Steiner symmetrization of Lebesgue measurable set

Question

I'm reading a proof in "Evans / Gariepy: Measure theory and fine properties of functions" of the following statemant:

Theorem: Let $A \subset \mathbb R^n$ be $\lambda^n$-measurable, $a \in \mathbb R^n$ with $\Vert a \Vert = 1$. Then $S_a(A)$ is $\lambda^n$-measurable and $\lambda^n(A) = \lambda^n(S_a(A))$.

Here $S_a(A)$ denotes the the Steiner symmetrization w.r.t the plane through $0 \in \mathbb R^n$ and perpendicular to $a$. This plane is denoted by $P_a$. We use the notation $L_b^a$ for the line through $b \in \mathbb R^n$ with direction $a$.

Now in the proof we set $a = (0, \ldots, 0, 1) \in \mathbb R^n$. Then we have $P_a = \mathbb R^{n-1}$. We define a function $f: \mathbb R^{n-1} \to \mathbb R$ by $$f(b) = \mathcal H^1(A \cap L_b^a) \; ,$$ where $\mathcal H^1$ denotes the "one-dimensional" Hausdorff measure. We know that $\mathcal H^1 = \lambda^1$ on $\mathbb R$. Now the author claims, that by Fubini the function $f$ is $\lambda^{n-1}$-measurable and $$\lambda^n(A) = \int_{\mathbb R^{n-1}} f(b) \; \mathrm db \; .$$ Why does this hold? Could maybe someone elaborate this?

Answer 1

Write points $x = (x_1,\ldots,x_n) \in \mathbb R^n$ as $x = (b,z)$ where $b = (x_1,\ldots x_{n-1}) \in \mathbb R^{n-1}$ and $z = x_n\in \mathbb R$. If $A \subset \mathbb R^n$ is measurable then Tonelli's theorem states that $$\lambda^n(A) = \int_{\mathbb R^n} \chi_A(x) \, dx = \int_{\mathbb R^{n-1}}\left(\int_{\mathbb R} \chi_A(b,z) \, dz \right) db,$$ where $z \mapsto \chi_A(b,z)$ is $\lambda^1$-measurable for $\lambda^{n-1}$ almost all $b \in \mathbb R^{n-1}$, and $b \mapsto \displaystyle \int_{\mathbb R} \chi_A(b,z) \, dz$ is $\lambda^{n-1}$ measurable.

Fix $b \in \mathbb R^{n-1}$. If $z \mapsto \chi_A(b,z)$ is $\lambda^1$-measurable then $$\int_{\mathbb R} \chi_A(b,z) \, dz = \lambda^1(\{z \in \mathbb R: (b,z) \in A\}) = {\cal H}^1(\{z \in \mathbb R: (b,z) \in A\}).$$

There is an (obvious?) isometry between the sets $\{z \in \mathbb R: (b,z) \in A\}$ and $\{(y,z) \in \mathbb R^n : y=b,\ (y,z) \in A\} = A \cap L_a^b$. Thus $${\cal H}^1(\{z \in \mathbb R: (b,z) \in A\}) = {\cal H}^1(A \cap L_a^b),$$ meaning that $$ \int_{\mathbb R} \chi_A(b,z) \, dz = {\cal H}^1(A \cap L_a^b)$$ for $\lambda^{n-1}$ almost every $b \in R^{n-1}$. It follows that $b \mapsto {\cal H}^1(A \cap L_a^b)$ is $\lambda^{n-1}$ measurable and that $$\lambda^n(A) = \int_{\mathbb R^{n-1}} {\cal H}^1(A \cap L_a^b) \, db.$$

Answer 2

$$ \lambda^n (A) = \int_{R^n} I_A (x_1,\ldots,x_n) d\lambda^n(x_1,\ldots,x_n)=\int_{R^{n-1}} \int_R I_A (x_1,\ldots,x_{n-1},x_n) d \lambda (x_n) d\lambda^{n-1} (x_1,\ldots,x_{n-1}) = \int_{R^{n-1}} f(x_1,\ldots,x_{n-1}) d\lambda^{n-1}(x_1,\ldots,x_{n-1}) = \int_{R^{n-1}} f(b) db. $$