11
$\begingroup$

Given some polish space $X$ and a probability measure $P$ on $X$ is it true that for any open set $A$ in $X$ one can construct a sequence of bounded continuous functions $\{f_{n}(x)\}_{n}$ such that $f_{n}$ converges in probability to the indicator function on $A$, $I_{A}$?

$\endgroup$
11
$\begingroup$

Yes. We can choose a compatible metric $d$ on $X$ such that $0 \leq d \leq 1$.

For a non-empty set $C \subset X$ let $$d(x,C) = \inf_{c \in C} d(x,c).$$ If $C$ is empty, set $d(x,C) = 1$ for all $x$. Then $x \mapsto d(x,C)$ is continuous since $\lvert d(x,C) - d(y,C)\rvert \leq d(x,y)$.

Assume that $A$ is a proper open subset. Let $F_n = \{x \in X \mid d(x, X \setminus A) \geq \frac1n\}$. Then $F_n \subseteq F_{n+1} \subseteq A$ and $A = \bigcup_{n=1}^{\infty} F_n$. The continuous functions $$ f_n(x) = \frac{d(x,X\setminus A)}{d(x,X \setminus A)+d(x,F_n)} $$ satisfy $0 \leq f_n \leq 1$ and converge pointwise everywhere and monotonically to $I_A$. In particular, $f_n \to I_A$ in probability for every probability measure $P$. Notice that $f_n(x) = 0$ for $x \in X \setminus A$ and that $f_n(x) = 1$ if and only if $x \in F_n$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.