= Solution
Use the standard <Hausdorff> state-space convention, so compact sets are closed. Let $F$ be any <closed set> and choose $K_M$ from <exponential tightness>. The <weak large deviation principle> applies to the compact set $F\cap K_M$, while
$$
\mathbb P(X^L\in F)\le\mathbb P(X^L\in F\cap K_M)+\mathbb P(X^L\notin K_M).
$$
The <principle of the largest exponential term> gives
$$
\limsup_La_L^{-1}\log\mathbb P(X^L\in F)
\le\max\{-\inf_{F\cap K_M}I,-M\}
\le\max\{-\inf_FI,-M\}.
$$
Let $M\to\infty$. This is the full closed-set upper bound; the open-set lower bound was already present.
For goodness, fix $r<\infty$ and choose $M>r$. The complement $K_M^c$ is open. Its lower bound and its exponential-tightness upper estimate imply
$$
-\inf_{K_M^c}I
\le\liminf_La_L^{-1}\log\mathbb P(X^L\notin K_M)
\le-M.
$$
Thus $\inf_{K_M^c}I\ge M$ and $\{I\le r\}\subset K_M$. Lower semicontinuity makes this <sublevel set> closed, so it is a closed subset of a compact set and therefore compact. This proves that <exponential tightness upgrades a weak large deviation principle> to the full principle with the same <good rate function>.
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