I am trying to solve a complicated probability problem related to Shannon Entropy.

Let $(E,p)$ be a finite set with a probability measure $p$ on $E$. $E^n$ is given the probability measure $p^n(x_1, ..., x_n) = p(x_1)...p(x_n)$.

For all $\epsilon \in ]0,1[$, denote $S_\epsilon(n) = \inf\{\textbf{Card}(A) | A \subset E^n \text{ such that } p^{ n}(A) \geq 1 - \epsilon\}$. Prove that $\frac{1}{n} \log(S_\epsilon(n))$ converges to a limit independent of $\epsilon$ when $n \to \infty$, and give an expression of this limit.

It is easy to obtain that in the case of a uniform measure this is $\log(|E|)$. It is also possible to obtain an asymptotic upper bound on $\frac{1}{n} \log(S_\epsilon(n))$ given by $-\log(p_1)$ where $p_1$ is the maximum probability of an element of $E$. However, I'm struggling to find the limit : a hint in the exercise says this is related to the Shannon entropy of the measure $p$. Would anyone have a clue on how to go further ? Many thanks !

I am trying to solve a complicated probability problem related to Shannon Entropy.

Let $(E,p)$ be a finite set with a probability measure $p$ on $E$. $E^n$ is given the probability measure $p^n(x_1, ..., x_n) = p(x_1)...p(x_n)$.

For all $\varepsilon \in \mathopen]0,1[$, denote $$S_\varepsilon(n) = \inf\{\textbf{Card}(A) | A \subset E^n \text{ such that } p^{ n}(A) \geq 1 - \varepsilon\}.$$ Prove that $\frac{1}{n} \log(S_\varepsilon(n))$ converges to a limit independent of $\varepsilon$ when $n \to \infty$, and give an expression of this limit.

It is easy to obtain that in the case of a uniform measure this is $\log(|E|)$. It is also possible to obtain an asymptotic upper bound on $\frac{1}{n} \log(S_\varepsilon(n))$ given by $-\log(p_1)$ where $p_1$ is the maximum probability of an element of $E$. However, I'm struggling to find the limit : a hint in the exercise says this is related to the Shannon entropy of the measure $p$. Would anyone have a clue on how to go further ? Many thanks !