Lemma 6.2 Lee's Introduction to Smooth Manifolds I am trying to understand how to apply Fubini's Theorem to the following lemma

Suppose $A \subset \mathbb{R}^n$ is a compact subset whose intersection with $\left\{ c \right\} \times \mathbb{R}^{n-1}$ has $n-1$-dimensional measure zero for every $c \in \mathbb{R}$. Then $A$ has $n$-dimensional measure zero.

The author claims this lemma follows easily from Fubini's theorem. I was trying trying to spot how to apply the theorem by observing
$$
A = \bigcup_{c \in \mathbb{R}} \left(\left\{ c \right\} \times \mathbb{R}^{n-1} \right) \cap A
$$
So $m(A) = m(\bigcup_{c \in \mathbb{R}} \left(\left\{ c \right\} \times \mathbb{R}^{n-1} \cap A\right))$ (here $m$ is the Lebesgue measure) but I don't know if this is any useful (I guess I was trying to write $A$ as product space so I could apply the theorem, but it doesn't look like I am going anywhere with this.
Can you maybe explain how to apply the theorem?
 A: I'll answer my own question, since I was encouraged to do so, however I think my use of notation isn't correct and it would be great if anyone would check. I'll denote with $m_n$ the $n$-dimensional Lebesgue measure and $\chi_A$ the characteristic function of the set $A$. Now we have
$$
m_{n}(A) = \int_{A} dm_n = \int_{\mathbb{R}^n} \chi_A dm_n = \int_{\mathbb{R}\times \mathbb{R}^{n-1}} \chi_A dm_n = \int_{\mathbb{R}} \int_{\mathbb{R}^{n-1}} \chi_A d(m \times m_{n-1}) = \int_{\mathbb{R}} \left( \int_{\mathbb{R}^{n-1}} \chi_A dm_{n-1} \right) \times dm = \int_{\mathbb{R}} \left( \int_{\mathbb{R}^{n-1}} \chi_A(x^1,\ldots,x^n) dm_{n-1}(x^2,\ldots,x^n) \right) \times dm(x^1) = \int_{\mathbb{R}} \left( \int_{A \cap \left\{ x^1 \right\} \times \mathbb{R}^{n-1}}  dm_{n-1}(x^2,\ldots,x^n) \right) \times dm(x^1) = \int_{\mathbb{R}} m_{n-1}(A \cap \left\{ x^1 \right\} \times \mathbb{R}^{n-1}) d_m(x^1)
$$
By hypothesis we have
$$
m_{n-1}(A \cap \left\{ x^1 \right\} \times \mathbb{R}^{n-1}) = 0
$$
So we have  $m_n(A) = 0$
My only doubt is the writing of
$$
\int_{\mathbb{R}^{n-1}} \chi_A d m_{n-1} = \int_{A \cap \left\{ x^1 \right\} \times \mathbb{R}^{n-1}} dm_{n-1}
$$
Is it correct?
I think I can justify by defining the family of functions $\left\{ \varphi_{x^1} \right\}$ where
$$
\varphi_{x^1} : \mathbb{R}^{n-1} \to A\cap\left\{ x^1 \right\} \times \mathbb{R}^{n-1}
$$
for every $x^1 \in \mathbb{R}$ this is defined as $\varphi_{x^1}(x^2,\ldots,x^n) = \left( x^1,x^2,\dots, x^n \right)$
With this definition I can write
$$
\int_{\mathbb{R}^{n-1}} \chi_A d m_{n-1} = \int_{\varphi_{x^1}^{-1} (A\cap\left\{ x^1 \right\} \times \mathbb{R}^{n-1})} \chi_A \circ \varphi_{x^1} (x^2,\ldots,x^n) d \left( m_{n-1} \circ \varphi_{x^1}\right)(x^1,x^2,\ldots,x^n) = \int_{\varphi_{x^1}^{-1} (A\cap\left\{ x^1 \right\} \times \mathbb{R}^{n-1})} \chi_{\varphi_{x^1}^{-1} (A\cap\left\{ x^1 \right\} \times \mathbb{R}^{n-1})} (x^2,\ldots,x^n) d \left( m_{n-1} \circ \varphi_{x^1}\right)(x^1,x^2,\ldots,x^n) = \int_{\varphi_{x^1}^{-1} (A\cap\left\{ x^1 \right\} \times \mathbb{R}^{n-1})} d \left( m_{n-1} \circ \varphi_{x^1}\right)(x^1,x^2,\ldots,x^n) = \left( m_{n-1} \circ \varphi_{x^1}^{-1} \right) (A\cap\left\{ x^1 \right\} \times \mathbb{R}^{n-1})
$$
I think this looks more correct, but I am still in doubt... I think I am getting closer anyway.
