For any $\theta \in \Theta$, let $g(t;\theta)$ be integrable in $t$ over, say, the half line $\mathbb{R}_+$. Then suppose $\Theta$ is a compact set, then can we say $\sup_{\Theta}g(t;\theta)$ is also integrable?
To me, intuitively speaking it should be but I am having a bit of trouble proving why? If it is not, what are some of the conditions on $g$ such that it is?
Suppose that $\Theta = [0,1]$. Consider $g(t,\theta) = 1$ for $t \in [n-1, n)$ where $n = [\theta^{-1}]$ and $g(t,\theta) = 0$ otherwise. Hence, for example $g(t,0)=0$ for all $t$. In this case $sup_{\theta}g(t,\theta)=1$ isn't integrable. According to this scheme, it is easy to make a counterexample with smooth functions $g$.
If $\Theta$ is infinite, then there is a subset $ \{ \theta_m \}$ which is infinite, so we may consider $g(t, \frac{1}{m})$ for $\theta = \theta_m$ and put all other $g(t,\theta) = 0$.
If $\Theta$ is finite then the statement holds true, because $g(t, \theta_0) \le sup_{\theta} g(t,\theta) \le \sum_{\theta} g(t,\theta)$.
One more trivial condition: If $ \int_M^{\infty} \sup_{\theta} g(t,\theta) dt < \infty$ for some $M$ and if $g(t,\theta)$ are unformly bounded then the statement holds true.