= Solution
Let epsilon decrease to zero in the upper estimate. In the lower estimate first take the <supremum> over all chosen points and then decrease epsilon to zero. The bounds coincide, proving <Varadhan lemma>:
$$
\boxed{\lim_{n\to\infty}\frac1n\log\mathbb E e^{nf(X_n)}=\sup_{x\in\mathcal X}\{f(x)-I(x)\}.}
$$
The value is finite: f is bounded, $I\geq0$, and the large-deviation bounds on the whole space force $\inf I=0$. This proof actually uses bounded <continuity> and the <large deviation principle>; compact sublevels are not additionally needed for this bounded version.
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