Is this stochastic Picard iterator well-defined? Preliminaries


*

*Let $x_0 \in \mathbb{R}^d$.

*Let $T \in (0, \infty)$.

*Let $$ \sigma: \mathbb{R}^d \rightarrow \mathbb{R}^{d \times d}$$ and
$$\mu: \mathbb{R}^d \rightarrow \mathbb{R}^{d}$$
be affine linear transformations (which implies, that they are smooth and globally Lipschitz).

*Let $(\Omega, \mathcal{G}, (\mathcal{G}_t)_{t \in [0,T]}, \mathbb{P})$ be a complete probability space with a complete, right-continuous filtration $(\mathcal{G}_t)_{t \in [0,T]}$.

*Let $B : [0,T] \times \Omega \rightarrow \mathbb{R}^d$, $(t,\omega) \mapsto B_t(\omega)$, be a standard $d$-dimensional $(\mathcal{G}_t)_{t \in [0,T]}$-adapted Brownian motion on $\mathbb{R}^d$, such that $B_0 = 0$ and such that, for every pair $(t,s) \in \mathbb{R}^2$ with $0 \leq t < s$, the random variable $B_s-B_t$ is independent of $\mathcal{G}_t$.
Question
Consider the set
    \begin{equation}
 \begin{gathered}
 \mathcal{S}_{T} :=  \{ S: [0,T] \times \Omega \rightarrow \mathbb{R}^d \mid \\ S \ \text{is} \ (\mathcal{G}_t)_{t \in [0,T]} \text{-adapted and has $\mathbb{P}$-a.s. continuous sample paths}  \} .
 \end{gathered}
 \end{equation} I would like to define a type of Picard iterator function via
    \begin{equation}
 \begin{gathered}
 \mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)} : \mathcal{S}_{T} \rightarrow \mathcal{S}_{T}, \\
 \mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}(S)_t = x_0 + \int_{0}^{t} \mu (S_s) ds + \int_{0}^{t} \sigma(S_s) dB_s.
 \end{gathered}
 \end{equation}
    But is $\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}$ truly well-defined for all $t$ and $\omega$? Are both integrals in the expression always well defined for all $t$ and $\omega$? And, most importantly, does $\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}$ really map $\mathcal{S}_{T}$ into itself? 
(I am still learning the basics of stochatic calculus.)
 A: In this answer, following your comment, I assume the Ito integral has been constructed as in Le Gall's book. Since we work on a bounded time interval, Brownian motion is a continuous $L^2$-bounded martingale there and so we only need that part of the construction. 
The main technicality is that the Ito integral there is really defined for equivalence classes of integrands. That is, Le Gall defines the Ito integral for integrands in the space $L^2(B):=L^2(\Omega \times [0,T], \mathcal{P}, d\mathbb{P} \otimes ds)$ where $\mathcal{P}$ is the progressive $\sigma$-algebra. This space consists of equivalence classes of $d\mathbb{P} \otimes ds$-a.e. equal processes (I can tell you haven't done this since your processes are only almost surely continuous). So in order to take an Ito integral of $S \in \mathcal{S}_T$ we first should identify $S$ with its $d\mathbb{P} \otimes ds$ equivalence class. Let's agree that we implicitly do this when we write the Ito integral. 
Another technicality is that the integral $\int_0^t \mu(S_s)(\omega) ds$ need not be defined for all $\omega$. For example if $d = 1, \mu = \operatorname{Id}$ this is essentially asking to define the Riemann integral for an arbitrarily irregular path, which we cannot do. So we need to agree what to do on the measure $0$ set of discontinuous paths. One way around this issue is to agree to integrate a representative of $S$ for which every sample path is continuous. Of course, there are many such representatives. However, the integrals we obtain this way can only differ on a measure $0$ set and so are versions of each other.
Having done this, since Le Gall's construction of the Ito integral against a continuous $L^2$-bounded martingale is a continuous $L^2$-bounded martingale and the Riemann integral is continuous, the process $\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}(S)$ is then defined as an equivalence class of continuous processes (all of which are versions of each other) and in particular are defined for all $t$. 
Finally, by approximating each integral by Riemann sum type approximations (which are adapted) we see that each member of this equivalence class is adapted.

The short version of this is that the question is a bit fiddly because often in this subject we write down a process when we really mean a whole equivalence class of processes. What the above says is that if you replace $\mathcal{S}_T$ by $\mathcal{S}_T / \sim$ where $S \sim \hat{S}$ if and only if $\hat{S}$ is a version of $S$, then 
$$\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}(S): \mathcal{S}_T \to \mathcal{S}_T.$$
If you don't do this, there's a bit of a type mismatch and you need to pick a member of an equivalence class in a somewhat arbitrary way. In truth, the usual way to deal with these issues is to just agree we've always identified processes in the right manner from the start.
