# Boole inequality, proof of

Let $\{B_{1},B_{2},\cdots\}$ be a sequence defined by:

 $B_{i}=A_{i}\setminus\bigcup_{k=1}^{i-1}A_{k}$

Clearly $B_{i}\in\mathcal{F},\forall i\in\mathbb{N}$, since $\mathcal{F}$ is $\sigma$-algebra, they are a disjoint family and :

 $\bigcup_{n=1}^{i}A_{n}=\bigcup_{n=1}^{i}B_{n},\forall i\in\mathbb{N}$

and since $P$ is a measure over $\mathcal{F}$ it follows that :

 $P(\bigcup_{n=1}^{i}B_{n})=\sum_{n=1}^{i}P(B_{n}),\forall i\in\mathbb{N}$

Clearly $B_{i}\subset A_{i}$ , then $P(B_{i})\leq P(A_{i})$ because measures are http://planetmath.org/node/4460monotonic, then it follows that :

 $P(\bigcup_{n=1}^{i}B_{n})\leq\sum_{n=1}^{i}P(A_{n}),\forall i\in\mathbb{N}$

finally taking $n\to\infty$ :

 $P(\bigcup_{n=1}^{\infty}A_{n})=P(\bigcup_{n=1}^{\infty}B_{n})\leq\sum_{n=1}^{% \infty}P(A_{n})$

the latter is valid because the measure continuity , and is the proof of the theorem

Title Boole inequality, proof of BooleInequalityProofOf 2013-03-22 15:47:18 2013-03-22 15:47:18 Bunder (13010) Bunder (13010) 6 Bunder (13010) Proof msc 60A99