# Measurability of an a.s limit

Let $$(\Omega,\mathcal{F},P)$$ be a probability space, $$(\mathcal{F}_n)_n$$ a decreasing sequence of sub-$$\sigma$$-algebra of $$\mathcal{F}.$$ Let $$(X_n)_n$$ be a sequence of r.v. converging a.s to $$X.$$ Prove that if for all $$n \in \mathbb{N}, X_n$$ is $$\mathcal{F}_n$$-measurable then $$X$$ is $$\bigcap_{n \in \mathbb{N}}\mathcal{F}_n$$-measurable.

We need to show that for a fixed $$n \in \mathbb{N},U \in B(\mathbb{R}),X^{-1}(U) \in \mathcal{F}_n.$$

Any suggestions is appreciated.

Note that for all $$n$$ $$\{X \in U\} = \bigcap_{q \in \mathbb N} \bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}.$$

Now, $$\{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}$$ is $$\mathcal F_n$$-measurable for all $$m \geq n$$ because $$(\mathcal F_n)_n$$ is decreasing. So, $$\bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}$$ is $$\mathcal F_n$$-measurable for all $$r \geq n$$. So, $$\bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}$$ is $$\mathcal F_n$$-measurable, and therefore so is $$\bigcap_{q \in \mathbb N} \bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}.$$

• Why do we have the first equality? (We have almost sure convergence, not pointwise)
– john
Nov 18, 2020 at 22:23
• @john In that case, I'm not sure it's true unless the spaces $(\Omega, \mathcal F_n, P)$ are complete because in general a.s. equality of random variables doesn't imply measurability without completeness.