Let $\Delta$ be the interval $[0,1]$, then we can consider the probability space $(\Delta , \mathcal{B}(\Delta),m)$, where $\mathcal{B}(\Delta)$ is the Borel $\sigma$-algebra and $m$ is the Lebesgue measure.
Then we can endow the space $\Delta^{\mathbb{N}}:= \{ (\omega_n)_{n\in \mathbb{N}};\ \omega_n \in \Delta, \ \forall \ n\in \mathbb{N}\}$ with the $\sigma$-algebr $\mathcal{B}(\Delta^{\mathbb{N}})$ (Borel $\sigma$-algebra of $\Delta^{\mathbb{N}}$ induced by the product topology) and the probability measuare $m^{\mathbb{N}}$ in the measurable space$(\Delta^{\mathbb{N}},\mathcal{B}(\Delta^{\mathbb{N}}))$, such that
$$m^{\mathbb{N}} \left(A_1\times A_2\times \ldots \times A_n \times \prod_{i=n+1}^{\infty} \Delta\right)=m(A_1) \cdot \ldots\cdot m(A_n). $$
Now, consider the Bernoulli shift map \begin{align*} \sigma: \Delta^{\mathbb{N}}&\to\Delta^\mathbb{N}\\ (\omega_n)_{n}&\to (\omega_{n+1})_n. \end{align*}
It is well known that the measure $m^{\mathbb{N}}$ on $(\Delta^{N}, \mathcal{B}(\Delta^{\mathbb{N}}))$ is invariant under $\sigma$. Moreover the pair $(\sigma,m^{\mathbb{N}})$ is ergodic. So, Birkhoff's theorem says that for any cilinder $$\Lambda = A_1\times A_2\times \ldots \times A_n \times \prod_{i=n+1}^{\infty} \Delta, $$ where all $A_i$'s are open subsets of $\Delta$, $$\lim_{n\to \infty}\frac{1}{n} \sum_{i=0}^{n-1} \chi_{\Lambda}(\sigma^{i}( (\omega_j)_j)) = m^{\mathbb{N}}(\Lambda),\ \text{$m^{\mathbb{N}}$- a.s.} $$
My Question: Birkhoff's theorem says that for almost every $(\omega_n)_n\in \mathbb{\Delta^{\mathbb{N}}}$, there exists $i = i((\omega_n)_n)\in \mathbb{N}$ such that $(\omega_{i+1},...,\omega_{i+n})\in A_1 \times ...\times A_n$. Is it possible calculate (or estimate) the following probability $$P_m(\Lambda) := m^{\mathbb{N}}\left(\left\{((\omega_n)_{n};\ \exists\ i \in \{0,...,m\}\ \text{such that }\sigma^{i}(\omega) \in \Lambda\right\}\right)\ ? $$ Moreover, given $\varepsilon>0$, are there $n_0 \in \mathbb{N}$, such that $$1- P_n(\Lambda) < \varepsilon, \forall n > n_0 ? $$