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.


1 Answer 1


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 $$\bigcap_{q \in \mathbb N} \bigcup_{r \geq n} \bigcap_{m \geq r} \{X_m + 1/q \in U\} \cup \{X_m - 1/q \in U\}.$$

  • $\begingroup$ Why do we have the first equality? (We have almost sure convergence, not pointwise) $\endgroup$
    – john
    Nov 18, 2020 at 22:23
  • $\begingroup$ @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. $\endgroup$
    – aduh
    Nov 18, 2020 at 22:37
  • $\begingroup$ They didn't mention completeness of the space $\endgroup$
    – john
    Nov 18, 2020 at 22:44

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