Boole inequality, proof of

Let $\{B_1,B_2,\mathrm{\cdots }\}$ be a sequence defined by:

$$B_i=A_i\setminus \bigcup _{k=1}^{i-1}{A}_k$$

Clearly $B_i\in ℱ,\forall i\in \mathbb{N}$, since $ℱ$ 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 $ℱ$ 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 $P(B_i)\le P(A_i)$ because measures are http://planetmath.org/node/4460monotonic, then it follows that :

$$P(\bigcup _{n=1}^i{B}_n)\le \sum _{n=1}^{i}P(A_n),\forall i\in \mathbb{N}$$

finally taking $n\to \mathrm{\infty }$ :

$$P(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n)=P(\bigcup _{n=1}^{\mathrm{\infty }}{B}_n)\le \sum _{n=1}^{\mathrm{\infty }}P(A_n)$$

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