Imagine you have a shift invariant ($\sigma$-invariant) probability measure $\eta$
in the Bernoulli space $\{0,1\}^{\mathbb{N}}$. Define


$\mathcal{P} = \{[0],[1]\}$;


$\mathcal{P}^{n} = \mathcal{P}\vee...\vee \sigma^{-n+1}(\mathcal{P})$ (cylinders of length $n$);


$k\_{n} = \#\{P \in \mathcal{P}^{n}:\eta(P)\leq\frac{1}{2^{n}}\}$.


For all $n$, enumerate all the cylinders of length $n$ in increasing order
with respect the weight that the measure $\eta$ gives to the cylinders. Say $\{P^{n}\_{1},...,P^{n}\_{2^{n}}\}$. Assume now a simple condition for $\eta$:


$(\sum\_{j=1}^{k\_{n}}\eta(P^{n}\_{j}) +\frac{2^{n}-k\_{n}}{2^{n}}) \rightarrow\_{n} 0$


Is it true that $h\_{\eta}(\sigma)=0$? 


Does anyone have any idea about how to prove this? Or, of course, an example the contradicts it? 


I could proof that if the entropy is zero, that condition above is satisfied. 
What I would like to prove is this:


$h\_{\eta}(\sigma)=0$ if and only if $(\sum\_{j=1}^{k\_{n}}\eta(P^{n}\_{j}) +\frac{2^{n}-k\_{n}}{2^{n}}) \rightarrow\_{n} 0$


Thanks for your attention



Entropy equals zero?
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.


