The random process $Y(t)$ indexed by $t\in \mathcal{T}$ denotes outcomes under treatment levels in $\mathcal{T}$ . In practice, one cannot observe $Y(t)$ for all $t\in \mathcal{T}$ . Rather, only a single $Y(t_0)$ can be observed, where $t_0$ is the realization of a random variable $T$. Thus, the observed outcome is the random variable $$Y=Y(T)=\int_{t\in\mathcal{T}}Y(t)dI\{t\ge T\}$$
My question is how to understand the expression $\int_{t\in\mathcal{T}}Y(t)dI\{t\ge T\}$. Why is this equal to $Y(T)$.
Note $t\mapsto I(t \geq T)=:F(t)$ is a random process of finite variation. So if we interpret the integral as the Lebesgue–Stieltjes integral with respect to the function $F(t)$ you get by the rules of the Lebesgue–Stieltjes integral that $$ \int_{t\in \mathcal{T}} Y(t) dI(t\geq T) =\int_{t\in \mathcal{T}} Y(t) dF(t) = Y(T)(F(T)-F(T-))=Y(T). $$
So the very short answer is that it follows from the properties of the Lebesgue–Stieltjes integral.