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I am having trouble trying to understand the topic of my question. For reference please use Virtual Laboratory of Probability and Statistics.

Let's start with limit superior: $$\limsup_{n \to \infty} A_n = \bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_i$$ It works for arbitrary sequences of events $A_n$, right? Let's assume we have independent replication of the same basic event $A$ whose probability is strictly positive $\mathbb{P}(A) = p \in (0, 1]$, so that each $A_n$ is just $A$. And since set union and set intersection are both idempotent operations, we get: $$\bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_i = \bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A = A$$ so that $$\limsup_{n \to \infty} A_n = A$$ and $$\mathbb{P} \left( \limsup_{n \to \infty} A_n \right) = \mathbb{P} \left( A \right) = p$$ So far, so good. But from the second Borel-Cantelli lemma, since $p > 0$, we get $$\mathbb{P} \left( \limsup_{n \to \infty} A_n \right) = 1$$ For me it's hard to accept, that arbitrarily chosen $p \in (0, 1]$ will always be equal to 1. Where's my mistake?

Anyway, what is that event $\limsup_{n \to \infty} A_n$? Since it is assigned probability 1, it must cover the entire sample space $\Omega$ of our probability space $(\Omega, \mathscr{F }, \mathbb{P})$, minus any countable subset. The first term in the definition of limit superior is $\bigcap_{n=1}^\infty$, and it specifies a decreasing (non increasing) sequence of events. From this it follows that any of it "tails" $\bigcup_{i=n}^\infty A_i$ must have probability 1 -- $\mathbb{P} \left( \bigcup_{i=n}^\infty A_i \right) = 1$. Putting it all together, any of the "tails" $\bigcup_{i=n}^\infty A_i$ must include every possible event from the probability space. Is this line of thinking is correct? I mean, do I use the right words to describe my intuition behind limit superior/inferior?

One last question, can limit superior/inferior of arbitrary sequence of events ever take any other probability different from 0 and 1?

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This is because independent replication refers to something completely different from what you describe. One is given a sequence $(A_n)_n$, say with common probability $\mathbb P(A_n)=p$ in $(0,1)$, and the key assumption is that the events $(A_n)_n$ are independent. This means that for every finite collection distinct indexes $(i_k)_{1\leqslant k\leqslant K}$, $$ \mathbb P(A_{i_1}\cap\cdots\cap A_{i_K})=\prod_{k=1}^K\mathbb P(A_k). $$ Then indeed (a special case of) Borel-Cantelli lemma ensures that the limsup of the sequence $(A_n)_n$ has probability $1$.

A sequence $(A_n)_{n\geqslant1}$ such that $A_n=A$ for every $n$, with $\mathbb P(A)=p$ in $(0,1)$, is never independent. Furthermore, to answer the last question of your post, note that in this case $$ \limsup\limits_nA_n=\liminf\limits_nA_n=A, $$ hence their common probability is $p$ as well, which can take every value in $(0,1)$ (a host of other examples exist, naturally).

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