Maybe you can look at the proof of BC's lemma with the previous examples in mind.

"For any n" => intersection

"there exits" => union

and you can use the equality

for independent events, and the inequality

in general.

Just an example. Translating into symbols, I suggested to first show that, for any

,

. In order to use independence, I must handle intersections, hence let's write

. And we have

by independence.

(If you aren't used to applying independence to infinitely many events at once, you can say

for all

, hence

. )

Thus

. What about

? Switching to the complement:

(using what we got before). In conclusion,

. This is what you need.

You can remember this (this was the last part): a countable intersection of almost sure events is almost sure.