Show that $f$ is upper semicontinuous

Question

Suppose $X$ is a completely metrizable space. Let $A$ and $B$ be closed subsets of $X$ such that $A \subseteq B.$ Let $f:X \rightarrow \mathbb{R}$ be a non-negative function such that $f(x) = h(x)$ for $x \in A$, $f(x) = g(x)$ for $x \in B \setminus A$ and $f(x)=0$ for $x \in X \setminus B.$

Note that $h$ is upper semicontinuous on $X$ while $g$ is continuous on $B \setminus A.$

$A$ and $B $ are assumed to be countable sets.

Question: Is $f$ an upper semicontinuous function?

DEFINITION: $f$ is upper semicontinuous if for all real number $c$, we have $f^{-1}[c,\infty)$ is closed.

Let $c$ be given. Note that $f^{-1}[c,\infty) =X$ if $c\leq0$ and $f^{-1}[c,\infty)=(h^{-1}[c,\infty) \cap A) \cup g^{-1}[c,\infty)$ if $c >0$.

If $c\leq 0,$ then clearly $f$ is upper semicontinuous as $X$ is closed in itself.

If $c>0,$ then $h^{-1}[c,\infty) \cap A$ is closed in $X$ as $h$ is upper semicontinuous on $X.$

Note that $g$ is continuous on $B \setminus A.$ So $g^{-1}[c,\infty) = D$ is closed in $B \setminus A.$ By definition of subspace topology, there exists a closed set $F$ in $X$ such that $D = F \cap (B \setminus A) = F \cap B \cap A^c.$ However, $A^c$ is open in $X.$ I have no idea how to proceed from here.

UPDATE: Note that $F \cap B$ is closed in $X.$ So $D$ is closed in $A^c.$ So $D^c$ is open in $A^c.$ Therefore, there exists an open set $O$ in $X$ such that $D^c = O \cap A^c.$ Hence, $D = O^c \cup A$ is closed in $X.$

I am not very confident about my last statement because complement should be taken at $A^c.$

Answer 1

It is not True. Take $A=\{0\}$, $B= \{ x |~ 0 \leq x \} $, $h(x) = 1$, and $g(x) = \frac{1}{x} +3$.

Therefore $ f^{-1} [2, + \infty) = (0, +\infty) $ which is not closed!

Answer 2

Let $h(x)=1$, $g(x)=2$, $A=\Bbb Z$, $B=\cup_{k=1}^{+\infty}\{\pm\frac{1}{k}\}\cup A$, $X=\Bbb R$. Then $$ f(x)= \begin{cases} 1 & \text{ if }x\in A,\\ 2 & \text{ if }x\in B\setminus A,\\ 0 & \text{ otherwise}. \end{cases} $$ We have $$ f(0)=1<2=\limsup_{x\to 0}\,f(x), $$ thus, the function is not upper-semicontinuous.