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.