Infimum of probability and probability of infimum I am studying the Borel Cantelli proof and there is the following step:
$$\Pr\left( \bigcap \limits_{N=1}^{\infty} \bigcup\limits_{n=N}^{\infty}E_n\right) \le \inf_{N\ge1} \Pr\left( \bigcup\limits_{n=N}^{\infty} E_n\right)$$
What happened here? I guess that:
$$\Pr\left(\bigcap \limits_{N=1}^{\infty}\bigcup\limits_{n=N}^{\infty} E_n\right) = \Pr\left(\inf_{N\ge1}\bigcup\limits_{n=N}^{\infty} E_n\right) \le \inf_{N\ge1} \Pr\left( \bigcup\limits_{n=N}^{\infty} E_n\right)$$
But why is this true?

Note that for any positive integer $K$ you have the inclusion $\bigcap \limits_{N = 1}^\infty \bigcup \limits_{n = N} E_n \subset \bigcup \limits_{n = K}E_n$. Using the monotonicity of measures we can deduce $P\left(\bigcap \limits_{N = 1}^\infty \bigcup \limits_{n = N} E_n \right) \le P\left(\bigcup \limits_{n = K}E_n\right)$. Now the left-hand side is a lower bound for the term on the right-hand side for all $K$, so taking the infimum over all $K$ yields the assertion.
