The key property here is that for probability measures:
$$\int_\Omega f\mathrm{d}\rho\in\overline{\langle f(\Omega)\rangle}$$
As the spectrum is closed it holds:
$$\int_{\sigma(N)}z\mathrm{d}\nu_{\hat{\varphi}}\in\langle\sigma(N)\rangle$$
(For a unit vector it is a probability measure.)
Concluding that the other inclusion holds, too:
$$\overline{\mathcal{W}(N)}\subseteq\langle\sigma(N)\rangle$$
By definition, for bounded $f\geq0$ we have
$$
\int_{\sigma(N)}f\,d\nu_\hat\varphi=\sup\left\{\sum_j \alpha_j\,\nu_\hat\varphi(\Delta_j):\ \bigcup_j\Delta_j=\sigma(N), \text{ and }\sum_j \alpha_j\,\chi_{\Delta_j}\leq f\right\}.
$$
So we can choose a sequence $\{s_n\}$ of simple functions with $s_n\nearrow f$ uniformly (this is achievable because $f$ is bounded). By choosing a subsequence if necessary, we may assume $f-s_n<2^{-n}$, say. If $s_n=\sum_j\alpha_j\,\chi_{\Delta_j}$, then $f(t)-\alpha_j<2^{-n}$ a.e. on $\Delta_j$. So if we choose one such a $t$ for each $\Delta_j$, we have that $f-\sum_jf(t_j)\,\chi_{\Delta_j}<2^{-n+1}$.
Note that $\sum_j\nu_\hat\varphi(\Delta_j)\,f(t_j)$ is a convex combination of $f(t_1),\ldots,f(t_r)$. So $\int_{\sigma(N)}f\,d\nu_\hat\varphi$ is a limit of convex combinations of points in $f(\sigma(N))$.
For complex $f$, we can write it as $f_1-f_2+i(f_3-f_4)$ with $f_1,\ldots,f_4\geq0$, and we define
$$
\int_{\sigma(N)}f\,d\nu_\hat\varphi=\int_{\sigma(N)}f_1\,d\nu_\hat\varphi-\int_{\sigma(N)}f_2\,d\nu_\hat\varphi+i\left(\int_{\sigma(N)}f_3\,d\nu_\hat\varphi-\int_{\sigma(N)}f_4\,d\nu_\hat\varphi\right).
$$
So
$
\int_{\sigma(N)}z\,d\nu_\hat\varphi(z)$ is a limit of convex combinations of convex combinations $\sum_jt_j\,\nu_\hat\varphi(\Delta_j)$, with $t_j\in\sigma(N)$.