Deterministic Integral of a Predictable Process is Predictable I was reviewing a proof of existence of solutions to stochastic evolution equations which takes the form of a fixed point argument on the space of predictable processes such that
$$
\sup_{t\leq T}\mathbb{E}[\|X(t)\|]<\infty.
$$
This space is certainly Banach under the above norm.  I noticed that in the contraction mapping argument, no one ever seems to comment on the deterministic integral that could appear, of the form
$$
\int_0^t f(s,X(s))ds
$$
for $f$, say, a globally Lipschitz continuous mapping. While this is certainly well defined in the above norm, to fully address the fixed point argument, one would need to know that this process, denoted $Y(t)$, is also predictable.  My question is, is this obvious?  The argument that springs to mind is to approximate it by a left Riemann sum which will certainly be predictable and in this Banach space, and then use the completeness of the space - is that correct? Is there another argument that one would make?
 A: I think you only need the Borel measurability of $f$ for the predictability of $Y_t=f(t,X_t)$.
$Y$ can be decomposed into
$$
[0,\infty)\times\Omega\overset{\phi_X}{\rightarrow}[0,\infty)\times\mathbb R
\overset{f}{\rightarrow}\mathbb R,
$$
so predictability follows once the $\mathcal P/\mathcal B([0,\infty)\times\mathbb R)$-measurability of $\phi_X$ is established, where $\phi_X(t,\omega):=(t,X_t(\omega))$. And to establish the last measurability it suffices to consider measurable sets of the form $[0,\tau]\times B$, $\tau\in[0,\infty)$, $B\in\mathcal B(\mathbb R)$, which generate $\mathcal B([0,\infty)\times\mathbb R)\equiv \mathcal B([0,\infty))\otimes\mathcal B(\mathbb R)$. So look at
$$
\{(t,\omega)\in[0,\infty)\times\Omega:
(t,X_t(\omega))\in[0,\tau]\times B
\}
=([0,\tau]\times\Omega)\cap X^{-1}(B).
$$
$X^{-1}(B)$ is in $\mathcal P$ by the assumption $X$ is predictable; and $([0,\tau]\times\Omega)$, too, is in $\mathcal P$ as $1_{t\in[0,\tau]}(t,\omega)$ is an adapted caglad process.
