= Solution
Write $b=\operatorname{ess\,sup}X_1$. The hypothesis $\mathbb P(X_1>a)>0$ implies $a<b$ and rules out a constant random variable. In particular $I(a)<\infty$. The <cumulant-generating function> is strictly convex, since its second derivative is the positive <variance> under <exponential tilting>, and $\psi'(0)=\mu$, $\psi'(\theta)\to b$ as $\theta\to\infty$.
If $a>\mu$, there is a unique $\theta_a>0$ with $\psi'(\theta_a)=a$, and the <Legendre transform> gives $I(a)=\theta_a a-\psi(\theta_a)$. Thus
$$
\theta_a(a+\varepsilon)-\psi(\theta_a)=I(a)+\theta_a\varepsilon>I(a).
$$
If $a=\mu$, use differentiability at zero: $\psi(\theta)=\mu\theta+o(\theta)$, so for some sufficiently small $\theta>0$,
$$
\theta(a+\varepsilon)-\psi(\theta)>0=I(a).
$$
In either case choose a parameter $\theta\geq0$ with $J=\theta(a+\varepsilon)-\psi(\theta)>I(a)$, and choose $0<\eta<J-I(a)$. The <Cramér theorem> gives, for all sufficiently large $n$,
$$
\mathbb P(S_n/n\geq a)\geq e^{-n(I(a)+\eta)}.
$$
The <Chernoff bound> at the chosen parameter gives $\mathbb P(S_n/n\geq a+\varepsilon)\leq e^{-nJ}$. The conditioning event has positive probability for every $n$, since all summands can simultaneously exceed $a$. Therefore
$$
\frac{\mathbb P(S_n/n\geq a+\varepsilon)}{\mathbb P(S_n/n\geq a)}
\leq e^{-n(J-I(a)-\eta)}\longrightarrow0.
$$
Taking the complement within the conditioning event proves the <conditional concentration at a large-deviation threshold>:
$$
\boxed{\mathbb P\left(S_n/n\in[a,a+\varepsilon)\mid S_n/n\geq a\right)\longrightarrow1.}
$$
\b[The farther tail has a strictly larger exponential cost.] The argument also works when $a+\varepsilon$ reaches or exceeds the upper support endpoint, without requiring an interior maximizer there.
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