Null set if intersection with preimage under a linear map is countable

Question

Let $T \subseteq \mathbb{R}^n$ be an measurable set and $$\varphi:\mathbb{R}^n \rightarrow \mathbb{R}^{n-1}$$ a surjective linear map such that for all $x \in \mathbb{R}^{n-1}$ the set $T \cap \varphi^{-1}(x)$ ist countable. Show $$\lambda^n(T) = 0.$$

It's obvious that $\lambda^n(T \cap \varphi^{-1}(x)) = 0$, since $T \cap \varphi^{-1}(x)$ is countable and we can write $T = \biguplus_{x\in \mathbb{R}^{n-1}} T \cap \varphi^{-1}(x)$, but the problem is that this is an uncountable union. Therefore we can't simply write $\lambda^n(T) = \lambda^n(\biguplus_{x\in \mathbb{R}^{n-1}} T \cap \varphi^{-1}(x)) = \sum_{x\in \mathbb{R}^{n-1}} \lambda^n(T \cap \varphi^{-1}(x)) = 0$, so there needs to be a different way solving this exercise (we haven't even used the linearity in this case).

Answer 1

1. Since $\varphi:\mathbb{R}^n \rightarrow \mathbb{R}^{n-1}$ a surjective linear map, then there is a linear bijective linear map $\psi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that for all $(x_1,x_2,...,x_{n-1}, x_n ) \in \mathbb{R}^n $, $ \varphi \circ \psi (x_1,x_2,...,x_{n-1}, x_n)= (x_1,x_2,..., x_{n-1})$

2. Since for all $x \in \mathbb{R}^{n-1}$ the set $T \cap \varphi^{-1}(x)$ is countable and $\psi$ is bijective, then, for all $x \in \mathbb{R}^{n-1}$, $ \psi^{-1}(T) \cap ( \varphi \circ \psi )^{-1}(x)= \psi^{-1}(T) \cap \psi^{-1}( \varphi^{-1}(x)) = \psi^{-1} (T \cap \varphi^{-1}(x) )$ is countable.

3. We can write $\mathbb{R}^n= \mathbb{R}^{n-1} \oplus \mathbb{R}$, and for any $v \in \mathbb{R}^n$, we can write it as $(w, r)$ where $w \in \mathbb{R}^{n-1}$ and $ r \in \mathbb{R}$. Note that $\varphi \circ \psi = \pi_1$, the projection of $\mathbb{R}^n$ onto the first factor (that is $\mathbb{R}^{n-1}$). Item 2 above means that, for each $w \in \mathbb{R}^{n-1}$, the set $\{r \in \mathbb{R} : (w,r) \in \psi^{-1}(T)\}$ is countable.

4. Note that, using Fubini Theorem, \begin{align} \lambda^n(\psi^{-1}(T)) &= \int \chi_{\psi^{-1}(T)}(w,r) \:d \lambda^n(w,r) = \\ & = \int_\mathbb{R} \left(\int_{\mathbb{R}^{n-1}}\chi_{\psi^{-1}(T)}(w,r) \: d\lambda^{n-1}(w) \right) \: d \lambda(r)= \\ &=\int_{\mathbb{R}^{n-1}}\left(\int_\mathbb{R}\chi_{\psi^{-1}(T)}(w,r) \: d \lambda(r) \right)\: d\lambda^{n-1}(w) \end{align} But, for all $w \in \mathbb{R}^{n-1}$, $\{r \in \mathbb{R} : \chi_{\psi^{-1}(T)}(w,r) \neq 0\}= \{r \in \mathbb{R} : (w,r) \in \psi^{-1}(T)\}$. So $\{r \in \mathbb{R} : \chi_{\psi^{-1}(T)}(w,r) \neq 0\}$ is countable. So, for all $w \in \mathbb{R}^{n-1}$, $\int_\mathbb{R}\chi_{\psi^{-1}(T)}(w,r) \: d \lambda(r)=0$. So, $\lambda^n(\psi^{-1}(T))=0$. Since $\psi$ is a linear bijective map, we have $\lambda^n(T)=0$.