How can we show that the time shift $t\mapsto x(s+\;\cdot\;)$ is measurable? This is a question of great importance and hence there should be an answer in any textbook on stochastic processes. However, I neither found it nor I know how to approach this:
Let $(E,\mathcal E)$ be a measurable space and $s\ge0$. Moreover, let $$\tau_s:[0,\infty)\to[0,\infty)\;\;\;t\mapsto s+t$$ and $$\theta_s:E^{[0,\:\infty)}\to E^{[0,\:\infty)}\;,\;\;\;x\mapsto x\circ\tau_s.$$

How do we show that $\theta_s$ is $\left(\mathcal E^{\otimes[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)}\right)$-measurable?

Moreover, what does happen if $E$ is a metric space and we restrict $\theta_s$ to

*

*the Skorohod space $D([0,\infty),E)$ of càdlàg functions from $[0,\infty)\to E$ equipped with the Skorohod metric?

*the space of continuous functions equipped with the supremum norm.

 A: A map $(F, \mathcal{F}) \to (E^{[0, \infty)}, \mathcal{E}^{\otimes [0, \infty)})$ into $E^{[0, \infty)}$ is $\mathcal{F}/\mathcal{E}^{\otimes[0, \infty)}$-measurable if and only if all of its (all $t \geq 0$) compositions with the canonical projections
$$
\pi_t \colon E^{[0, \infty)} \to E,
\pi_t(x) = x(t),
$$ are $\mathcal{F}/\mathcal{E}$ measurable.
That's more or less the definition of the product sigma algebra.
With your map $\theta_s$ it is
$$
(\pi_t \circ \theta_s)(x) = \pi_t(x(\cdot + s)) = x(t+s) = \pi_{t+s}(x),
$$
i.e.
$$
\pi_t \circ \theta_s = \pi_{t+s}
$$ a canonical projection again.
Therefore $\pi_t \circ \theta_s$ is measurable for all $t, s \geq 0$.
This shows the measurability of $\theta_s$ as map  $(E^{[0, \infty)}, \mathcal{E}^{\otimes [0, \infty)}) \to  (E^{[0, \infty)}, \mathcal{E}^{\otimes [0, \infty)})$.
For the Skorokhod space and the space of continuous functions the same proof works because the corresponding Borel sigma algebras are generated by the canonical projections too.
