# Rigorous construction of the pointwise limit of a sequence of random variables

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 $$\omega \in \Omega$$.

• 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. – Sangchul Lee Apr 18 at 10:08

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.
• 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. – Kavi Rama Murthy Apr 18 at 10:16
• 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 $$X(\omega) := h(\limsup X_n(\omega)) \ ?$$ – Joker123 Apr 18 at 10:20