# Show that $P\left(\bigcup_{k=1}^\infty E_k\right)\le \sum_{k=1}^\infty P(E_k)$

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?

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

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

• thank you so much
– user829028
Jan 24, 2021 at 14:20