What does it mean to integrate over a random measure? I'm currently looking at a paper of Dirichlet process random effects model and the model specification is as follows:
$$ \begin{align*}y_{i} &= X_{i}\beta + \psi_{i} + \epsilon_{i}\\ \psi_{i} &\sim G \\ G &\sim \mathcal{DP}\left(\alpha, G_{0}\right) \end{align*}$$
where $\alpha$ is the scale parameter and $G_{0}$ is the base measure. Later on in the paper, it suggests that we integrate a function over the base measure $G_{0}$ such as
$$ \int f\left(y_{j}|\theta, \psi_{j}\right)\, dG_{0}\left(\psi_{j}\right).$$ Is the base measure in Dirichlet process a c.d.f. or is it a p.d.f.? What happens if the base measure is a Gaussian?
 A: Denote by $\mathcal{M}$ a measurable space of probability measures, containing the realisations of the Dirichlet process. The random probability measure $G$ is a measurable function
$$
G : \omega \mapsto G_\omega \in \mathcal{M}
$$
and the integral with respect to $G$ is the random variable
$$
\int f(\,\cdot\,|\, \psi) dG(\psi) : \omega \mapsto \int f(\,\cdot\,|\,\psi) dG_\omega(\psi).
$$
Thus $\int f(\,\cdot\,|\, \psi) dG(\psi)$ is itself a random p.d.f. (if $f(\cdot| \psi)$ is a p.d.f.).

The idea is that $\psi_i$ follows some unknown distribution $G$. In some cases, you may have reasons to believe that $\psi_i$ is normally distributed and then put a prior on the mean and variance. In other cases, you don't want to make such parametric assumptions. In your model, for instance, the prior on $G$ is a Dirichlet process.


Is the base measure in Dirichlet process a c.d.f. or is it a p.d.f.?

The base measure is any probability measure, usually taken to have full support. In some cases, it can be represented by a probability density function. This is not very important.
