Let $X$ be a compact Hausdorff space. Let $B(X)$ be the C*-algebra of bounded Borel measureable functions on $X$ (under the supremum norm). I am curious whether the (say unital) $*$-representations of $B(X)$ are completely classified. One way to get these is from spectral measures on $X$. See IX.1.12 in A Course in Functional Analysis by John B. Conway. I'm not sure whether this is all of them, though. What this all boils down to is a question about the following continuity condition.
Let $\pi : B(X) \to B(H)$ be a unital $*$-representation of $B(X)$ on a Hilbert space $H$. Suppose that $E_1,E_2,\ldots$ is a countable pairwise disjoint collection of Borel measurable subsets of $X$. Let $E = \bigcup_{n=1}^\infty E_n$. Is it necessarily true that $$\pi(\chi_E) = \sum_{n=1}^\infty \pi(\chi_{E_n})$$ with the sum converging in the strong operator topology?
Not all representations of $B(X)$ satisfy the continuity condition mentioned by the OP.
To exhibit an example, let us first notice that $B(X)$ is a unital commutative C*-algebra, and hence it is isomorphic to $C(K)$, where $K$ is its spectrum.
Given any Borel set $E\subseteq X$, we have that the characteristic function $\chi_E$ is an idempotent element in $B(X)$. So, once $\chi_E$ is viewed as an element of $C(K)$ via Gelfand's transform, it is necessarily given by the characteristic function of some clopen subset of $K$, which we will denote by $\bar E$.
To make things more concrete, let us suppose that $X=[0,1]$, although similar counter-examples can be given in any infinite compact space.
Letting $E_n=(0,1/n)$, we have that $$ \chi_{E_n}\chi_{E_{n+1}}=\chi_{E_{n+1}}, $$ so $$ \chi_{\bar E_n}\chi_{\bar E_{n+1}}=\chi_{\bar E_{n+1}}, $$ which means that $\bar E_{n+1}\subseteq \bar E_n$, for every $n$. Since the $\bar E_n$ are nonempty and compact, their intersection is nonempty, so we may choose a point $\omega $ lying in every $\bar E_n$.
We then consider the character $\pi $ of $C(K)$ given by point evaluation at $\omega $, namely $$ \pi :g\in C(K)\mapsto g(\omega )\in \mathbb C, $$ which we view as a *-representation by identifying $\mathbb C$ with the algebra of all operators on a one-dimensional Hilbert space.
Viewing $\pi $ as a representation of $B(X)$ by composing it with the Gelfand transform, we then have that $\pi (\chi_{E_n})=1$, for every $n$.
We next claim that the alleged continuity condition fails for $\pi $, regarding the following pairwise disjoint union of measurable sets: $$ E_1 = \bigcup_{n=1}^\infty E_n\setminus E_{n+1}. $$ Indeed, $$ \pi \left(\chi_{E_n\setminus E_{n+1}}\right) = \pi \left(\chi_{E_n}(1-\chi_{E_{n+1}})\right) = \pi \left(\chi_{E_n}\right) - \pi \left(\chi_{E_{n+1}}\right) = 1 - 1 = 0, $$ while clearly $\pi \left(\chi_{E_1}\right) = 1$.
Combining IX.1.12 and IX.1.13 in Conway's book, one has that every *-representation $\pi $ of $C(X)$ on $H$ extends to a *-representation $\tilde\pi $ of $B(X)$ on $H$, which moreover satisfies the following continuity property: if $\{f_n\}_n$ is a uniformly bounded sequence in $B(X)$, converging pointwise to some $f$, then $\tilde\pi (f_n)$ converges strongly to $\tilde\pi (f)$ (see Lemma 3.5.5 in Sunder's "Functional Analysis - Spectral Theory"). In particular $\tilde\pi $ satisfies the desired continuity condition. The counter-example described therefore shows that not all representations of $B(X)$ arise as the extension of a representation of $C(X)$, as above.