Representations of a C*-algebra of bounded Borel functions 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?

 A: 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.
