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Suppose $\lim_{n\to\infty}X_n=X$ a.s. and $|X|<\infty$ a.s. Let $Y=\sup_n|X_n|$, Show that $Y<\infty$ a.s.

If $\lim_{n\to\infty}X_n=X$ a.s. then

$S_1:=\{w:\lim_{n\to\infty}X_n=X\}$ has probability $1$ and

$S_2:=\{|X|<\infty\}$ has also probability $1$

so take an element from their intersection, since $\lim_{n\to\infty}X_n=X$ there exists an $N$, such that $\forall n\ge N$, $|X_n-X|\le\epsilon$, therefore all $X_n$ starting from $N$ are finite, but what about the rest, is it not possible that for example $X_1=\infty$ ?

Thanks in advance.

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    $\begingroup$ $X_1=\infty$ is not a R.V. in my book. I think the definition of a R.V. means $X_1:\Omega\rightarrow\mathbb{R}$? $\endgroup$
    – Lost1
    Jul 1 '14 at 13:12
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    $\begingroup$ If you allow infinite random variables, a trivial counterexample is just define $X_1=\infty$ and $X_i=0$ for all $i>1$. Then $X_n\rightarrow 0$ (surely) but $\sup_n X_n = \infty$. But I agree with @Lost1 that this would be kind of silly. So your argument above is sufficient to prove $\sup_n |X_n| < \infty$, given that any finite collection of random variables must have a finite supremum magnitude. $\endgroup$
    – Michael
    Jul 1 '14 at 13:21
  • $\begingroup$ @Michael Ok that's because of the definition then $\endgroup$
    – OBDA
    Jul 1 '14 at 13:23
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    $\begingroup$ Yes. The importance of $\sup_n |X_n| < \infty$ is in showing that, as we consider all infinite values $\{X_1, X_2, \dots\}$, the sequence of random variables does not have an unbounded supremum (such as $X_n=n$ for all $n$). This is automatically true if we only consider a finite collection of random variables (unless we allow them to take infinite values). $\endgroup$
    – Michael
    Jul 1 '14 at 13:25
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    $\begingroup$ Well, $\sup_{n\geq 1} |X_n| \leq \sup_{n \in \{1, \ldots, N-1\}} |X_n| + \sup_{n\geq N} |X_n|$. $\endgroup$
    – Michael
    Jul 1 '14 at 13:33
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I assume the usual definition of random variables, which take only finite values.

First, a fact from real analysis: For any convergent sequence of real numbers $\{a_n\}$, $\sup_n a_n < \infty \Leftrightarrow \lim_n a_n < \infty$ (Think about it, every $a_j$ is finite, so the only way to "get to" infinity is via the limit).

Now, note that for a random variable $Y$, $Y<\infty$ a.s. $\Leftrightarrow Y(\omega)<\infty$ for (almost) all $\omega$.

Hence, $\sup_n |X_n|<\infty$ a.s. $\Leftrightarrow \sup_n |X_n(\omega)|<\infty$ for (almost) all $\omega$.

Hence using the real analysis fact, $\sup_n |X_n|<\infty$ a.s. $\Leftrightarrow \lim_n |X_n(\omega)|<\infty$ for (almost) all $\omega$.

That is the same as, $\sup_n |X_n|<\infty$ a.s. $\Leftrightarrow \lim_n |X_n|<\infty$ a.s.

We will show the RHS to be true: $$ X_n \rightarrow X \text{ a.s.} \\ \Rightarrow |X_n| \rightarrow |X| \text{ a.s.} < \infty \\ \Leftrightarrow \lim_n |X_n| < \infty \text{ a.s.} $$

The proof is now complete.

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    $\begingroup$ The problem was clear, but your answer makes me unsure, are you sure that $\lim|X_n|=\sup_n|X_n|$ always holds ? $\endgroup$
    – OBDA
    Jul 1 '14 at 14:03
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    $\begingroup$ I see your point. I will make a modification. $\endgroup$
    – ved
    Jul 1 '14 at 14:06

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