$\sup_nX_nA)<\infty$ Let $(X_n)_{n\in\mathbb N}$ be a sequence of independent random variables. Show that $\sup_nX_n&lt;\infty$ almost surely iff there exists $A&gt;0$ such that, $\sum_nP(X_n&gt;A)&lt;\infty$


By Borel-Cantelli we have

$$\sum_nP(X_n&gt;A)=\infty\Longleftrightarrow P(\limsup_nX_n&gt;A)=1$$

Hence $\forall A&gt;0$;

$\sum_nP(X_n&gt;A)=\infty\Rightarrow 1=P(\limsup_nX_n&gt;A)&lt;P(\sup_nX_n&gt;A)$ 

Can you give a hint for the other direction, Thanks.
I'll assume the random variables are real-valued.

'direction 1'

Let $A_n^c := \{X_n &gt; A\}$. By BCL1, we have

$$P(\limsup A_n^C) = 0$$

$$\to P(\liminf A_n) = 1$$

$$\to P(\lim A_n) = 1$$

$$\to \lim P(A_n) = 1$$

$$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$

$$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$

$$\to \forall n \in \mathbb N, P(A_n) = 1$$

$$\to \forall n \in \mathbb N, P\{X_n \le A\} = 1$$

$$\to P( \sup_{n \ge 1} (X_n) &lt; \infty) = 1$$

'direction 2'

Show the contrapositive:

By BCL2, we have

$$P(\limsup A_n^C) = 1 \ (\forall A &gt; 0)$$

$$\to P(\liminf A_n) = 0$$

$$\to \lim_{m \to \infty} P(\bigcap_{n=m}^{\infty} A_n) = 0$$

$$\to \lim_{m \to \infty} \prod_{n=m}^{\infty} P(A_n) = 0$$

$$\to \sup_{m \ge 1} \prod_{n=m}^{\infty} P(A_n) = 0$$

$\to \forall m \ge 1, \exists n \ge m$ s.t. $P(A_n) = 0$

$$\to P(X_n \le A) = 0 \tag{*}$$

Now suppose on the contrary that $P[\sup X_n &lt; \infty] = 1$.

$$\to \sup[X_1, X_2, ...] &lt; \infty \ \text{a.s.}$$

$\to \exists N \ge 0$ s.t.

$$P(\bigcap_{n=1}^{\infty} X_n \le N ) = 1$$

$$\to \prod_{n=1}^{\infty} P(X_n \le N ) = 1$$

$$\to P(X_n \le N ) = 1 \forall n \ge 1$$

Choose $A = N$ in $(*)$. ↯

$$\therefore, \forall A &gt; 0, \sum_{n=1}^{\infty} P(X_n &gt; A) =\infty \to P[\sup X_n &lt; \infty] &lt; 1$$

QED
I'll assume the random variables are real-valued.

'direction 1'

Let $A_n^c := \{X_n &gt; A\}$. By BCL1, we have

$$P(\limsup A_n^C) = 0$$

$$\to P(\liminf A_n) = 1$$

$$\to P(\lim A_n) = 1$$

$$\to \lim P(A_n) = 1$$

$$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$

$$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$

$$\to \forall n \in \mathbb N, P(A_n) = 1$$

$$\to \forall n \in \mathbb N, P\{X_n \le A\} = 1$$

$$\to P( \sup_{n \ge 1} (X_n) &lt; \infty) = 1$$

'direction 2'

Show the contrapositive:

By BCL2, we have

$$P(\limsup A_n^C) = 1 \ (\forall A &gt; 0)$$

$$\to P(\liminf A_n) = 0$$

$$\to \lim_{m \to \infty} P(\bigcap_{n=m}^{\infty} A_n) = 0$$

$$\to \lim_{m \to \infty} \prod_{n=m}^{\infty} P(A_n) = 0$$

$$\to \sup_{m \ge 1} \prod_{n=m}^{\infty} P(A_n) = 0$$

$\to \forall m \ge 1, \exists n \ge m$ s.t. $P(A_n) = 0$

$$\to P(X_n \le A) = 0 \tag{*}$$

Now suppose on the contrary that $P[\sup X_n &lt; \infty] = 1$.

$$\to \sup[X_1, X_2, ...] &lt; \infty \ \text{a.s.}$$

$\to \exists N \ge 0$ s.t.

$$P(\bigcap_{n=1}^{\infty} X_n \le N ) = 1$$

$$\to \prod_{n=1}^{\infty} P(X_n \le N ) = 1$$

$$\to P(X_n \le N ) = 1 \forall n \ge 1$$

Choose $A = N$ in $(*)$. ↯

$$\therefore, \forall A &gt; 0, \sum_{n=1}^{\infty} P(X_n &gt; A) =\infty \to P[\sup X_n &lt; \infty] &lt; 1$$

QED