I'm trying to show that if $E \subset \mathbb{R}^d$ is Borel, then the set $\{ (x,\omega) \in \mathbb{R}^d \times S^{d-1} \; : \; x + [0,1]\omega \subset E \}$ is Borel, where I define $x + [0,1]\omega := \{x + t \omega \; : \; t \in [0,1]\}$.
My idea is to considet the function $s(x,\omega,t) := x + t\omega$ and then consider the Borel set $A := s^{-1}(E)$. The set I'm looking for will then be
$\{(x,\omega) \in \mathbb{R}^d \times S^{d-1} \; : \; (x,\omega,t) \in A \; \forall t \in [0,1]\}$. Finally, letting $N := 2d - 1$, is enough to show that $\{ x \in \mathbb{R}^N \; : \; (x,t) \in B \forall t \in [0,1]\}$ is Borel whenever $B \subset \mathbb{R}^{N+1}$ is Borel, but I have no idea of how to prove this.
