Let \(\beta >1\) and consider a \(\beta\)-shift, \(T_\beta : [0,1] \rightarrow [0,1]\), given by \(T_\beta(x) = \beta x- \lfloor \beta x \rfloor\). The map induces \(\beta\)-expansions of \(x \in [0,1]\), \[ x = \sum_{n=1}^\infty \frac{\varepsilon_n(x,\beta)}{\beta^n}, \quad \varepsilon_n(x,\beta) = \lfloor \beta T_\beta^{n-1}(x)\rfloor. \] The sequences \((\varepsilon_n(x,\beta))\) in \(\{0,1, \dots, \lceil \beta \rceil - 1\}^\mathbb{N}\) arising in this way from some \(x\in [0,1]\) are called admissible. The closure of the set in the product topology of the discrete topology on \(\{0,1, \dots, \lceil \beta \rceil - 1\}\) is denoted by \(S_\beta\).
It is first shown that for a set \(Z \in S_\beta\), the Hausdorff dimension of \(Z\) as a subset of \(\{0,1, \dots, \lceil \beta \rceil - 1\}^\mathbb{N}\) is equal to the Hausdorff dimension of \[ \left\{ \sum_{n=1}^\infty \frac{\varepsilon_n(x,\beta)}{\beta^n} : (\varepsilon_n(x,\beta)) \in Z\right\} \] as a subset of \([0,1]\).
Suppose now that \(\beta > 1\) is such that the \(\beta\)-expansion of \(1\) terminates. It is show that in order to calculate the Hausdorff dimension of the frequency set \[ F_{\beta, a} = \left\{x \in [0,1) : \lim_{n\rightarrow \infty} \frac{\# \{1 \le k \le n : \varepsilon_k(x,\beta) = 0\}}{n} = a \right\}, \] one needs only find the maximal entropy of certain Markov measures on the symbolic shift, with the set of measures depending on the length of the expansion of \(1\), and then divide by \(\log \beta\). Finally, for \(\beta \in (1,2)\) with \((\varepsilon_n(1,\beta)) = 1^m 0^\infty\), it is deduced from the above that for \(a < 1/m\), \(F_{\beta, a} = \emptyset\); and that in order to calculate the Hausdorff dimension of \(F_{\beta, a}\), one just needs to calculate \[ \frac{1}{\log \beta} \max f_a(x_1, \dots, x_{m-2}), \] where \(f_a\) is an explicit function depending on \(a\) and \(m\), and the maximum is taken over an explicit region.