EDIT - The answer below deals with an ergodic m.p.s

As this question got up-voted, I've decided to fuly write a solution, based on the sketch I've made in the comments.

Fix some $\varepsilon>0$ small, and $n \gg _\varepsilon 0$, and denote by $C_{n}$ to be the cylinders of length $n$.
Let $h=h_{\mu}(\sigma)$ be the metric (Kolmogorov-Sinai) entropy of the system $(\Sigma,\sigma,\mu)$. Moreover, assume $\mu$ is $\sigma$-ergodic measure!

By the Shannon-McMillan-Breiman theorem, there exists a partition of $C_n$ into two sets - $G_{n}, B_{n}$ where $G_{n}$ are the ''good'' cylinders, namely for every $g\in G_{n}$ we have $\mu(g_{n})\approx 2^{-n(h\pm\varepsilon)}$.
Now for the ''bad'' cylinders, we have that $\sum_{b\in B_{n}}\mu(b) < \varepsilon$.

Moreover, define the set $S_{n}$ to be the ''small'' cylinders, namely $s\in S_{n}$ iff $\mu(s)\leq 2^{-n}$.

From now on, assume $h<1$.

We see that $G_{n} \cap S_{n} = \emptyset$, as $\mu(g)\geq 2^{-n(h+\varepsilon)}>2^{-n}$ for suitably chosen $\varepsilon$ and $g\in G_{n}$, hence $\sum_{s\in S_{n}} \mu(s) \leq \sum_{s\in B_{n}}\mu(s) <\varepsilon$.

Now we want to estimate $2^{n} - |S_{n}|$, as $|G_{n}|\leq 2^{n} - |S_{n}|$, we first bound $|G_{n}|$.

By a crude packing bound we get $|G_{n}|\approx (1-\varepsilon)2^{n(h\pm\varepsilon)}$.

Now we need to estimate $B_{n}\setminus S_{n}$. The atoms in $B_{n}$ are of two types - large atoms (more than a typical one of the atoms in the good set), and small atoms which are not tiny (namely between $2^{-n}$ and $2^{-n(h+\varepsilon)}$).
The number of the larger ones is at-most $\varepsilon 2^{n(h+\varepsilon)}$, and the number of the smaller ones is at-most $\varepsilon 2^{n}$, by a simple union bound and again a volume packing argument.

Therefore, the total number of those atoms is $\leq \varepsilon 2^{n}+o(2^{n})$, which translates to $\frac{2^{n}-|S_{n}|}{2^{n}} \lesssim \varepsilon$.

Hence $\lim_{n} \sum_{s\in S_{n}}\mu(s)+\frac{2^{n}-|S_{n}|}{2^{n}} =0 $ for any measure with $h<1$.

Notice that for $h=1$ (recall I normalize entropy to $log_{2}$ basis), you get simply the Lebesgue measure, as we have uniqueness of measure of maximal entropy in this system, and in that case, $\sum_{s\in S_{n}}\mu(s)+\frac{2^{n}-|S_{n}|}{2^{n}} =1$ by a simple computation.

`s a normal computation, but it`

s true... I don`t know the other side... $\endgroup$ – Bruno Brogni Uggioni Oct 7 '15 at 13:271more comment