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Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a probability space and let $$X_1,X_2,X_3,... \: \Omega \rightarrow \mathbb{R} $$ be a sequence of random variables. Moreover, let there be an event $A \subseteq \Omega$ with $\mathbb{P}(A) = 1$ such that, for all $\omega \in A$, it holds that $$ \lim_{n \rightarrow \infty} X_n(w) $$ exists and is finite.

From all of this, how can I rigorously construct a random variable $$ X: \Omega \rightarrow \mathbb{R}$$ for which it holds that $$ X_n \rightarrow X $$ almost surely, without setting $$ X := X_n \mathbb{1}_{A} \quad? $$ Note that, since $X$ is a random variable, it of course needs to be $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable. Furthermore, note that the range of $X$ should stay finite and should exclude the possibility, that $\lvert X(\omega) \rvert = \infty $ for some $\omega \in \Omega$.

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    $\begingroup$ Set $X' = \liminf_{n\to\infty} X_n$, which always exists in $[-\infty, \infty]$, and $X = X'\mathbf{1}_{\{X'\in\mathbb{R}\}}$. This $X$ always exists for any sequence of random variables and $X'(\omega) = \lim_{n\to\infty} X_n(\omega)$ whenever the limit exists. $\endgroup$ – Sangchul Lee Apr 18 at 10:08
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Let $X(\omega)=\lim\sup X_n(\omega)$ if $\lim\sup X_n(\omega) \in \mathbb R$ and $X(\omega)=0$ otherwise. This $X$ has the desired properties.

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  • $\begingroup$ Thank you. How can one prove that the function you suggest is measurable ? $\endgroup$ – Joker123 Apr 18 at 10:14
  • $\begingroup$ That is very standard. $\lim \sup$ of a sequence of measurable functions is always measurable so the set of points where $\lim \sup$ is finite is also in the sigma algebra. That makes $X$ measurable. $\endgroup$ – Kavi Rama Murthy Apr 18 at 10:16
  • $\begingroup$ Thanks, Would it also work if I defined a function $$ h :[- \infty, \infty] \rightarrow (- \infty, \infty) $$ with $h(x) = x$ for $x \in \mathbb{R}$ and $h(x) = 0$ otherwise, and then set $$ X(\omega) := h(\limsup X_n(\omega)) \ ?$$ $\endgroup$ – Joker123 Apr 18 at 10:20
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    $\begingroup$ Yes, that works. $\endgroup$ – Kavi Rama Murthy Apr 18 at 10:21

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