Large deviation principle continues to hold when rate function is replaced by its lower semicontinuous regularization

Question

Consider

Lemma

Suppose $I$ is a function such that (2.1) holds for all measurable sets A. Then (2.1) continues to hold if $I$ is replaced by $I_{\text{lsc}}$

proof:

$I_{lsc}\le I$ and the upper bound is immediate. For the lower bound observe that $\inf_G I_{\text{lsc}}=\inf_G I$ when G is open.

I am having trouble understanding the last sentence of the proof. Can someone please explain why is it true for G open.?

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Relevant definitions:

Answer 1

By definition, for any $y \in G$, $I_{lsc}(y) = \sup_{G'}\inf_{z \in G'} I(z)$ where the supremum is taken over open sets $G'$ such that $y \in G'$. Selecting $G = G'$, this means that $I_{lsc}(y) \geq \inf_{z \in G} I(z)$. This implies that $\inf_{x \in G} I_{lsc}(x) \geq \inf_{x\in G} I(x)$.

However, $I_{lsc}(x) \leq I(x)$ for all $x$ so $\inf_{x \in G} I_{lsc}(x) \leq \inf_{x\in G} I(x)$.

Combining these yields $\inf_{x \in G} I_{lsc}(x) = \inf_{x\in G} I(x)$.