Let $\mathcal{A}$ be unital $\mathrm{C}^*$-algebra, and $\mathcal{A}^{**}$ its bidual.
Let $a\in \mathcal{A}_{\text{s.a.}}$ be self-adjoint with spectrum $\sigma(a)\subset \mathbb{R}$. For each Borel set $E\subseteq \sigma(f)$, let $\mathbf{1}_E(a)\in \mathcal{A}^{**}$ be the associated spectral projection. Denote the set of Borel sets of $\Omega\subset \mathbb{R}$ by $\mathcal{B}(\Omega)$. Denote by $\mathcal{P}\subseteq \mathcal{A}^{**}$ the set of spectral projections: $$\mathcal{P}:=\{\mathbf{1}_E(a):a\in \mathcal{A}_{\text{s.a.}},\, E\in \mathcal{B}(\sigma(a))\}.$$
Let $\mathcal{Q}$ be the set of projections in $\mathcal{A}^{**}$.
Do we have $\mathcal{Q}=\mathcal{P}$? That is, is every projection $p\in \mathcal{A}^{**}$ of the form $\mathbf{1}_E(a)$ for some self-adjoint $a\in\mathcal{A}_{\text{s.a.}}$ and Borel set $E\subset\sigma(a)$?
Let $\pi$ be a faithful representation on $\mathcal A$.
The universality of the universal representation consists in that there exists a normal representation $\tilde\pi$ on $\mathcal A^{**}$ such that $\tilde\pi$ extends $\pi$ and $$\tilde\pi(\mathcal A^{**})''=\pi(\mathcal A)''.$$ Then there exists a central projection $p$ such that $$ p\mathcal A^{**}=\ker\tilde\pi. $$ Suppose that $p=1_E(a)$. Let $K$ be compact with $K\subset E$ and $m(E\setminus K)<\varepsilon$. Since $E\setminus K$ is Borel, we can find a second compact set $K'$ with $K\subsetneq K'\subset E$. Applying Urysohn's Lemma to $K$ and $\mathbb C\setminus K'$, we can find $g$ continuous with $g|_K=1$ and $g|_{E\setminus K'}=0$. We have $$ \tilde\pi(1_K(a))\leq \tilde\pi(g(a))\leq \tilde\pi(1_E(a))=\tilde\pi(p)=0. $$ Then, since $g(a)\in\mathcal A$, $$ \pi(g(a))=\tilde\pi(g(a))=0. $$ As $\pi$ is faithful, $g(a)=0$, and then $1_K(a)\leq g(a)=0$. If we do this for an increasing sequence $K_1\subset K_2\subset\cdots\subset E$ with $m(E\setminus K_n)<\frac1n$, we get (the limits are wot, using monotone convergence on the scalar spectral measures $\langle E(\Delta)\xi,\xi\rangle$) $$ p=1_E(a)=\int_E1\,dE_a(\lambda)=\lim_n\int_{K_n}1\,dE_a(\lambda)=\lim_n1_{K_n}(a)=0. $$ This would imply that $\tilde\pi$ is faithful and thus $\pi$ is the universal representation of $\mathcal A$. But with most (all?) simple infinite-dimensional C$^*$-algebras, it is possible have non-equivalent faithful representations, and they cannot be both the universal representation.