Let $X$ be a topological space, let $B_X$ be its Borel $\sigma$-algebra, and let $B$ be the completion of $B_X$ under some regular measure $\mu $. We say that the family of probability measures $\mu_\epsilon$, defined on $(X,B)$ and absolutely continuous with respect to $\mu $, satisfies the large deviation principle with the rate function $I$ if
$$\limsup_{\epsilon \to 0} \epsilon \log \left ( \mu_\epsilon(F) \right ) \leq -\inf_{x \in F} I(x) \\ \liminf_{\epsilon \to 0} \epsilon \log \left ( \mu_\epsilon(G) \right ) \geq -\inf_{x \in G} I(x)$$
whenever $F$ is closed and $G$ is open. (One can generalize somewhat, but I don't think it is necessary for this context.) I have seen examples that show that some "natural" situations where, if we allowed $F,G$ to be arbitrary measurable sets, we would not have an LDP, yet we "should" have one.
For instance, if $\mu_\epsilon$ is the uniform distribution on $(-\epsilon,\epsilon)$, then the large deviation principle should tell us that $\mu_\epsilon$ "collapses to $0$", and the rate function should be $0$ at $0$ and $+\infty$ elsewhere. But if we drop the openness requirement in the lower bound, then we can take $G=\{ 0 \}$, in which case the inequality reads $-\infty \geq 0$, which is not true.
I imagine there is a similar example if we drop the closedness requirement in the upper bound. So I understand logically that these topological constraints are necessary. But these counterexamples don't give me an intuition for why the topological constraints should be the way they are.
With that in mind: why, intuitively, should the large deviation lower bound hold for open sets, and the upper bound for closed sets? For instance, is this a manifestation of the idea that open sets are "large" while closed sets are "small"?