[proofplan] We express $\mathcal{L}^n(E)$ and $\mathcal{L}^n(F)$ as integrals of the cross-sectional measures $\mathcal{L}^{n-1}(E_t)$ and $\mathcal{L}^{n-1}(F_t)$ using [Tonelli's Theorem](/theorems/3017) applied to the indicator functions $\mathbb{1}_E$ and $\mathbb{1}_F$ on the product space $\mathbb{R}^{n-1} \times \mathbb{R}$. Since the integrands agree $\mathcal{L}^1$-a.e., the integrals are equal. The Lebesgue measure $\mathcal{L}^n$ factors as the product $\mathcal{L}^{n-1} \otimes \mathcal{L}^1$, which is the structural fact underlying the entire argument. [/proofplan] [step:Express $\mathcal{L}^n(E)$ as an iterated integral of cross-sectional measures] Since $\mathcal{L}^n = \mathcal{L}^{n-1} \otimes \mathcal{L}^1$ on $\mathbb{R}^n = \mathbb{R}^{n-1} \times \mathbb{R}$, the measure spaces $(\mathbb{R}^{n-1}, \mathcal{B}(\mathbb{R}^{n-1}), \mathcal{L}^{n-1})$ and $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{L}^1)$ are both $\sigma$-finite (each can be written as a countable union of bounded intervals, which have finite Lebesgue measure). The function $\mathbb{1}_E : \mathbb{R}^{n-1} \times \mathbb{R} \to \{0, 1\}$ is $\mathcal{B}(\mathbb{R}^{n-1}) \otimes \mathcal{B}(\mathbb{R})$-measurable because $E$ is $\mathcal{L}^n$-measurable and hence Borel-measurable up to a null set (since $\mathcal{L}^n$ is the completion of the Borel measure; the indicator of the Borel representative has the same integral). The function $\mathbb{1}_E$ is non-negative, so [Tonelli's Theorem](/theorems/3017) applies with $\mu = \mathcal{L}^{n-1}$, $\nu = \mathcal{L}^1$, and $f = \mathbb{1}_E$. The hypotheses are satisfied: both measure spaces are $\sigma$-finite, and $\mathbb{1}_E \geq 0$ is measurable. Tonelli gives: \begin{align*} \mathcal{L}^n(E) = \int_{\mathbb{R}^n} \mathbb{1}_E\, d\mathcal{L}^n = \int_{\mathbb{R}} \left(\int_{\mathbb{R}^{n-1}} \mathbb{1}_E(x, t)\, d\mathcal{L}^{n-1}(x)\right) d\mathcal{L}^1(t). \end{align*} For each fixed $t \in \mathbb{R}$, the inner integral is \begin{align*} \int_{\mathbb{R}^{n-1}} \mathbb{1}_E(x, t)\, d\mathcal{L}^{n-1}(x) = \mathcal{L}^{n-1}(\{x \in \mathbb{R}^{n-1} : (x, t) \in E\}) = \mathcal{L}^{n-1}(E_t). \end{align*} Therefore: \begin{align*} \mathcal{L}^n(E) = \int_{\mathbb{R}} \mathcal{L}^{n-1}(E_t)\, d\mathcal{L}^1(t). \end{align*} [/step] [step:Apply the same decomposition to $F$ and compare] The identical argument applied to $F$ gives: \begin{align*} \mathcal{L}^n(F) = \int_{\mathbb{R}} \mathcal{L}^{n-1}(F_t)\, d\mathcal{L}^1(t). \end{align*} By hypothesis, $\mathcal{L}^{n-1}(E_t) = \mathcal{L}^{n-1}(F_t)$ for $\mathcal{L}^1$-a.e. $t \in \mathbb{R}$. The integrands in the two formulas agree $\mathcal{L}^1$-almost everywhere, so their Lebesgue integrals are equal: \begin{align*} \mathcal{L}^n(E) = \int_{\mathbb{R}} \mathcal{L}^{n-1}(E_t)\, d\mathcal{L}^1(t) = \int_{\mathbb{R}} \mathcal{L}^{n-1}(F_t)\, d\mathcal{L}^1(t) = \mathcal{L}^n(F). \end{align*} This completes the proof. [/step]