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.