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...



Conditions for pointwise convergence of indicators precomposed with uniformly continuous sequence
$\newcommand\de\delta\newcommand\N{\mathbb N}\newcommand\R{\mathbb R}$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.


