Let $I$ be an interval and $\omega$ be a probability density function so that $\int_{x \in I} \omega(x) \mathrm{d}x = 1$.
Moreover, let $c:\mathbb{R} \mapsto \mathbb{R}_+$ be a Borel function, and $\gamma(\cdot |t)$ be a probability measure supported on $\mathbb{R}$ so that $\int_{x \in \mathbb{R}} \mathrm{d} \gamma(x | t)= 1, \ t \in I$.
Now, I am dealing with the following integral: $$ \displaystyle \int_{t \in I} \omega(t) \left[ \int_{x \in \mathbb{R}} c(x) \mathrm{d} \gamma(x | t) \right] \mathrm{d} t.$$
What is the most general condition I need on $\gamma$ so that the above integral is well-defined? Is there a more general way than assuming each $\gamma(\cdot | t)$ admits a density function (wrt the Lebesgue measure)?