Infimum of probability and probability of infimum

Question

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?

In the proof that you refer, it is $\cap \cup$ and not $\cup \cap$ as you have it here — Jimmy R., Nov 21 '16 at 15:04
Oh right, I had the reverse order. I have edited it to correct. — mechanical_fan, Nov 21 '16 at 15:24
Look here http://math.stackexchange.com/questions/242920/what-are-some-tricks-to-remember-fatous-lemma the answer with the pictures — Jimmy R., Nov 21 '16 at 15:31

score 3 · Accepted Answer · answered Nov 21 '16 at 15:51

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.

Infimum of probability and probability of infimum

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