Confusion about the ability to extend a state on a C*-algebra to a normal state on a von Neumann algebra Sorry for the potentially circuitous post, but this is the basic sketch of it. I found a result which seems, at first glance, to be counterintuitive to the point of being wrong. I sketch below the (short) proof of this dubious claim, and then explain my guess for why the result I found isn't necessarily a contradiction. I then ask you, the reader: Is my proof of this claim correct, and if so, is my explanation for why the conclusion makes sense correct?
My claim is as follows.

Claim:Let $\mathfrak{A}$ be a unital C*-algebra, and $\phi$ a faithful state on $\mathfrak{A}$. Then there exists a von Neumann algebra $\mathfrak{M}$ on a Hilbert space $\mathscr{H}$, a normal state $\rho$ on $\mathfrak{M}$, and a faithful representation $\pi$ of $\mathfrak{A}$ in $\mathscr{B}(\mathscr{H})$ such that $\pi(\mathfrak{A})$ is weakly dense in $\mathfrak{M}$, and $\rho \circ \iota = \phi$.


Proof: Let $\pi : \mathfrak{A} \to \mathscr{B}(\mathscr{H})$ be the GNS representation corresponding to $\phi$, where $\mathscr{H}$ is the completion of $\mathfrak{A}$ in the inner product $\left< a , b \right> = \phi \left( b^* a \right)$. Let $\mathfrak{M}$ be the weak closure of $\pi(\mathfrak{A})$ in $\mathscr{B}(\mathscr{H})$, and define $\rho$ on $\mathfrak{M}$ by
$$\rho(x) = \left< x 1 , 1 \right> .$$
Then $\rho(\pi(a)) = \left< \pi(a) 1 , 1 \right> = \left< a, 1 \right> = \phi(1 a) = \phi(a)$. Further, because $\rho$ is of the form $x \mapsto \sum_n \left< x \xi_n, \xi_n \right>$ for a sequence of vectors $(\xi_n)_{n = 1}^\infty$ in $\mathscr{H}$ such that $\sum_{n = 1}^\infty \left< \xi_n, \xi_n \right> = 1$, we conclude that $\rho$ is a normal state.

Now, my reason for concern is that this seems to suggest the clearly absurd idea that every state is normal. But this is false even in the commutative case. Let $\mathfrak{A} = L^\infty([0, 1], \lambda)$, where $\lambda$ is the standard Lebesgue measure on $[0, 1]$. Then we know that not every state on $\mathfrak{A}$ is normal in the vN algebra $\mathfrak{A}$. This seems at first glance like a contradiction, since it would seem that $\mathfrak{M}$ from the previous claim is just $\mathfrak{A}$.
I think the resolution to this problem is that the closure operation I described in my proof of the claim won't necessarily fix every vN algebra. A state $\phi$ on a vN algebra $\mathfrak{A}$ is called normal if for every bounded monotone increasing net $(x_i)_{i \in I}$ of positive elements in $\mathfrak{A}$, we have that $\phi \left( \sup_i x_i \right) = \sup_i \phi(x_i)$. But I know that in some ordered settings (e.g. in the case of Boolean subalgebras), passing to a subset can actually expand the suprema. Here's an "example" to draw out what I need.
Let $\mathcal{A}$ be the family of all closed subsets of $\mathbb{R}$, and let $\mathcal{B}$ be the power set of $\mathbb{R}$. Let $\{ q_n \}_{n \in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$, and define a monotone sequence $(E_n)_{n = 1}^\infty$ in $\mathcal{A}$ by setting $E_n = \{q_1, q_2, \ldots, q_n\}$. Then the supremum of $(E_n)_{n = 1}^\infty$ in $\mathcal{A}$ is $\overline{\bigcup_n E_n} = \overline{\mathbb{Q}} = \mathbb{R}$, but the supremum in $\mathcal{B}$ is exactly $\mathbb{Q}$.
My point here is that the supremum of a sequence isn't just an intrinsic property of that sequence. It's dependent on the ambient space, since if you have more upper bounds for a net, then you might have better upper bounds for that net, meaning that the sup taken in a larger ambient space can end up strictly smaller than the sup taken in a smaller ambient space.
So my hope is that if I have a faithful non-normal state $\phi$ on a vN algebra $\mathfrak{A}$, then the weak closure of $\pi (\mathfrak{A})$ under the GNS representation will be strictly bigger than $\pi(\mathfrak{A})$. This means that if I had a net $(x_i)$ in $\pi ( \mathfrak{A} )$ such that $\phi \left( \sup_{i, \mathfrak{A}} x_i \right) > \sup_i \phi(x_i)$, then $\mathfrak{M} \supseteq \pi(\mathfrak{A})$ is a proper extension of $\pi(\mathfrak{A})$, and $\rho \left( \sup_{i, \pi (\mathfrak{A})} \pi(x_i) \right) > \left( \sup_{i, \mathfrak{M}} \pi(x_i) \right) = \sup_i \rho( \pi(x_i ) )$.
Am I on the right track?
 A: Everything you say is right. I'm just curious where you learnt C$^*$-algebras and von Neumann algebras that this appears to be an issue; I'm also impressed by your insight.
What you have noticed is that a von Neumann algebra can be represented on a Hilbert space in such a way that it is not a von Neumann algebra in the new space. An obvious way to do this is to take a von Neumann algebra $M\subset B(H)$ and an irreducible representation $\pi:M\to B(K)$. Then $\pi(M)$ is sot-dense in $B(K)$; if $M$ is not type I, it cannot be everything.
When you take all faithful states instead of just one and do your construction on a C$^*$-algebra $A$, you get the Universal Representation $\pi_A$, and the sot-closure $\pi_A(A)''$ is the enveloping von Neumann algebra of $A$. Even if you start with a von Neumann algebra $M$ its enveloping von Neumann algebra is strictly bigger, precisely because of the reason you state: every state in $M$ becomes normal in the enveloping von Neumann algebra.
The enveloping von Neumann algebra has two particularly interesting properties:

*

*If $\pi:A\to B(H)$ is any representation, then there exists a projection $P$ in the centre of $\pi_A(A)''$ such that $\pi_A(A)P\simeq\pi(A)$ via a $*$-isomorphism that lifts to the von Neumann algebra level: $\pi_A(A)''P\simeq \pi(A)''$.


*If we consider $\pi_A(A)''$ with the $\sigma$-weak topology, it is isometrically homeomorphic to $A^{**}$ considered with the weak$^*$-topology.
Because of the latter property, the enveloping von Neumann algebra of $A$ is simply called the double dual of $A$.
