Let $(X,\tau)$ be a topological space, let $\mathcal{B}$ be its Borel $\sigma$-algebra, and $\mu_\epsilon$ be a family of probability measures on $(X,\mathcal{B})$. Suppose also that $\mu_\epsilon$ satisfies the large deviation principle with rate function $I$, i.e. for any $A \in \mathcal{B}$:
$$-\inf_{x \in \operatorname{Int}(A)} I(x) \leq \liminf_{\epsilon \to 0} \epsilon \log(\mu_\epsilon(A)) \leq \limsup_{\epsilon \to 0} \epsilon \log(\mu_\epsilon(A)) \leq -\inf_{x \in \operatorname{Cl}(A)} I(x).$$
We say that $A$ is regular if $\lim_{\epsilon \to 0} \epsilon \log(\mu_\epsilon(A)) = -\inf_{x \in A} I(x)$, or equivalently if $\inf_{x \in \operatorname{Int}(A)} I(x) = \inf_{x \in \operatorname{Cl}(A)} I(x)$. This is a fairly mild requirement, and indeed, in some sense regular sets are "typical".
On the other hand, irregular sets exist. One example is when $\mu_\epsilon$ is the uniform measure on $(-\epsilon,\epsilon)$. Then $I(0)=0$ and otherwise $I(x)=+\infty$. Thus if $A=\{ 0 \}$ then $\inf_{x \in \operatorname{Int}(A)} I(x)=+\infty$ and $\inf_{x \in \operatorname{Cl}(A)} I(x) = 0$.
A more important example arises when $\mu_n$ is the probability distribution of the sample mean of $n$ iid Bernoulli(p) variables (with $p \in (0,1)$). In this case there is an exponentially decaying but positive probability of attaining the value $1$, while there is no probability of attaining a value any higher than $1$. Consequently $I(1)$ is finite while $I(x)$ is infinite whenever $x>1$. Thus $(1,\infty)$ is an irregular set. Exactly the same thing would happen if we replaced Bernoulli(p) with any non-constant distribution which is bounded and has positive probability to attain its maximum.
What are some other instructive examples of irregular sets? Examples where $X=C([0,T])$ would be especially helpful.
Let $(B_t)_{t \geq 0}$ be a Brownian motion. It is well-known that the scaled process $X_t^{\epsilon} := \sqrt{\epsilon} B_t$ satisfies a large deviation principle in $C[0,1]$ as $\epsilon \to 0$ with (good) rate function
$$I(f) := \begin{cases} \frac{1}{2} \int_0^1 |f'(s)|^2 \, ds, & f(0)=0, \, \text{$f$ absolutely continuous}, \\ \infty, & \text{otherwise} \end{cases}$$
If we set $$A := \{f: [0,1] \to \mathbb{R}; f(0) \neq 0\},$$ then $A$ is open in $C[0,1]$ and therefore
$$\inf_{f \in A} I(f) = \inf_{f \in \text{int} \, A} I(f) = \infty.$$
On the other hand, it is not difficult to see that
$$\inf_{f \in \text{cl} \, A} I(f) = 0.$$