." EDIT: I understand why this comment is true, because if $A_0 \subset A $ has measure zero, the integral with $A_0$ will be $\leq |A_0| = 0$. I just dont understand if or how it helps for step 3.
- Using step 2, and the fact that for $E = \{ x: J_f(x) = 0 \}$ we have $H^n(f(E)) = 0$, we can directly assume that $A \subset F$ where $F = \{ x: 0 < J_f(x) < \infty \}$. The proof continues.
My question is how does the comment in step 2 help, and how do we get step 3 from step 2?