This is Exercise 3.11.
Let $\mu$ be a measure on $\mathbb{R}^+$ such that $\mu[a,b] < \infty$ for every finite interval. Suppose that $X$ is a Brownian motion. Let $$Y_t = \int_{0}^{t} X_s \mu(ds)$$ The first part of this exercise shows that $Y$ is a Gaussian Process.
Using the relation $$ (\mu(r, t])^2 = 2 \int_{r}^{t} \mu(r, u] d\mu(u)$$ prove that $Y_t$ has a centered Gaussian law with variance $\sigma_t^2 = \int_{0}^{t} \mu(s,t]^2 ds$ and compute $E[Y_tY_s].
I'll present the given solution to this exercise and then mention my trouble with it.
It is easy to show $E[Y_t] = 0$. Suppose $s \leq t$.
We have $$ E[Y_tY_s] = E[\int_{0}^{t} X_u d\mu(u) \int_{0}^{s} X_v d\mu(v)] = \int_{0}^{t} d\mu(u) \int_{0}^{s} E[X_uX_v] d\mu(v)$$ but this is brownian motion, so $$E[Y_tY_s] = \int_{0}^{t} d\mu(u) \int_{0}^{s} u \wedge v\, d\mu(v) = \underbrace{\int_{0}^{s} d\mu(u) \int_{0}^{s} u \wedge v\, d\mu(v)}_{I_1} + \int_{s}^{t} d\mu(u) \int_{0}^{s} u \wedge v\, d\mu(v) $$
My first issue is here, with the notation of $\int_{s}^{t}$ itself. Since $\mu$ is not specified, it may be the case that $\mu(s) > 0$ so $\mu(s,t]$ and $\mu[s,t]$ need not be equal. I assume here that the first integral is $(0,s]$ and the second is $(s,t]$, but I am not sure.
Let us continue. To compute the first term, we can write
$$ I_1 = \int_{0}^{t} d\mu(u) \int_{0}^{u} v d\mu(v) + \int_{0}^{t} d\mu(u) \int_{u}^{t} ud\mu(v)$$ Again, I need to assume that this is $(0,u]$ and $(u,t]$.
For the second term, $$ \int_{0}^{t} d\mu(u) \int_{u}^{t} ud\mu(v) = \int_{0}^{t} d\mu(v) \int_{0}^{v} ud\mu(u) = \int_{0}^{t} d\mu(u) \int_{0}^{u} vd\mu(v)$$ which is just another copy of the first term and so we only need to compute the right term to compute $I_1$.
Again, I am having trouble with the limits here. Let me try to write it the above equation explicitly. $$\begin{align*} \int_{\mathbb{R}^+} \mathbb{1}_{(0,t]}(u)\mu(du) \int_{\mathbb{R}^+} \mathbb{1}_{(u,t]}(v) d\mu(v) &= \int_{\mathbb{R}^+ \times \mathbb{R}^+} \mathbb{1}_{(0,t]}(u)\mathbb{1}_{(0,t]}(v) \mathbb{1}_{(u < v)}(u,v) \\ &= \int_{\mathbb{R}^+} \mathbb{1}_{(0,t]}(v)\mu(dv) \int_{\mathbb{R}^+} \mathbb{1}_{(0,v)}(u) \mu(du)\end{align*}$$ This is not necessarily equal to $$\int_{\mathbb{R}^+} \mathbb{1}_{(0,t]}(v) \mu(dv) \int_{\mathbb{R}^+} \mathbb{1}_{(0,v]}(u) \mu(du)$$ as $$\mu({v}) may be positive and ${v : \mu(v) > 0}$ clearly won't have negligible measure.
There are multiple other instances in this proof where the notation $\int_{0}^{t}, \int_{s}^{t}$ etc. are used and it seems to obfuscate this difference. One particular case is in the relation the question asks to use. To me, that relation is only valid if $\mu(\delta) = 0$ where $\delta = \{((x,x) \mid x \in (r,t] \}$ is the diagonal.
Can you help me either understand why the proof might still go through? Is it easy to fix this proof or does further assumptions of $\mu$ need to be made? If so what assumptions suffice?