Uniform LLN Assumptions, Weak vs. Strong A theorem from Jennrich (1969) is of a strong uniform law of large numbers,
Let $g_n$ be an iid sequence of random functions $\Theta \rightarrow R$ with $E[g_1(\theta)]=0$ for each $\theta \in \Theta$. Assume that $\Theta \subset R^k$ is compact, that $g_1$ is continuous almost everywhere, and that $E\left(\sup\limits_{\theta \in \Theta}|g_1(\theta)|\right) < \infty$. Then $\frac{1}{n}\sum_{i=1}^ng_i\rightarrow^{a.s.}0 $   uniformly over $\Theta$.
Wikipedia has assumptions for a weak uniform law of large numbers (as seen with the convergence in probability notation). To me, the assumptions on Wikipedia are equivalent to the assumptions made by Jenrich, but the conclusion is weaker.
My question is, what assumptions are needed for a strong uniform law of large numbers and how do those assumptions compare to those needed for a weak law?
 A: Note that for any particular $\theta \in \Theta$ the sequence $g_i(\theta)$ form a sequence of iid random variables with mean $0$.
By the SLLN,
$$\frac1n \sum g_i(\theta) \xrightarrow{a.s.}0\;\;\forall \theta\in\Theta$$
This is the "pointwise" version. For uniform convergence to $0$ we need to show
\begin{equation}w.p. 1\;\; \forall \epsilon > 0\;\; \exists N_{\epsilon}>0: \sup_{\theta \in \Theta}\left|\frac1n \sum g_i(\theta)\right|<\epsilon\;\forall n>N_{\epsilon} \tag{1}\end{equation}
This is where the second piece helps us:
$$E\left(\sup_{\theta \in \Theta}\left|g_i(\theta)\right|\right) <\infty$$
Let $G_i =\sup_{\theta \in \Theta}\left|g_i(\theta)\right|$ then from $(1)$ the above, we know that $\frac1n \sum G_i \xrightarrow{a.s.} c < \infty$
This means that
$$w.p. 1\;\forall \epsilon>0\;\;\exists N_{\epsilon}: \left|\frac1n \sum G_i-c\right|<\epsilon\;\;\forall n>N_{\epsilon}$$
By definition
$$|g_i(\theta)| \leq G_i \implies \left|\frac1n \sum g_i(\theta)\right| \leq \frac1n\sum\left|g_i(\theta)\right| \leq \frac1n \sum G_i\;\;\forall \theta \in \Theta$$
By the dominated convergence theorem and the compactness and continuity of $g_i$ we know:
$$\lim_{n\to \infty} \sup_{\theta \in \Theta} \left|\frac1n \sum g_i(\theta)-0\right|=0$$
$$\implies \forall \epsilon>0\;\exists N_{\epsilon}:  \sup_{\theta \in \Theta} \left|\frac1n \sum g_i(\theta)\right| \leq \epsilon\;\;w.p. 1\;\square$$
As for your question of how we get to $a.s.$ -- the Wikipedia article starts with convergence in probability only. However, if we can make the case for $a.s.$ convergence, then we can move to $a.s.$ uniform here too.
