# Conditions for pointwise convergence of indicators precomposed with uniformly continuous sequence

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?

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...

$$\newcommand\de\delta\newcommand\N{\mathbb N}\newcommand\R{\mathbb R}$$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.