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The Shannon-McMillan-Breiman theorem, in its dynamical systems formulation, contains (or implies) the following statement: If $T$ is the map, $\mu$ is the ergodic $T$-invariant measure, $\mathcal{P} = \{P_1,\ldots,P_r\}$ is the partition of the state space, $\Sigma_{n,\epsilon}$ is a typical set of length $n$ itineraries, i.e., $\Sigma_{n,\epsilon} = \{ a \in \{1,\ldots,r\}^n : 2^{-n(h+\epsilon)} \leq \mu(x: i_n(x) = a) \leq 2^{-n(h-\epsilon)} \}$ with $h = h_{\mu}(T;\mathcal{P})$ and $i_n(x) = (i_n^0(x),\ldots,i_n^{n-1}(x))$, $T^j(x) \in P_{i_n^j(x)}$, then $\mu(x: i_n(x) \notin \Sigma_{n,\epsilon}) \rightarrow 0$ for $n\rightarrow\infty$. My question is: Are there any results on the rate of convergence for this limit, maybe under specific, but not too restrictive assumptions on the dynamical system?

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    $\begingroup$ As you probably know, there are negative results, showing that you cannot hope to have rates of convergence in any ergodic theorem (including Shannon-Macmillan-Breiman), even for the “in measure” variants that you are looking for. If you want a positive result, I would impose Gibbs or exponential mixing conditions. Keller’s book on equilibrium states is quite closely related to what you are asking. $\endgroup$ – Anthony Quas Oct 11 '17 at 20:26
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In the case of (weak) Gibbs measures you can find exponential large-deviation bounds as well, as almost sure estimates for the error term in the convergence to entropy given by the Shannon–McMillan–Breiman formula for both uniformly and non-uniformly expanding shift spaces [ETDS, 37:7, (2017), 2313-2336]. For broader classes of invariant measures, in opposition to the case of Birkhoff averages, I belive it is not known reasonably mild assumptions under which e.g. some L^1 convergence holds.

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  • $\begingroup$ I think you may have accidentally deleted the link you meant to post; you should be able to edit it back in. $\endgroup$ – user44191 Feb 7 '19 at 19:37
  • $\begingroup$ Hi @paulo welcome to MO. It appears you have a broken link above. What reference were you hoping to including for the Shannon–McMillan–Breiman formula? $\endgroup$ – Neil Hoffman Feb 7 '19 at 19:38

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