I have 2 questions on integrated Brownian motion and would appreciate any guidance on them.
Question 1
Let $\mathcal F_t = \sigma(W_t)$, the $\sigma$-field generated by $W_t$, is $$Z_t = \int_{0}^{t} W_s \,ds$$ $\mathcal F_t$-measurable? Why so?
I asked the above because I am trying to prove:
For $\mathcal F_t = \sigma(W_t)$ and $Z_t = \int^{t}_{0}\,e^{W_u}\,du$ show $$E[Z_T|\mathcal F_t]=Z_t+W_t(T-t)\quad\forall \,t<T$$
The solution provided from the book I am using https://www.worldscientific.com/worldscibooks/10.1142/9620 is \begin{align} E[Z_T|\mathcal F_t] &= E\left[\int^{t}_{0}W_u\,du|\mathcal F_t\right] + E\left[\int^{T}_{t}W_u\,du|\mathcal F_t\right] \\ &= Z_t + E\left[\int^{T}_{t}W_u - W_t+W_t\,du|\mathcal F_t\right] \tag1 \\ &= \cdots \end{align}
Wouldn't $(1)$ imply that $Z_t$ is $\mathcal F_t$-measurable? But I am not sure why.
Question 2
In an attempt to prove $cov(Z_t, W_t) = \frac{t^2}{2}$, consider
\begin{align} cov(Z_t, W_t) &= E[Z_tW_t] - E[Z_t]E[W_t] \\ &= E[Z_tW_t] \\ &= E\left[W_t\int_{0}^{t}W_s\,ds\right]\tag1 \\ &=E\left[\int_{0}^{t}W_tW_s\,ds\right] \tag2 \\ &= \cdots \end{align}
What is the reason for $(1)$ to $(2)$? Does it have to do with Fubini's theorem? Gordan's answer in Correlation between stochastic processes may be helpful for context.
Thanks!