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Question

Consider the following sets: $$ A = \Big\{ x\in X: \Pr\bigg(\lim_{n \to \infty}d\big(p_n, [\ell(x), u(x) ] \big)= 0\bigg)=1 \Big\}, $$ and $$ A_n = \Big\{ x\in X: d\big(p_n, [\ell(x), u(x) ] \big)= 0\Big\}, $$ where:

• $\Pr$ denotes probability.
• $X\subseteq \mathbb{R}$.
• $(p_n)_n$ is a sequence of random variables taking values in $[0,1]$.
• $\ell(\cdot)$ and $u(\cdot)$ are real function taking values in $[0,1]$.
• $d\big(p_n, [\ell(x), u(x) ] \big):= \inf \big\{|p_n - y| : y \in [\ell(x), u(x) ] \big\}$.

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Could you help me to show that $$ d_H(A, A_n)\rightarrow_{a.s.} 0, $$ where $$ d_H(A, A_n)\equiv \max\{\sup_{x\in A_n}d(x,A), \sup_{x\in A}d(x, A_n)\}, $$ is the Hausdorff distance.

Answer 1

$\newcommand\de\delta\newcommand\ep\varepsilon$You wrote

Could you help me to show that under Ass1 and Ass2 $$d_H(A, A_n)\rightarrow_{a.s.} 0$$

Of course, this is not true in such generality. For instance, suppose that $X=[0,1)$, $\ell(x)=u(x)=x(1-x)$ for $x\in X$, and $p_n=u(\ep_n)$, where $\ep_n:=\sqrt{(\ln2)/(2n)}$.

Then $$A_n=\{x\in X\colon p_n=u(x)\}=\{\ep_n,1-\ep_n\},$$ $$A=\{x\in X\colon p_n\to u(x)\}=\{0\}\ne\emptyset,$$ $$A_n(\de)=\{x\in X\colon|p_n-u(x)|\le\de\}.$$ So, for each $x\in A$ and each real $\de>0$ we have $x=0$ and hence $$\Pr(x\in A_n(\de))=\Pr(0\in A_n(\de))=\Pr(|p_n|\le\de) \\ =\Pr(p_n\le\de)\ge\Pr(\ep_n\le\de)=1(\ep_n\le\de) \\ =1(\sqrt{(\ln2)/(2n)}\le\de)\ge1-\frac{2}{\exp(2n\de^2)}.$$

So, all your assumptions hold. However, $$d_H(A,A_n)\to1\ne0.$$

Answer 2

EDIT: That whole answer assumes the $p_n$'s are uniformly and independently distributed on $[0,1]$. I have no idea on how to extend it to the general case of other distributions.

It gets elementary after you look at $A$ in simple cases. For example when $\ell(x) = 0$ and $u(x) = \frac{1}{2}$ for all $x$ and the $p_n$'s are uniformly and independently distributed on $[0,1]$. Then almost surely both $p_n\in[0,\frac{1}{2}]$ for infinitely many $n$'s and $p_n\in[\frac{3}{4},1]$ for infinitely many other $n$'s. Therefore $d\big(p_n, [0, \frac{1}{2}]\big)$ keeps oscillating too widely and $$a := \lim_{n \to \infty}d\big(p_n, [\ell(x), u(x) ]\big)$$ almost surely doesn't exist and $A$ is almost always empty in that special case of functions.

Now let's go back to the general case of any real functions $\ell()$ and $u()$. With the same reasonning, $a$ almost surely doesn't exist for any $x$ with $0 < \ell(x)$ or $u(x) < 1$. Conversely, $a$ is always $0$ for any $x$ with $\ell(x) \le 0$ and $1 \le u(x)$. Therefore $$ A = \{ x\in X: \ell(x) \le 0 < 1 \le u(x)\} $$ and trivially $$A\subseteq A_n\quad \forall n.$$

EDIT: This proves (2) is true iff $\ell(x) \le 0 < 1 \le u(x)$ for some $x\in X$.

(1) and (3) are false when for example $X=[0,1]$ and $\ell(x) = 0$ on $[0,\frac{1}{2}]$, $\ell(x)=\frac{3}{4}$ on $(\frac{1}{2}, 1]$ and $u(x) = 1$ on $[0,1]$. Because then $A = [0,\frac{1}{2}]$ and $A_n$ almost surely contains $\frac{3}{4}$ for infinitely many $n$'s hence $d_H(A,A_n)>\frac{1}{4}$ infinitely often and cannot go to zero.

Along these lines, you should have no trouble showing (1) and (3) hold if the functions $\ell()$ and $u()$ are continuous on $X$, $\ell(x) \le 0 < 1 \le u(x)$ for some $x\in X$ and $\ell(x) > 0$ for some other $x\in X$.