Suppose $|X_n|\leqslant Y$ a.s. for all $n\in\Bbb N$. Show that $\sup|X_n|\leqslant Y$ a.s. too

Question

Since $|X_n| \leq Y$ a.s. for all $n$, we have that $\mathcal{P}\left(\bigcap_{n=1}^{\infty}\{|X_n|\leqslant Y\}\right) =1$

So now the solution asses that: $$\textstyle\mathcal{P}\left(\bigcap^\infty_{n=1}\{|X_n|\leqslant Y\}\right)=1=\mathcal{P}\big(\sup\{|X_n|\leqslant Y\}\big)$$

and clearly, the $\,\sup|X_n|\leqslant Y$ almost surely.

What is not clear to me and so my doubt is: Why can I rewrite the intersection as the supremum?

Answer 1

There is something wrong with the notation.

The following statements are equivalent:

• $\omega\in\bigcap_{n=1}^{\infty}\{|X_n|\leq Y\}$
• $\forall n\in\mathbb N_+[X_n(\omega)\leq Y(\omega)]$
• $\sup_n|X_n(\omega)|\leq Y(\omega)$
• $\omega\in\{\sup_n|X_n|\leq Y\}$

This makes us conclude that:$$\bigcap_{n=1}^{\infty}\{|X_n|\leq Y\}=\{\sup_{n\in\mathbb N_+}|X_n|\leq Y\}$$The RHS does not agree with the expression in your question.

I am not quite familiar with expressions like $$\sup_{i\in I}A_i$$where ever $A_i$ is a set.

It can only be meaningful if there is an order on these sets.

Inclusion for instance, but then I would prefer "union" above "sup".