Riemann integral over Itô integral? let's say I have the Itô integral
$I(t) = \int_{0}^{t} f(s)dW_{s} $
How do I then calculate
$I_{2}(u) = \int_{0}^{u} I(v)dv = \int_{0}^{u} (\int_{0}^{t} f(s)dW_{s})dv$
?
Is it going to become $0$ because $dW_{t} \cdot dt = 0$, or is there some kind of a Fubini theorem that has to be applied?
 A: Warning: the formula $dW_t \cdot dt = 0$ means that you will get zero when you integrate the product of these two forms with respect to the same variable $t$. You can't conclude from this that the integral in your question is going to be $0$, as you have a different variable of integration for each differential.
Intuition
To get some intuition for what is going on, take for example $f$ equal to the constant function $1$. Then you have $I(t)=W_t$ and  $I_2 (u)= \int _0 ^u W_s ds$, an integrated Wiener process.
Thus, here you are integrating the values of the normally distributed r.v.s $W_t$ over time (if you plotted an instance of the Wiener process, then the "area below the curve" and you could approximate its value by Riemann sums). Of course, that integrated value is a random variable as well (not a.s. equal to 0).
You know the process $W$ is continuous, so the integral makes sense. However, as these r.v.s are not independent (even if their increments are), it is probably going to be difficult to get closed-form expression for it using the definition of the Riemann integral. However, after some playing with Itô's lemma, you can find that $\int _0 ^u W_s ds = uW_u - \int_0^us dW_s.$ You can rewrite this as $\int_0^u (u-s)dW_s$.
As you may know (otherwise, I recommend trying to prove it) an integral of the form $\int_0^u f(s)dW_s$ is normally distributed with zero mean and $\int_0^u |f(s)|^2ds  $ variance. Furthermore, it defines an a.s. continuous process with independent increments. I assume this is what you are looking for, so let's go back to the general case.
The general case
We can try to express this double integral as a single Itô integral. For this, use the integration by parts formula: $$MdN=d(MN)-NdM-dMdN.$$
In your case,  $$M_v=\int_0^v f(s)dW_s, \quad dN_v=dv, \\ dM_v=f(v)dW_v,\quad N_v=v.$$
Then $dM_vdN_v=0$ (because $dW_v \cdot dv = 0$, notice that the integration variable in both differentials is the same) and we get
$$I_2(u)= \int_0^u M_vdN_v =\int _0^u d\left( v\int_0^v f(s)dW_s \right)-\int _0^uvf(v)dW_v.$$
The first term is simply $u\int_0^u f(s)dW_s-0 \cdot 0$, so after rearranging we have $$I_2(u)=\int_0^u(u-s)f(s)dW_s,$$
another Itô integral! Again, this defines a normally distributed process with independent increments, zero mean and $\int_0^u |(u-s)f(s)|^2ds  $ variance.
If you apply integration by parts once more, and supposing $f$ to be differentiable, you can get a representation of $I_2$ as an integral of a stochastic process over time, like the integrated Wiener process introduced before.
