If $\sum P(A_n) \leq 1$, is the probability of being in at least $k$ of the $A_n$'s bounded by $1/k$?

Question

Let $(\Omega, \mathcal{F},P)$ be a probability space and $A_1, A_2, \ldots$ a sequence of events such that $\sum P(A_n) \leq 1$. Is the probability of being in at least $k$ of the $A_n$'s bounded by $1/k$? I.e. does the following hold: $$ P \bigg( \bigcup_{n_1,\ldots, n_k} A_{n_1} \cap \cdots \cap A_{n_k} \bigg) \leq \frac{1}{k}? $$

I know this is true when the $A_n$'s are open subsets of $[0,1]$; and I suppose the proof works in a more general context (e.g. open sets in $\mathbb{R}^n$ and more generally in Polish spaces), but it doesn't seem to work in this, most general, one.

Note that this would give a 'constructive' proof of the Borel-Cantelli lemma.

Answer 1

Let $N=\sum\limits_n\mathbf 1_{A_n}$, then the event that one is in at least $k$ of the events $A_n$ is $[N\geqslant k]$. Furthermore, $E[N]=\sum\limits_nP[A_n]\leqslant1$ hence Markov inequality yields $P[N\geqslant k]\leqslant E[N]/k\leqslant1/k$.