How to argue that $\int_{\mathbb{R}^{n-m}}\mathcal{X}_A(x,y)d\mathcal{L}^{n-m}(x)=\mathcal{H}^{n-m}(A\cap P^{-1}(y))$? I have a problem giving a complete argument of this statement from the page 126, Measure theory and fine properties of functions by Evans, 2e.

$A$ is a $\mathcal{L}^n-$measurable subset of $\mathbb{R}^n$. $P$ is an orthogonal projection of $\mathbb{R}^n$ onto $\mathbb{R}^m$ where $n\ge m$. Then for each $y\in\mathbb{R}^m$, $P^{-1}(y)$ is the inverse image of $y$ under $P$ and here is an $(n-m)-$dimensional affine subspace of $\mathbb{R}^n$. By Fubini's theorem,
  $$y\mapsto \mathcal{H}^{n-m}(A\cap P^{-1}(y))\text{ is $\mathcal{L}^m$ measurable}$$
  and
  $$\int_{\mathbb{R^m}}\mathcal{H}^{n-m}(A\cap P^{-1}(y))d\mathcal{L}^{m}(y)=\mathcal{L}^n(A).$$ 

$\mathcal{H}^{n-m}$ is Hausdorff measure. Someone may see that this is part of proof of co-area formula. For the part after 'By Fubini's theorem', I can only give a partial argument

I know $\mathcal{L}^n(A)=\int_{\mathbb{R}^n}\mathcal{X}_A(z)d\mathcal{L}^n(z)$. By Fubini's theorem (the version for non-negative functions), the right-hand side is equivalent to
  $$\int_{\mathbb{R}^n}\mathcal{X}_A(z)d\mathcal{L}^n(z)=\int_{\mathbb{R}^m}\int_{\mathbb{R}^{n-m}}\mathcal{X}_A(x,y)d\mathcal{L}^{n-m}(x)d\mathcal{L}^m(y),$$where $y\mapsto\int_{\mathbb{R}^{n-m}}\mathcal{X}_A(x,y)d\mathcal{L}^{n-m}(x)$ is $\mathcal{L}^m-$measurable. Then the remaining step is to prove that for a fixed $y$, $$\int_{\mathbb{R}^{n-m}}\mathcal{X}_A(x,y)d\mathcal{L}^{n-m}(x)=\mathcal{H}^{n-m}(A\cap P^{-1}(y)).$$ This is where I need help. I know $\mathcal{X}_A(x,y)=\mathcal{X}_{A\cap P^{-1}(y)}(x,y)$ for a fixed $y$. And I know Hausdorff measure is equivalent to Lebesgue measure when the dimension is the same and I feel I'm close to the conclusion but I can't give a rigorous argument. Thanks for any tip.

 A: We can use a lemma in the book of Evans and Gariepy (Lemma 1 in the section "The Area Formula"):
Lemma: Suppose $L: \mathbb{R}^p \to \mathbb{R}^q$ is linear, $p \leq q$. Then,
$$ \mathcal{H}^p(L(B)) = [[ L]] \mathcal{L}^p(B), \; \forall B \subset \mathbb{R}^p.
$$ Here $[[L]]$ is the determinant of the map $S$ in the decomposition $L = O \circ S$, where $O: \mathbb{R}^p \to \mathbb{R}^q$ is orthogonal and $S: \mathbb{R}^p \to \mathbb{R}^p$ is symmetric. In particular, if $L$ is orthogonal, then $[[L]] = \det \mathrm{Id}_{\mathbb{R}^p} = 1$.
Now we prove the equation that you pointed out. Applying the lemma with
$p = n - m$; $q = n$; $L: \mathbb{R}^{n - m} \to \mathbb{R}^n, x \mapsto (x, 0)$ (observe that $L$ is orthogonal); and $B = \{x \in \mathbb{R}^{n - m}: (x, y) \in A \}$, we obtain
\begin{align} \int_{\mathbb{R}^{n - m}} \chi_{A}(x, y) d \mathcal{L}^{n - m} (x) &= \mathcal{L}^{n - m} (\{x \in \mathbb{R}^{n - m}: (x, y) \in A\}) \\
&= \mathcal{H}^{n - m}(\{(x, 0) | x \in \mathbb{R}^{n - m}, (x, y) \in A \}) \text{ (using the lemma)} \\
&= \mathcal{H}^{n - m} ( (A \cap P^{-1}(y)) - \{(0, y)\}) \\
&= \mathcal{H}^{n - m} (A \cap P^{-1}(y)) \text{ (by translation-invariance of Hausdorff measures)},
\end{align}
and your equation follows.
