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