Integrated Brownian motion $\int_{0}^{t} W_s \, ds$ properties

Question

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!

Answer 1

To expand on my comment for Question 1: $Z_t$ is not measurable with respect to $\mathcal{F}_t$ as you have currently written it, as it depends on values of $W_s$ for all $s\leq t$, not just $W_t$. It is, however, measurable with respect to $\sigma(\cup_{s\leq t} W_s)$, that is, the sigma-algebra generated by all previous values of $W_s$. To understand why, note that the map $s\mapsto W_s$ is a.s. continuous, so the integral $Z_t$ exists and moreover is almost surely equal to the limit of Riemann sums. The Riemann sums for this integral are just linear combinations of $W_s$ for $s\leq t$, so each Riemann sum is measurable with respect to $\sigma(\cup_{s\leq t} W_s)$, and therefore the same is true of the limit.

Answer 2

Question 1: Yes. According to Ito product rule $d(tW)=W dt+tdW$, then we could get $$ \int_{0}^t W(s)ds=tW(t)-\int_{0}^{t}s dW(s) $$ For the right-hand side of first term, it is $\mathcal{F}_t$-measurable;for the second term, it is also $\mathcal{F}_t$-measurable. For how to determine whether a random variable is $\mathcal{F}_t$-measurable, there is an sentence from Oksendal in his book named " Stochastic Differential Equations" which may be helpful. It says,

Intuitively, that $h$ is $\mathcal{F}_t$-measurable means that the value of $h(w)$ can be decided from the values of $B_s(w)$ for $s\leq t$. For example, $h_1(w)=B_{t/2}(w)$ is $\mathcal{F}_t$-measurable, while $h_2(w)=B_{2t}(w)$ is not.

Question 2: since the integral $\int_{0}^t W_s ds$ is for variable $s$, therefore you could recognize $W_t$ as a constant. The integral has no influence on $W_t$.