For a real-valued $f$ on a topological space $X$, the upper limit of $f$ at $x\in X$ is
defined as follows:
$
f^{\ast }\left( x\right) =\inf \left\{ \sup \left\{ f\left( y\right) :y\in
U\right\} :U\in \mathcal{N}\left( x\right) \right\} 
$,
where $\mathcal{N}\left( x\right) $ is the neighborhood system at $x$. The
lower limit of $f$ at $x$ is defined dually and denoted by $f_{\ast }$. The $%
f^{\ast }$ and $f_{\ast }$ are extended real-valued functions on $X$ are
respectively upper semi-continuous and lower semi-continuous.

A real-valued function is called normal lower semi-continuous if $\left(
f^{\ast }\right) _{\ast }=f$ at each point of $X$.

We know the next theorem

**Theorem:** An lower semi-contionuous function $f$ on $X$ is normal iff for
each real number $\lambda $, $\left\{ x\in X:f\left( x\right) <\lambda
\right\} $ is a union of regular closed sets. (A set equals the closure of
its interior is called regular closed)

It well known that the characteristic function of a set is lower
semi-continous if and only if the set is open.

Suppose that there exists a decreasing sequence $\left\{
U_{i}\right\} $ of open sets. Define

$f_{n}\left( x\right) =\left\{ 
\begin{array}{c}
1, \\ 
0,
\end{array}%
\right. 
\begin{array}{c}
x\in X\backslash cl\left( U_{i}\right)  \\ 
x\in cl\left( U_{i}\right) 
\end{array}%
$

Then each $f_{n}$ is normal lower semi-contionuous function. Set $f\left(
x\right) =\sum\limits_{n\geq 1}2^{-n}f_{n}\left( x\right) $. Then $f$ is
also normal semi-contionus function.

Being semi-contionus function of $f$ is clear. But I can't prove to be
normal of $f$.

It is clear that $\left\{ x\in X:f\left( x\right) >\lambda \right\}
=\bigcup\limits_{n\geq 1}\left\{ x\in X:g_{n}\left( x\right) >\lambda
\right\} $, where $g_{n}\left( x\right) =2^{-1}f_{1}\left( x\right)
+2^{-2}f_{2}\left( x\right) +...+2^{-n}f_{n}\left( x\right) $.

Moreover, we write $$\left\{ x\in X:f\left( x\right) <\lambda \right\} =\bigcup _{\lambda ^{\prime
}<\lambda }\left\{ x\in X:f\left( x\right) \leq \lambda ^{\prime }\right\} $$

Since $$\left\{ x\in X:f\left( x\right) \leq \lambda ^{\prime }\right\}
=\bigcap _{n\geq 1}\left\{ x\in X:g_{n}\left( x\right) \leq \lambda ^{\prime
}\right\} $$, we have
$$
\left\{ x\in X:f\left( x\right) <\lambda \right\} =\bigcup _{\lambda ^{\prime
}<\lambda }\bigcap _{n\geq 1}\left\{ x\in X:g_{n}\left( x\right) \leq \lambda
^{\prime }\right\} 
$$
Therefore, I need to show that $\bigcap _{n\geq 1}\left\{ x\in X:g_{n}\left(
x\right) \leq \lambda ^{\prime }\right\} $ is regular closed set or a union
of regular closed sets.

How can I prove this?