Let be $(\Omega,\mathcal{F},P)$ a probability space and $$ \limsup\limits_{n\to\infty} A_n:=\bigcap\limits_{n\geq 1}\bigcup\limits_{m=n}^{\infty}A_m,\text{ where } A_m\in\mathcal{F}\text{ for all } m\in\mathbb{N}.$$ If all $A_n$ are stochastically independent, then (Borel-Cantelli lemma): $$ \sum\limits_{n=1}^{\infty}P(A_n)=\infty\implies P\left(\limsup\limits_{n\to\infty} A_n\right)=1. $$
We consider a Bernoulli experiment with probability $p$ and length $n$ where the random variable $X_j:\Omega\to \{0,1\}$, with $1\leq j\leq n$, denotes the $j$-th step. Further, we define the set $$ D_{m,i}:=\{X_{mi+k}=1\text{ for all }k=1,\dots, m\}, $$ where $0\leq i$ denotes the $i$-th array of $m$-consecutive $1$'s.
If we define for some $0<\epsilon<1$ the array-length $m_n:=\lceil(1-\epsilon)\log_{\frac{1}{p}}(n)\rceil$, then show that $P\left(\limsup\limits_{n\to\infty}D_{m_n,i}\right)=1$.
In our lecture the professor conducted the proof as follows:
Let be $T_{m}:=\sum\limits_{i\geq 0} 1_{D_{m,i}}$, i.e. a summation over all possible arrays of length $m$. We have at most $\left(\frac{n}{m}-1\right)-$ many arrays. Then, it follows $$ \mathbb{E}(T_{m_n})\geq\left(\frac{n}{m_n}-1\right)p^{m_n}=\dots=\frac{p}{m_n}n^{\epsilon}-pn^{-(1-\epsilon)}. $$ We see that $\lim\limits_{n\to\infty}\mathbb{E}(T_{m_n})=\infty$, so by Borel-Cantelli lemma it follows $P\left(\lim\sup\limits_{n\to\infty} D_{m_n,i}\right)=1.$
I am highly skeptical that this proof is correct.
1.) What does $P\left(\lim\sup\limits_{n\to\infty} D_{m_n,i}\right)=1$ mean? If I simply apply the aforementioned definition, then I have to fix the $i$ and it yields $\limsup\limits_{n\to\infty} D_{m_n,i}:=\bigcap\limits_{n\geq 1}\bigcup\limits_{m_n=n}^{\infty}D_{m_n,i}$. But if I fix $i$, then the sets $D_{m_n,i}$ are not necessarily independent. So this makes no sense!?
2.) Instead if I fix the $m_n$ and consider the sets $D_{m_n,i}$ with $0\leq i$, then they would be stochastically independent. But this would only allow us to apply Borel-Cantelli lemma at the sequence $(D_{m_n,i})$ where $i\to\infty$ and $m_n$ is arbitrary but fixed.
Maybe someone is more familiar with this issue and can confirm that this proof is false or help me if I have misunderstood something?