Conditions for pointwise convergence of indicators precomposed with uniformly continuous sequence

Question

Let $X$ be a compact metric space, $\{\delta_n\}_{n=1}^{\infty}$ be a strictly monotonically decreasing sequence in $[0,1]$ converging to $0$, and $\{h_n\}_{n=1}^{\infty}$ be a uniformly convergence sequence of continuous functions on $X$ converging to $h:X\rightarrow [0,1]$. Distinguish a non-empty compact subset $A\subseteq X$. If:

• $h^{-1}[0]=A$,
• $h_n(A)\subseteq [0,\delta_n)$ (for every $n$),

Can we guarantee that $$ I_{[0,\delta_n)}\circ h_n \mbox{ converges point-wise to } I_{\{0\}}\circ h? $$ If not, what additional conditions am I missing?

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It seems that $I_{[0,\delta_n)}\circ h_n(A)=1$ for all $n$; which is good. So the convergence is uniform on $A$. However, I'm most worried about the pointwise convergence on $X-A$...this I can't manage to control...

Answer 1

$\newcommand\de\delta\newcommand\N{\mathbb N}\newcommand\R{\mathbb R}$The answer is yes. Indeed, fix any $x\in X$. We need to show that $$l_n:=I\{0\le h_n(x)<\de_n\}\to r:=I\{h(x)=0\}\tag1$$ as $n\to\infty$. Let $$N:=\{n\in\N\colon0\le h_n(x)<\de_n\}.$$

If $N\ni n\to\infty$, then $l_n=1$, $h_n(x)\to0$, and hence $h(x)=\lim_n h_n(x)=0$, so that $r=1$ and $l_n=1\to1=r$, i.e., (1) holds.

If $N\not\ni n\to\infty$, then $l_n=0$ and, by the condition $h_n(A)\subseteq [0,\de_n)$, we have $x\notin A$, so that, by the condition $h^{-1}(\{0\})=A$, we have $h(x)\ne0$ and hence $r=0$, so that $l_n=0\to0=r$, i.e., (1) again holds.