For an infinite sequence of events $E_1,E_2,E_3,\ldots$ show that $$P\left(\bigcup_{k=1}^\infty E_k\right)\le \sum_{k=1}^\infty P(E_k)$$

This is the question given to us. Now I know we can solve this by induction but is there any other method to solve this?

  • 1
    $\begingroup$ Do you mean $\mathbb P\left(\bigcup\limits_{k=1}^{\infty} E_k \right) \le \sum \limits_{k=1}^{\infty} \mathbb P\left( E_k \right)$ ? $\endgroup$
    – Henry
    Jan 24, 2021 at 14:08
  • $\begingroup$ yes I meant that $\endgroup$
    – user829028
    Jan 24, 2021 at 14:11
  • $\begingroup$ Use MathJax for typesetting math. $\endgroup$ Jan 24, 2021 at 17:28

1 Answer 1


I suppose you are talking about $\mathbb P\left(\bigcup\limits_{k=1}^{\infty} E_k \right) \le \sum \limits_{k=1}^{\infty} \mathbb P\left( E_k \right)$

This is true when the $E_k$ are mutually disjoint (countable addition of probability measure) and you get equality.

Define $F_k$ by $F_1=E_1$ and $F_k = E_k \backslash \bigcup\limits_{j=1}^{k-1} E_j$ so the $F_k$ are mutually exclusive. Then

  • $\mathbb P\left( F_k \right) \le \mathbb P\left( E_k \right)$ since $F_k \subset E_k$
  • $\bigcup\limits_{k=1}^{\infty} F_k = \bigcup\limits_{k=1}^{\infty} E_k $ since any element of one is an element of the other
  • so $\mathbb P\left(\bigcup\limits_{k=1}^{\infty} E_k \right) =\mathbb P\left(\bigcup\limits_{k=1}^{\infty} F_k \right) = \sum \limits_{k=1}^{\infty} \mathbb P\left( F_k \right) \le \sum \limits_{k=1}^{\infty} \mathbb P\left( E_k \right)$

Which of these steps implicitly use induction, I leave to you

  • $\begingroup$ thank you so much $\endgroup$
    – user829028
    Jan 24, 2021 at 14:20

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