imagine you have a sequence $\eta\_{n}$ of (shift) invariant measures in the Bernoulli space $\{0,1\}^{\mathbb{N}}$ that satisfy the following: there are a $0<\delta <1$ and an $N$ such that $$n > N \Rightarrow \sum\_{i=1}^{2^{m}}|\eta\_{n}(C\_{i}^{m}) - \frac{1}{2^{m}}|^{2}<\delta^{2n},$$ for all $m$ where 


$C\_{1}^{1} = [0], C\_{2}^{1} = [1]$, 


$C\_{1}^{2} =[0,0],...,C\_{4}^{2}=[1,1]$,


$C\_{i}^{m}$ is the $i^{th}$ cylinder of length $m$, with $i \in \{1,2,...,2^{m}\}$.


So, of course you have much more than $\eta\_{n} \rightarrow \lambda$ in the weak\* topology, where $\lambda$ is the Bernoulli uniform probability $(\frac{1}{2},\frac{1}{2})$. For instance, you have that all your cylinders are converging (in measure) to $\frac{1}{2^{m}}$ at the same time!


My question is: $h(\eta\_{n})\rightarrow \log2$? Are the entropy of the measures growing to $\log2$?


Thanks for your attention


Bruno



Entropy, Convergence
I think you can make the measures have entropy as small as you like (even zero if you want, but this is slightly more complicated): regard $\{0,1\}^{\mathbb N}$ as a group (with coordinatewise addition modulo 2). Now you can define convolution of measures: $\int f\,d(\mu\*\nu):=\int f(x+y)\,d\mu(x)\,d\nu(y)$. Let $P\_n$ denote the collection of points fixed by the $n$th power of the shift and let $\nu\_n$ be the uniform measure on $P\_n$ so that $\nu\_n$ is invariant. Finally pick your favourite $\delta<1$, let $\mu$ be any Bernoulli measure and let $\eta\_n=\mu\*\nu\_{k\_n}$ where $(k\_n)$ is chosen below. Now $\eta\_n(C^k\_i)=2^{-k}$ for all $i$ and all $k\le k\_n$. Now to estimate the sum: the sum is 0 for $m\le k\_n$. For $m>k\_n$, the sum is bounded above by $2\sum\_i \eta\_n(C^m\_i)^2 + 2\sum\_i (2^{-m})^2\le 2^{-m+1} + 2\sum\_i \eta\_n(C^m\_i)^2\le 2^{-k\_n}+\max\_i \eta\_n(C^m\_i)$. Now just choose the $k\_n$ to make sure this is at most $\delta^{2n}$. 


