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?


1 Answer 1


Since $\left\{ x\in X:f\left( x\right) <\lambda \right\} =\left\{ \begin{array}{cl} \emptyset, & \lambda\le 0, \\ X, & \lambda>1 \\ \overline{U}_k, & \frac 1{2^k}<\lambda\le \frac1{2^{k-1}} \end{array} \right. $

we have $f$ is a normal function.


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