Borel Cantelli Lemma Lim-sup Question I am stuck on this one step in the proof of the first Borel-Cantelli Lemma.

*

*We have infinite $A_1, A_2, \ldots$ where the sum of their probabilities is finite. (convergence)

*Let $B$ be the event that infinitely many of the events $A_1,A_2,\ldots$ occur.

*Let $B_n=\cup_{k\ge n}^\infty A_k$.

*$B$ occurs if and only if $B_n$ occurs for all $n$, so $B=\cap_{n=1}^\infty B_n=\cap_{n=1}^\infty\cup_{k\ge n}^\infty A_k$. (lim sup)

*etc

How in the world does (4) work??
Going up to for example 3 events, we have $(A_1\cup A_2\cup A_3) \cap (A_2\cup A_3) \cap A_3=A_3$, no? Since the sequence of $B_k$ is decreasing, and since $A_n$ terms approach $\varnothing$ as required by convergence, then $B=\varnothing$ -- quite the opposite of infinitely many events. But as far as I'm aware, this isn't supposed to be a proof by contradiction.
Clarification: I'm following an existing proof not trying to do it myself.
 A: Here is a counter example for the statement

... and  since $A_n$ terms approach $\varnothing$ as required by convergence, then $B = \varnothing$

We consider a uniform distribution on $[0, 1]$ (i.e. Lebesgue measure on this set), and say
$$
A_i = \left[\frac{1}{2} - \frac{1}{2^n}, \frac{1}{2} + \frac{1}{2^n}\right] = \left\{ x: \left|x - \frac{1}{2} \right| \leq \frac{1}{2^n}\right\},
$$
then by simple calculation, we have
$$
\sum_{i=1}^\infty P(A_i) = \sum_{i=1}^\infty \frac{1}{2^{n-1}} = 2 < \infty
$$
which satisfies the condition of Borel-Caltelli lemma.
And then we calculate $B$:
On the one hand, from the definition, $B$ means "infinitely many of events in $\{A_n\}$ happens, which means $x\in B$ satisfies: for infinitely many $n\in \mathbb{N}$, we have
$$
\left|x - \frac{1}{2} \right| \leq \frac{1}{2^n},
$$
the only real number $x$ satisfying above condition is $1/2$, so
$$
B = \{1/2\}.
$$
On the other hand, according to the formula in step 4, we have
$$B=\cap_{n=1}^{\infty} B_n=\cap_{n=1}^{\infty} \cup_{k \geq n}^{\infty} A_k,$$
and firstly, it is not hard to verify that
$$
B_n = \cup_{k \geq n}^{\infty} A_k = A_n,
$$
(this is true only for this specific example!) and therefore
$$
B = \cap_{n=1}^{\infty} B_n = \cap_{n=1}^{\infty}B_n  = \{1/2\}
$$
from a direct calculation.
In conclusion, with both definitions, we can find that $B = \{1/2\}$. This means that the statement that $B = \varnothing$ is not correct. On the other hand, the Borel-Cantelli lemma predicts that $P(B)=0$, and in fact this is correct since finite sets in $[0,1]$ have zero Lebesgue measure.
It is often advantageous to look for counterexamples when learning analysis.
