# Function series of normal lower semi-continuous functions

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-continuous if and only if the set is open.

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

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

Then each $$f_{n}$$ is normal lower semi-continuous 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-continuous 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?

Since $$\left\{ x\in X:f\left( x\right) <\lambda \right\} =\left\{ \begin{array}{c} \emptyset, \\ X, \\ \overline{U_{k}}, \end{array} \right. \begin{array}{c} \lambda \leq 0 \\ \lambda >1 \\ \frac{1}{2^{k}}<\lambda \leq \frac{1}{2^{k-1}}% \end{array}%$$
we have $$f$$ is a normal function.