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How to prove Boole’s inequality

The set of events $\mathcal{A}$ is an collection subsets of $\Omega$ where:

D1: $\Omega \in \mathcal{A}$

D2: $A\in\mathcal{A}\implies A^c\in\mathcal{A}$

D3: $A_1,A_2,...\in\mathcal{A}\implies\bigcup_{i=1}^{\infty}A_i\in\mathcal{A}$

The probability measure $P:A\to\mathbb{R}$ is an image from $\mathcal{A}$ to $\mathbb{R}$ where:

(D4) $\forall A\in\mathcal{A},(0\leq P(A)\leq 1)$

(D5) $P(\Omega)=1$

(D6) $P(\bigcup_{i=1}^{\infty}A_i) = \sum_{i=1}^{\infty}P(A_i)$$ \text{ when } A_1,A_2,... \in \mathcal{A}\text{ are disjunct}$

Prove that: $$P(\bigcup_{i=1}^{\infty}A_i)\leq \sum_{i=1}^{\infty}P(A_i)$$ when $A_1,A_2,... \in \mathcal{A}$

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marked as duplicate by Stefan Hansen, Jonas Meyer, Sasha, rschwieb, Hagen von Eitzen Jan 10 '13 at 18:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can find a disjoint collection $\{B_i\}$ such that $\cup B_i = \cup A_i$ and such that $B_i\subseteq A_i$ for each $i$. You can show that if $B\subseteq A$ then $P(B)\leq P(A)$. – Jonas Meyer Jan 10 '13 at 18:11
@JonasMeyer How do you know that ? – Kasper Jan 10 '13 at 18:12
Kasper: For example if you have two sets $A_1$ and $A_2$, you could take $B_1=A_1$ and $B_2=A_2\setminus A_1$. – Jonas Meyer Jan 10 '13 at 18:13
aaaah, bingo, thanks – Kasper Jan 10 '13 at 18:16
up vote 1 down vote accepted

First note that $P(A)\leq P(B)$ if $A\subset B$, by using the disjoint composition $B=(B\setminus A)\cup A$, $(D6)$ and $P\geq 0$.

Then, compose $\bigcup_{i=1}^{\infty}A_{i}$ to a disjoint union $\bigcup_{i=1}^{\infty}B_{i}$, where $B_{i}=A_{i}\setminus\Big(\bigcup_{k=1}^{i-1}A_{k}\Big)$ for all $i\geq 2$ and $B_{1}=A_{1}$. Can you see how this together with the above condition and $(D6)$ implies the result?

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So $ B_i \subset A_i$ for all $i$, which gives: $P \left(\bigcup_{i=1}^\infty A_i\right)=P\left(\bigcup_{i=1}^\infty B_i\right)=\sum_{i=1}^\infty P(B_i)\leq \sum_{i=1}^\infty P(A_i)$ – Kasper Jan 10 '13 at 18:22
@Kasper. Yes, without the cup though and $\subset$ instead of $\leq$. I.e. $B_{i}\subset A_{i}$ for all $i$. And also, note that in the union we have an equality, i.e. $\cup A_{i}=\cup B_{i}$. – T. Eskin Jan 10 '13 at 18:23
thanks !${}{}{}{}{}{}{}{}{}$ – Kasper Jan 10 '13 at 18:24

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