The sample paths of a measurable stochastic process are Borel

Let $\displaystyle (\Omega,\mathcal{F},P)$ be a probability space. Let $\displaystyle (X(t))_{t\geq 0}$ be a stochastic process (so the time domain is $\displaystyle [0,\infty)$).

Suppose that the stochastic process considered as a map $\displaystyle [0,\infty)\times\Omega\rightarrow\mathbb{R}$ given by $\displaystyle (t,\omega)\mapsto X(t,\omega)$ is measurable with respect to the product sigma-algebra $\displaystyle \mathcal{B}([0,\infty))\otimes\mathcal{F}$. Why then does it follow that for fixed $\displaystyle \omega\in\Omega$, the function $\displaystyle t\mapsto X(t)$ is a Borel function?

I'm thinking that the restriction of a measurable function to a measurable set is measurable, but the point set $\displaystyle \{\omega\}$ need not be measurable.