Every normal operator is a continuous function of a Hermitian operator

Question

For normal operator $N$, prove that there is a hermitian operator $A$ and continuous function $f$ such that $N=f(A)$

I don't know how to start, just I know For normal operator $N$ there is a unique spectral measure $E$ such that $N=\int\lambda\,dE(\lambda)$

Thanks in advance