Given $\{A_n\}$ with $P(A_n)\to 0$ show that for every $\alpha>0$ there exists a subsequence $\{A_{n_k}\}$, such that $P(\cup_k A_{n_k})\leq\alpha$

Question

Given $\{A_n\}$ with $P(A_n)\to 0$ show that for every $\alpha>0$ there exists a subsequence $\{A_{n_k}\}$, such that

$$P(\cup_k A_{n_k})\leq\alpha$$

I found this variation of the problem I am trying to solve. I figure the solution to my problem is similar, but I don't exactly see through it.

Because $P(A_n)\to 0$ then I figure there exists a subsequence $A_{n_k}$ such that $\sum_{k=1}^\infty P(A_{n_k})< \infty$ and from Borel-Cantelli I have $P(\lim\sup A_{n_k})=0$. Moreover I get that $\lim_{k\to\infty} P(\cup_k A_{n_k})=0$ from the properties of limit superior. Does this reasoning help to get to the needed result?

I am grateful for any suggestions.

Answer 1

I think the standard trick here is to find some absolutely convergent series$\sum_{k=1}^\infty a_k$ and then a subsequence $\{ A_{n_k}\}_{k=1}^\infty $ such that $\mathbb{P}(A_{n_k})\leq a_k$. I claim that this should imply your result from the properties of a measure. Can you see why?

Answer 2

Your problem is pretty much the same as the variation that you quoted. See this excerpt from the accepted solution, which avoids the Borel-Cantelli lemma:

Now because $P(A_n^c)\to0$, we can choose a subsequence $n_k$ such that $$ P(A_{n_k}^c)< 3^{-(k+1)}$$

You can use this same idea. Since $P(A_n)\to0$, you can choose $n_k$ so that $P(A_{n_k})$ is as small as you please -- so small that the sum of $P(A_{n_k})$ is less than $\alpha$.