I am playing around with Borel functional calculus to try to understand it, and made the following argument:
Let $T\in B(H)$ (bounded operator on Hilbert space) be normal. Let $f\in C(\sigma(T)) $ (continuous function on the spectrum of T). Let $\mathcal{A}=\text{Alg}_{C^*}(I,T)$ be the C*-algebra generated by $T$ and the identity operator I.
Using the continuous functional calculus we can write $T=\Psi(\text{Id}_{\sigma(T)}) $ for $\Psi$: $C(\sigma(T))\to \mathcal{A}$ an isometric $*$-isomorphism. Further we can give meaning to $f(T)$ for any $f\in C(\sigma(T))$, consistently with how polynomials would be defined, by setting $f(T)=\Psi(f(\text{Id}_{\sigma(T)}))$.
We can expand this to $BB(\sigma(T))$ (the bounded borel functional calculus). Now if we set let $g \in BB(\sigma(T))$ and define $S=\Psi(g)$ since $\Psi$ is now a $*$-homomophism and $$S^*S = \Psi(g)^*\Psi(g)=\Psi(g^*g)=\Psi(gg^*)=SS^*$$
Since a bounded borel function commutes with its complex conjugate. So S is normal. Ergo any Borel function of a normal operator is normal.
I think the conclusion seems a bit strong, so I wonder if there is a flaw in my reasoning? If so, when is it true?
What I wan't to use the property for is proving that $\sigma(S)=g(\sigma(T))$, so if the above is wrong an argument for that would be a nice addition :)
The functional calculus works as stated. However, the spectral mapping result that you want to prove is not true in general. For example, consider the operator $M$ of multiplication by $x$ on $L^{2}[0,1]$. The spectrum of $M$ is $[0,1]$. If you let $g(x)=x$ for $x \in [0,1/2)\cup(1,2,1]$ and $g(1/2)=50$, then $g(M)$ does not have $50$ in its spectrum. On the other hand, if you consider $M$ on $L^{2}_{\mu}[0,1]$ where $\mu$ is Lebesgue measure with an additional atom at $1/2$, then $g(M)$ does have $50$ in its point spectrum. Look into essential range: http://en.wikipedia.org/wiki/Essential_range , and see if you can formulate something in terms of the spectral measure.