Understanding the limits of the Ito Process Defintion I would like to understand what kind of stochastic process are Ito Processes. According to Kuo[p. 102] an Ito Process is a stochastic process of the form
$$dX_t=g(t)dt+f(t)dW(t),$$
where $W(t)$ is a wiener process, $f\in \mathcal{L}_\text{ad}(\Omega,L^2[a,b])$ and $g \in \mathcal{L}_\text{ad}(\Omega,L^1[a,b])$.
$\mathcal{L}_\text{ad}(\Omega,L^2[a,b])$ is explained in in Notation 5.1.1 (Kuo[p. 61]). I have tried to understand it too a degree to describe it here but failed.
The same is true for $\mathcal{L}_\text{ad}(\Omega,L^1[a,b])$, which is explained in Notation 7.4.1 (Kuo[p. 102]).
I would like to know if there is an easier equivalent set of conditions. For all the examples of Ito Processes that I have seen, $g(t)=h(t)X_t$ with $h:\mathcal{X} \rightarrow \mathbb{R}^{n \times n}$, with  $\mathcal{X}$ being the index set of $X_t$, and $f:\mathcal{X} \rightarrow \mathbb{R}^{n \times n}$. In particular, I am interested if $h(t)$ can be a stochastic process.
 A: A Mathematician's definition of the Ito process will probably be too technical. I myself adopt a more "pragmatic" approach to the problem.
So, can $g(t)$ be a stochastic process? My answer is, definitely yes. A trivial example will be $dX_t = \exp(W_t) dt$. Here, $g(t) = \exp(W_t)$. There is nothing wrong with this definition because $W(t)$ is continuous almost everywhere (in the probability space). 
It's always useful to attempt to understand the operations by discretising on a uniform grid and using pre-visibility etc. 
For example, the statement $dW_t^2 = dt$ (correct, if with an appropriate proportionality constant) should be understood as $\int dW_t^2 = \int dt$ in $L2$-norm or $\sum_i (W_{(i+1)h}-W_{ih})^2 = t$ in $L2$-norm. One thus needs to show $E\left[ \left(\sum_i (W_{(i+1)h}-W_{ih})^2 - t\right)^2 \right] = 0$, where $E[\cdots]$ denotes the expectation operator. The latter can be easily shown to be true using normality and independence of increments on non-overlapping intervals.
It's always said that what's difficult in understanding the stochastic calculus thing is to make sense of $dW_t$, which is undefined "classically", since $W_t$ is non-differentiable. In stochastic calculus,  it's understood as convergence under the $L2$-norm. The previous statement may be a bit confusing, so I shall give an example. The equation $dX_t = h(t, W_t)dW_t$ should be understood as $X_t - X_0 = \int_0^t h(s, W_s) dW_s$, where the equality sign is understood as "$\lim_{n \rightarrow \infty} E[(LHS_n-RHS_n)^2] = 0$ when the grid size approaches 0 (i.e. $n \rightarrow \infty$)". I'm avoiding some "technicalities" such as the limit being independent of the partition (i.e. grid) one adopts. In the integral form it becomes immediately obvious why one needs to adopt a new definition (non-classical) for $dW_t$; the integral does not exist in the usual sense.
A: Unfortunately, simple answer to your question does not exist. This has two reasons. The first one is only technical but stochastic processes are  more complicated than simple functions which you use in analysis for ordinary differential equations. A stochastic process $X: \Omega \times [0,t] \to \mathbb{R}^d $ is on the product space defined. The $[0,t]$-component is the "normal" time-dependence which you also have in deterministic functions used in (deterministic) ODEs: $f : [0,t] \to \mathbb{R}^d$. The $\Omega$-component is the stochastic "add-on". You have your filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_s)_{0 \leq s \leq t}, \mathbb{P})$ and for every fixed sample point $\omega \in \Omega$, the function $X_{\cdot}(\omega) : [0,t] \to \mathbb{R}^d $ is the sample path (realization, trajectory) of the process associated with $\omega$.
Hence, you (often) get additional conditions for stochastic processes, like $\mathcal{F}_s$-adaption or measurability in comparison to deterministic functions.
The second issue why there is no simple answer to your question is that the Wiener process $W$ is nowhere differentiable or more precisely, for almost every $\omega \in \Omega$, the Wiener sample path $W_{\cdot} (\omega)$ is nowhere differentiable (theorem by Paley, Wiener and Zygmund [1933]). Thus you have to give a sense to the expression $dW_s$ which should be something like a differential. This is done by constructing the stochastic (Ito) integral. More precisely, the differential notation like
\begin{equation}
dX_s = g(X_s, s)ds + f(X_s,s) dW_s \quad \text{ for } s \in [0,t],
\end{equation}
is only a commonly used shorthand for the integral notation
\begin{equation}
X_s = X_0 + \int_0^s g(X_r, r) dr + \int_0^s f(X_r, r) dW_r \quad \text{ for } s \in [0,t].
\end{equation}
This means if you want condition such that your Ito process $X$ is well-defined, you need the two integrals to be well-defined. Note that due to simplification mathematicians like to avoid too much notation. So dependence of all the stochastic processes (like $X, f, g, W$) on $\omega \in \Omega$ is omitted. For example $a : \Omega \times (\mathbb{R}^d \times [0,t]) \to \mathbb{R}^d $ and $b: \Omega \times (\mathbb{R}^d \times [0,t]) \to \mathbb{R}^d $ are a stochastic processes like the Wiener process. In fact you can choose them deterministic (no dependence on $\omega$) if you want to.
Now to the question, when are the integrals well-defined. Consider only the first integral $\int_0^s g(X_r, r) dr$. This is a (deterministic) Lebesgue-integral. As you want your Ito-process $X$ to be $\mathcal{F}_s$-adapted (this is most probably part of your definition of an Ito-process, I guess), your integrand $g$ has to be an $\mathcal{F}_s$-adapted process which trajectories are Lebesgue-integrable (this means $L^1([0,t])$).
For the second integral $\int_0^s f(X_r, r) dW_r$, it is more complicated as this is an stochastic integral with respect to a Wiener process (or equivalently to a Brownian motion). I will skip this theory but you can find it in the literature: for a more practical construction using something like "simple" processes which reminds you to the construction of Lebesgue integral see Karatzas and Shreve "Brownian Motion and Stochastic Calculus" and for a more theoretically construction using brute-force theory see Revuz and Yor "Continuous Martingales and Brownian Motion". In the end you get the same stochastic integral definition which tells you that you need your integrand (which is a stochastic process) to be $\mathcal{F}_s$-adapted and for almost every $\omega \in \Omega$ the trajectories $f(\cdot, \omega): [0,t] \to \mathbb{R}^d$ need to be Lebesgue-square-integrable (this means $L^2([0,t])$).
What does this mean in practice? Your integrands $f$ and $g$ can be stochastic processes as long as they are adapted (or more precisely $\mathcal{F}_s$-adapted) and their trajectories have to fulfill a integrability condition. First question: when is a stochastic process adapted (to the generic filtration of the Wiener process)? An easy solution: every deterministic process (no dependence on $\omega$) is adapted. If you look at non-deterministic integrands, you have to ensure adaptedness.
Second question: Under which conditions are the trajectories Lebesgue-(square-)integrable? Here again, if the trajectories are (left-)continuous on the interval $[0,t]$, they are Lebesgue-(square-)integrable on this finite time-interval.
To answer your both questions:

*

*An easier and equivalent set of conditions is not possible. But you can look at easier conditions in the sense that every deterministic and continuous function $f: \mathbb{R}^d \times [0,t] \to \mathbb{R}^d$ fulfills the integrability condition.


*Your $g(s) := h(s) X_s $ can be a stochastic process. As long it is adapted and the trajectories of $ X_s$ are in $L^2([0,t])$ (by Hölders's inequality).
