Edit: I have worked on the problem a bit, and thus have refocussed. The first part has now been asked at MO.
Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra. Let $f\in \mathcal{A}$ self-adjoint with spectrum $\sigma(f)$. The bidual $\mathcal{A}^{**}$ contains the spectral projections of $f$, denoted $\mathbf{1}_{E}(f)$, for Borel $E\subseteq \sigma(f)$. Let $\lambda\in\sigma(f)$, $\varepsilon>0$ and define $$p_\varepsilon(f):=\mathbf{1}_{(\lambda-\varepsilon,\lambda+\varepsilon)}(f).$$
Suppose that $\varphi$ is a state on $\mathcal{A}$, with an extension to a state on $\mathcal{A}^{**}$ denoted $\omega_\varphi$. Suppose further, and this is to be a key assumption, that for all $\varepsilon>0$, $$\omega_\varphi(p_\varepsilon(f))>0\qquad(\star).$$
If the following is defined, it defines a state. But is it well-defined?
Question: Let $g\in\mathcal{A}$. Under the assumption $(\star)$, does the following limit exist:
$$\lim_{\varepsilon\to 0^+}\frac{\omega_\varphi(p_\varepsilon(f)\,g\,p_\varepsilon(f))}{\omega_\varphi(p_\varepsilon(f))}.$$
If this is well-defined, it defines a state $\varphi_\lambda$ on $\mathcal{A}$ with extension to the bidual $\omega_\lambda$. I want to show that: $$\omega_\lambda(\mathbf{1}_{\{\lambda\}}(f))=1.$$ In a naive way, if the $\varphi_\lambda$ is well-defined, this seems easy. Roughly: $$\varphi_\lambda(\mathbf{1}_{\{\lambda\}}(f))=\lim_{\varepsilon\to 0^+}\frac{\omega_\varphi(\mathbf{1}_{\{\lambda\}}(f))}{\omega_\varphi(p_\varepsilon(f))},$$ and then what can be helpful is that facts about Borel functional calculus, and the $\sigma$-weak continuity of $\omega_\lambda$, shows that: $$\lim_{\varepsilon\to 0^+}\omega_\varphi(p_\varepsilon(f))=\omega_\varphi(\mathbf{1}_{\{\lambda\}}(f)).$$ But the problem here is that $\varphi_\lambda$ is a state on $\mathcal{A}$ and that spectral projection is in the bidual. So the natural thing would be to take $(\phi_\alpha)\subset \mathcal{A}$ converging $\sigma$-weakly to $\mathbf{1}_{\{\lambda\}}(f)$, and then write down something like: $$\omega_\lambda(\mathbf{1}_{\{\lambda\}}(f))=\lim_\alpha\left[\lim_{\varepsilon\to 0^+}\frac{\omega_\varphi(p_{\varepsilon}(f)\phi_\alpha p_{\varepsilon}(f))}{\omega_\varphi(p_\varepsilon (f))}\right]$$ If I can exchange these limits, I can use $\sigma$-continuity, and spectral projection stuff to get back to the naive picture: $$\omega_\lambda(\mathbf{1}_{\{\lambda\}}(f)))=\lim_{\varepsilon\to 0^+}\frac{\omega_\varphi(\mathbf{1}_{\{\lambda\}}(f))}{\omega_\varphi(p_\varepsilon(f))}$$
Question: Can these limits be exchanged?