I am trying to show, that for fixed $v$ and $y \in \mathbb{R}^n$ the intersection of $L_y = \{y+tv ; t \in \mathbb{R}\}$ and a set E with n-dimensional measure 0, has 1-dimensional measure 0 for almost every $y \in \mathbb{R}^n$. Since $L_y$ is a line the intersection $L_y \cap E$ is one dimensional. I am supposed to use Fubini's theoreme for the proof.
My Idea was to define $\varphi : \mathbb{R}^n \to \mathbb{R}$ with $\varphi(y) := \lambda^1[L_y \cap E ] = \int_\mathbb{R} \chi_{E}(y+tv) dt$ and $\psi:\mathbb{R}^{n-1} \to \mathbb{R}$ with $ \psi(y) := \lambda^1[L_y \cap E] = \int_\mathbb{R} \chi_{E }(y+tv) \ dt.$
Than we know: $ 0=\lambda^n[E] = \int_{\mathbb{R}^n} \chi_{E } \ d\lambda^n = \int_{\mathbb{R}^{n-1}}\int_\mathbb{R} \chi_{E}(y+tv) dtdy = \int_{\mathbb{R}^{n-1}} \psi(y) dy$.
Now we know that $\psi = 0$ almost everywhere on $\mathbb{R}^{n-1}$.
I want to use Fubini again to show that $\varphi = 0$ almost everywhere on $\mathbb{R}^{n}$
Edit:
Therefore I define a function $h :\mathbb{R}^{n-1} \times \mathbb{R}$ by $h(x, z):= \varphi(y)$ for $(x, z) =y$
Now $\int_{\mathbb{R}^n} \varphi \ d\lambda^n = \int_{\mathbb{R}^n} h(x,z) \ d\lambda^n\overset{\text{Fubini}}{=} \int_\mathbb{R}\int_{\mathbb{R}^{n-1}} h(x,z) \ dx dz = 0$ Because for fixed z $h(x,z) = \psi(x)$.