Here is an image of the proposition:
An highlighted below is the part of the proof I'm having trouble with.
Why is the equation true for every n? $f_n(T) = \int_{\sigma(T)} f_n \: dE$, but I don't see how if you take some $x \in \ker(\lambda I-f(T))$ that then $f_n(T)x=0$. I tried using the fact that: $$\lVert f_n(T)x \rVert^2 = \int_{\sigma(T)} f^2_n \: d\mu_{x,x}$$ but I'm failing to see how this integral is $0$, given that all I know about it is that $\mu_{x,x}(Y) = \langle E(Y)x,x \rangle$ on measurable subsets of the spectrum.
Note that $f_n = g_n(\lambda - f)$ where $g_n(t) = (|t| \wedge 1)^{1/n}$. $g_n$ is a continuous function with $g_n(0) = 0$, so it can be uniformly approximated by a sequence of polynomials $(g_{nm})_{m \in \mathbb{N}}$ with zero constant term. Clearly, when $x \in \mathrm{ker}(\lambda I - f(T))$, $[g_{nm}(\lambda I - f(T))](x) = 0$, since $g_{nm}$ is a polynomial with no constant term and you can directly evaluate $g_{nm}(\lambda I - f(T))$ to see that this is the case. As $g_{nm} \to g_n$ uniformly, we see that $g_{nm}(\lambda I - f(T)) \to g_n(\lambda I - f(T)) = f_n(T)$, so $[f_n(T)](x) = 0$.