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.

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Aug 11 at 5:46
  • $\begingroup$ 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. $\endgroup$
    – unwissen
    Aug 11 at 7:07
  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Aug 11 at 17:57
  • $\begingroup$ No, $E^{[0, \infty)}$ consists of all functions $[0, \infty) \to E$. $\endgroup$
    – unwissen
    Aug 11 at 18:09

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