# 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

1. the Skorohod space $$D([0,\infty),E)$$ of càdlàg functions from $$[0,\infty)\to E$$ equipped with the Skorohod metric?
2. the space of continuous functions equipped with the supremum norm.

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 $$(E^{[0, \infty)}, \mathcal{E}^{\otimes [0, \infty)}) \to (E^{[0, \infty)}, \mathcal{E}^{\otimes [0, \infty)})$$.
• Thank you for your answer. Can we weven show that $$E^{[0,\:\infty)}\times[0,\infty)\to E^{[0,\:\infty)}\;,\;\;\;(x,s)\mapsto\theta_s(x)$$ is $\left(\mathcal E^{\otimes[0,\:\infty)}\otimes\mathcal B([0,\infty)),\mathcal E^{\otimes[0,\:\infty)}\right)$-measurable? Aug 11 at 5:46
• I dont think that is true in general. Denote your map by $\psi$. Fix a non-measurable $x_0 \in E^{[0, \infty)}$. Consider the measurable map $i \colon [0, \infty) \to E^{[0, \infty)} \times [0, \infty), s \mapsto (x_0, s)$. Then we have $(\pi_0 \circ \psi \circ i)(s) = \pi_0(x_0(\cdot + s)) = x_0(s)$. This map is not measurable, but it had to be if $\psi$ were measurable. Aug 11 at 7:07
• Hm, first of all: Is $x_0\in\mathcal E^{\otimes[0,\:\infty)}$ if and only if $x_0:[0,\infty)\to E$ is $(\mathcal B([0,\infty)),\mathcal E)$-measurable? Aug 11 at 17:57
• No, $E^{[0, \infty)}$ consists of all functions $[0, \infty) \to E$. Aug 11 at 18:09