Let $S$ be a finite set and $\nu$ be a probability measure on $S$ with full support (that is, $\mu(s)>0$ for all $s\in S$). Let $\mu$ be the uniform probability measure on $S$. Endow each $S^n$ with the product measures $\mu^n$ and $\nu^n$. Suppose that for each $n\geq 1$ there is a subset $B_n$ of $S^n$ such that
1) $\displaystyle \mu^n(B_n)\leq \frac{1}{2^n}$.
2) $\lim_{n\to \infty} \nu^n(B_n)\to 1$ as $n\to \infty$.
(1) says that $B_n$ is exponentially smaller in size compared to $S^n$ and (2) says that $B_n$ is big according to $\nu^n$.
Problem. I am trying to show that the above cannot happen.
Attempt. Let $|S|=k$. Choose integers $p, q>1$ such that such that $1/p+1/q=1$. Let $F_n:S^n\to \mathbb R$ be defined as $F_n(x)= k^n\nu^n(x)$. Thus $F_n$ is the Radon-Nikodym derivative $d\nu^n/d\mu^n$. Now $$\nu^n(B_n) = \int_{S^n} \chi_{B_n}F_n\ d\mu^n\leq \|\chi_{B_n}\|_{p}\|F_n\|_q$$ where we have used Holder's inequality (the norms are with respect to $\mu^n$). Noting that $\sum_{x\in S^n}\nu^n(x)^q=(\sum_{x\in S}\nu(x)^q)^n$, we get that $$\|F_n\|_q = \left[k^{1/p}\left(\sum_{x\in S} \nu(x)^q\right)^{1/q}\right]^{n}$$ and $$\|\chi_{B_n}\|_p \leq \frac{1}{2^{n/p}}$$ So when $k=2$ we get the desired result (for which we could argue directly) by choosing $p=q=2$. We don't seem to get anything for $k>2$. If one could choose $p$ and $q$ such that $(k/2)^{1/p}\|\nu\|_q<1$ then we are done but this is artificial.