Definition of measurability of a stochastic process 
Definition: A stochastic process $(X_t)_{t \in [0,\infty)}$ on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \in [0,\infty)},\mathbb{P})$ with values in a measurable space $(E,\mathcal{E})$ is called measurable if the map
  $$
f_{\infty} \colon \Omega \times [0,\infty) \rightarrow E \colon (\omega,s) \mapsto X_s(\omega)
$$
  is $\mathcal{F} \otimes \mathcal{B}_{[0,\infty)}$-measurable.

Sometimes it is also required that the map 
$$
f_t \colon \Omega \times [0,t) \rightarrow E \colon (\omega,s) \mapsto X_s(\omega)
$$
is $\mathcal{F} \otimes \mathcal{B}_{[0,t)}$-measurable for every $t \in [0,\infty)$. 
Now my question is, are those two definitions equivalent? If not, is there one which implies the other?
 A: The first definition is a measurable stochastic process. The second definition is a progressively measurable stochastic process. All progressively measurable processes are measurable. The converse, that is suggested in the comments is not true. All measurable processes are not progressively measurable.
I'm getting this from the Karatzas & Shreve text which defines a measurable stochastic process in 1.6 and a progressively measurable stochastic process in 1.11. Karatzas & Shreve references Chung & Doob for this naming convention.
I will prove that progressively measurability implies measurability.
The definition of a measrable process means that for any set in $\mathcal{E}$, the pre-image under $f_\infty$ is in $\mathcal{F} \otimes \mathcal{B}_{[0,\infty)}$.
The definition of a progressively measurable process means that for any set in $\mathcal{E}$, and any $t \ge 0$, the pre-image under $f_t$ is in $\mathcal{F}_t \otimes \mathcal{B}_{[0,t)}$.
It is clear that $\mathcal{F}_t \otimes \mathcal{B}_{[0,t)}$ is a subset of $\mathcal{F} \otimes \mathcal{B}_{[0,\infty)}$, so any pre-image in the former is also in the latter.
