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Let $(\Omega,\mathcal{A}, P)$ be a probability space and $X$ be a Borel measurable and separable map.

(i) $X_{n}\stackrel{\text{ as }}{\rightarrow}X$ and $\left(d\left(X_{n},X\right) \right)$ is asymptotically measurable;

(ii) $X_{n}\stackrel{\text{ as }}{\rightarrow}X$ and $\left( X_{n} \right)$ is asymptotically measurable.

How can we show that (i) implies (ii)?

(This is a part of Problem 1.9.1 on van der Vaart and Wellner, 1996)

Here, $X_{n}\stackrel{\text{ as }}{\rightarrow}X$ means that $P_{*}\left(\lim d\left(X_n, X\right)=0\right)=1$.

$\left( X_{n} \right)$ is asymptotically measurable if and only if $\mathrm{E}^{*} f\left(X_n\right)-\mathrm{E}_{*} f\left(X_n\right) \rightarrow 0, \quad \text { for every } f \in C_{b}(\mathbb{D})$.

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  • $\begingroup$ I think before you can get an answer to your question you should go into some details. First in (ii) you have to replace $X_n$ by $(X_n)$, I think. Then you should explain what is meant by almost sure convergence of not necessarily measurable functions (I think it has to do with the probability measure space $(\Omega,\mathcal{A},\mathbb{P})$ (inner probability measure $\mathbb{P}^*$?). Since the terminology seems to be not known generally, such elaborations are important. Otherwise you reach only specialists. $\endgroup$
    – Dieter Kadelka
    Commented Jan 17, 2021 at 14:01
  • $\begingroup$ Thank you for your comment, Dieter. I added some definitions and made notations clearer. $\endgroup$
    – Geor11
    Commented Jan 18, 2021 at 3:12
  • $\begingroup$ What is a separable map and where does it act to? $\endgroup$
    – Fedor Petrov
    Commented Jan 18, 2021 at 10:01
  • $\begingroup$ Thanks, I should have mentioned it. We say a map is separable when there is a separable (in the topological meaning), measurable set $A$ with probability 1. That is, $P(X \in A) = 1$. It may work in somewhere, but I have no idea. $\endgroup$
    – Geor11
    Commented Jan 18, 2021 at 13:42

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