= Solution
Write $Ph=\mathbb E h(X)$ and $P_nh=n^{-1}\sum_i h(X_i)$ for the <expectation> and <empirical measure>. A sufficient <bracketing of a function class> condition is that, for every $\varepsilon>0$, finitely many brackets $[l_j,u_j]$ cover $\mathcal H$, with measurable integrable endpoints and $P(u_j-l_j)<\varepsilon$. The <uniform strong law from finite L1 bracketing> then gives
$$
\boxed{\sup_{h\in\mathcal H}|P_nh-Ph|\longrightarrow0\quad\text{almost surely}.}
$$
Here the observations are <independent and identically distributed>. For an uncountable class, one either assumes a measurable supremum, for example through a <pointwise separable function class>, or formulates the conclusion as a pathwise bound on a common probability-one event.
For the parameterized class, the <Heine-Borel theorem> makes $\Theta$ <compact>. Put $M(x)=\sup_{\theta\in\Theta}|q(\theta,x)|$. A countable dense subset of $\Theta$ gives the same supremum because of <continuity>, so $M$ is measurable. Define the modulus
$$
w_\delta(x)=\sup_{\substack{\theta,\theta'\in\Theta\\|\theta-\theta'|\le\delta}}|q(\theta,x)-q(\theta',x)|.
$$
The supremum is measurable by taking a countable dense subset of the <compact> set of admissible pairs. For each $x$, <uniform continuity> on $\Theta$ gives $w_\delta(x)\to0$. Also $w_\delta\le2M$, and $PM<\infty$. Thus the <dominated convergence theorem> gives $Pw_\delta\to0$.
Choose a <finite net> of radius $\delta$ $\theta_1,\ldots,\theta_N$ in $\Theta$. The lower and upper envelopes of $q(\theta,\cdot)$ over each closed ball $\Theta\cap\overline B(\theta_j,\delta)$ are measurable integrable <function brackets>; the same countable-dense-set argument applies within each such <compact> ball. Their widths are at most $w_{2\delta}$. They cover the whole class, so its <bracketing of a function class> condition follows. Moreover, <dominated convergence> shows that $\theta\mapsto Pq(\theta,\cdot)$ is continuous. Both empirical and population functions therefore have a supremum over a common countable dense parameter set. Applying the <uniform strong law from finite L1 bracketing> proves \b[uniform almost-sure convergence over the entire <compact> parameter set], rather than merely convergence at each fixed parameter.
For the <exponential family>,
$$
\log f(\theta,x)=\theta x-K(\theta)+\log f_0(x),\qquad
\sup_\theta|\log f(\theta,X)|\le A|X|+B+|\log f_0(X)|,
$$
where $A=\sup_\Theta|\theta|$ and $B=\sup_\Theta|K(\theta)|$. Thus a weak sufficient condition is \b[bounded $K$ on $\Theta$, $f_0(X)>0$ almost surely, and <integrability> of $|X|+|\log f_0(X)|$ under the sampling law]. No positive lower bound on $f_0$ over the whole real line is required. Zeros off the sampling support are harmless: choose arbitrary finite versions of the log-densities on that common null set when applying the function-class theorem.
If the sampling law is $f(\theta_0,\cdot)$, one convenient assumption is that $\Theta$ is a <compact> subset of the interior of the finite domain of the <cumulant function of an exponential family>, and
$$
\int e^{\theta_0x}f_0(x)|\log f_0(x)|\,dx<\infty.
$$
On that interior, $K$ is continuous, hence bounded on $\Theta$. Finiteness of $K$ at $\theta_0\pm a$ for some $a>0$ implies $\mathbb E_{\theta_0}e^{a|X|}<\infty$, hence $\mathbb E_{\theta_0}|X|<\infty$. The displayed integral gives the remaining <integrability>. If the observations come from an arbitrary unrelated law, assumptions on $K$ and $f_0$ alone cannot control that law's tails; the sampling-law <integrability> must then be stated explicitly.
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