ボレル・カンテリの補題
o.}}\}^{c}\right)\\&=\Pr \left(\left(\bigcap _{N=1}^{\infty }\bigcup _{n=N}^{\infty }E_{n}\right)^{c}\right)=\Pr \left(\bigcup _{N=1}^{\infty }\bigcap _{n=N}^{\infty }E_{n}^{c}\right)\\&=\Pr \left(\liminf _{n\rightarrow \infty }E_{n}^{c}\right)=\lim _{N\rightarrow \infty }\Pr \left(\bigcap _{n=N}^{\infty }E_{n}^{c}\right)\end{aligned}}}
ここで独立性より
Pr ( ⋂ n = N ∞ E n c ) = ∏ n = N ∞ Pr ( E n c ) = ∏ n = N ∞ ( 1 − Pr ( E n ) ) ≤ ∏ n = N ∞ exp  ( − Pr ( E n ) ) = exp  ( − ∑ n = N ∞ Pr ( E n ) ) = 0 {\displaystyle {\begin{aligned}\Pr \left(\bigcap _{n=N}^{\infty }E_{n}^{c}\right)&=\prod _{n=N}^{\infty }\Pr(E_{n}^{c})\\&=\prod _{n=N}^{\infty }(1-\Pr(E_{n}))\\&\leq \prod _{n=N}^{\infty }\exp(-\Pr(E_{n}))\\&=\exp \left(-\sum _{n=N}^{\infty }\Pr(E_{n})\right)\\&=0\end{aligned}}}