### Introduction to von Neumann algebras, Lecture 3 (some more generalities, projection constructions, commutative von Neumann algebras)

In this lecture we will describe some projection construction in von Neumann algebras, and we will classify commutative von Neumann algebras.

So far (the first two lectures and in this one), the references I used for preparing these notes are Conway (A Course in Operator Theory) Davidson (C*-algebras by Example), Kadison-Ringrose (Fundamentals of the Theory of Operator Algebras, Vol .I), and the notes on Sorin Popa’s homepage. But since I sometimes insist on putting the pieces together in a different order, the reader should be on the look out for mistakes.

#### 1. *-isomorphisms are isometric

We begin by proving an interesting rigidity property of *-homomorphisms from C*-algebras: they are automatically contractive, and if they are injective then they are isometric.

Theorem 1: Let $A$ be a (concrete) C*-algebra possessing a unit, and let $\pi :A \to B(K)$ be a *-homomorphism (meaning that $\pi$ is a linear map that also satisfies $\pi(ab) = \pi(a)\pi(b)$ and $\pi(a^*) = \pi(a)^*$). Then $\|\pi(a)\| \leq \|a||$ for all $a \in A$.  Moreover, if $\pi$ is injective, then $\|\pi(a)\| = \|a\|$ for all $a \in A$

In the above theorem, a C*-algebra $A \subseteq B(H)$ possessing a unit is simply an algebra that has a multiplicative identity element, not necessarily $I_H$. If we want to say that the unit of $A$ is actually equal to $I_H$, then we will say that $A$ is a unital C*-subalgebra of $B(H)$.

For the proof, we require the following lemma, which is sometimes referred to as the spectral permanence theorem. If $A$ is a unital Banach algebra, and $a \in A$, then the spectrum of $a$ relative to $A$ is the set

$\sigma_A(a) = \{\lambda \in \mathbb{C} : a - \lambda 1$ has no inverse in $A\}$.

Thus, the spectrum of an operator $\sigma(a)$ as we defined it in the first lecture, is the spectrum relative to $B(H)$, i.e., $\sigma_{B(H)}(a)$. It is conceivable that if $a$ is an element of a unital C*-algebra $A \subseteq B(H)$, then $\sigma_A(a)$ is bigger than $\sigma_{B(H)}(a)$ (in other words, it is possible, that an element $a$ in $A$ has an inverse $a^{-1}$ in $B(H)$ which is not contained in $A$). One of the remarkable properties of C*-algebras is that this does not happen.

Lemma: If $a$ is an operator in a unital C*-subalgebra $A \subseteq B(H)$, then $a$ is an invertible operator if and only if $a$ has an inverse $a^{-1} \in A$. Consequently, for every unital C*-subalgebras $A \subseteq B \subseteq B(H)$

$\sigma_A(a) = \sigma_B(a)$.

Proof of the lemma: Consider first the case of a selfadjoint operator $a$. Then by the spectral theorem, we may as well assume that $a$ is a multiplication operator $M_f \in B(L^2(X, \mu))$. A moment of thought reveals that if $M_f$ is a bounded invertible operator, then the inverse $(M_f)^{-1}$ must be equal to $M_{f^{-1}}$. Since by the bounded inverse theorem $(M_f)^{-1}$ is bounded, we find that there is some $\epsilon > 0$ such that $|f| \geq \epsilon$ almost everywhere. This implies that $\sigma(M_f) \subseteq [-\|f\|_\infty, -\epsilon] \cup [\epsilon, \|f\|_\infty]$ Now, the function $\phi(x) = \frac{1}{x}$ is continuous on $\sigma(M_f)$, and by the continuous functional calculus we have that $M_{f^{-1}} = M_{\phi\circ f} = \phi(M_f)$ is contained in the C*-algebra generated by $M_f$ and $1$.

If $a$ is a general invertible element in $B(H)$, then $a^* a$ is also invertible, contained in $C^*(a)$, and is selfadjoint. Thus, by the previous paragraph, $(a^* a)^{-1} \in C^*(a)$, so $a^{-1} = (a^*a)^{-1} a^* = a^{-1} (a^*)^{-1} a^*$ is in $C^*(a)$.

Finally, the assertion regarding spectrum follows immediately from the assertion about invertible operators.

Exercise A: Give an example of an operator $a$ and a unital norm closed operator subalgebra $A \subseteq B(H)$ containing $a$ such that

$\sigma_A(a) \neq \sigma(a)$.

Proof of Theorem 1: We may assume that $A$ is a unital C*-subalgebra of $B(H)$ and that $\pi(I_H) = I_K$ (because otherwise, $1_A$ and $\pi(1_A)$ are projections, and everything orthogonal to the ranges of these projections is irrelevant). For every $a \in A$, if $a$ is an invertible operator, then it is invertible in $A$, and it follows that $\pi(a)$ is invertible in $\pi(A)$ (with inverse $\pi(a^{-1})$). Thus, $\sigma(\pi(a)) \subseteq \sigma(a)$. Then, for every $a \in A$,

$\|\pi(a)\|^2 = \|\pi(a^*a)\| = \sup\{|\lambda| : \lambda \in \sigma(\pi(a^*a))\}$

but – as $\sigma(\pi(a^*a)) \subseteq \sigma(a^*a)$ – the right hand side is less than $\sup\{|\lambda| : \lambda \in \sigma(a^*a) \} = \|a^*a\| = \|a\|^2$. We find that $\|\pi(a)\| \leq \|a\|$ for every $a \in A$.

In particular, we find that every *-homomorphism is continuous.

Now, suppose that $\pi$ is injective, and assume for contradiction that $\pi$ is not isometric. Then there exists some $a \in A_+$ such that $\|a\| = 1$ and $\|\pi(a)\| < 1$ (the fact that we can assume that there is a positive element on which norm preservation fails follows from the C* identity). Let $f \in C_\mathbb{R}([0,1])$ be such that $f \equiv 0$ on $[0,\|\pi(a)\|]$, and $f(1) = 1$. Since $p(\pi(a)) = \pi(p(a))$ for every polynomial, and since $\pi$ is continuous, we have that $f(\pi(a)) = \pi(f(a))$. By the continuous functional calculus, $f(\pi(a)) = 0$, so $\pi(f(a)) = 0$, but this is a contradiction to injectivity, because $f(a) \neq 0$.

Remark: In these notes we are working with concrete C*-algebras. We have reached a point that nicely exemplifies the deficiency in this concrete approach. If we were using abstract C*-algebras, we would easily be able to show that the image of a C*-algebra under a *-homomorphism is a C*-algebra (the key issue is that it is closed). This is done by noting that, since a *-homomorphism is continuous, its kernel $\ker \pi$ is a closed ideal. Therefore, we would be able to take the quotient $A/ \ker \pi$, and we get an injective *-homomorphism $\dot{\pi} : A / \ker \pi \to B(K)$ which must be isometric, hence its image is closed. But the image of $\dot{\pi}$ is equal to the image of $\pi$, so the image of $\pi$is a C*-algebra. Thus, we see that the abstract approach has significant advantages, even at the early parts of the theory.

#### 2. Some constructions involving projections

Definition 2: For every $T \in B(H)$, the projection onto $\overline{Im T}$ is called the range projection of $T$ (also called the left support of $T$). The range projection of $T$ is denoted $R(T)$. Our first goal is to show that von Neumann algebras are closed under the operation of “passing to the range projection”. In solving the exercise below (as well as some following exercises), Theorem 11 from Lecture 2 will be useful.

Exercise B: Prove that if $A$ is a von Neumann algebra and $T \in A$, then $R(T) \in A$. (Hint: First, assume that $T \geq 0$, and without loss of generality assume $\|T \|\leq 1$, and consider the monotone sequence $T^{1/n}$ (note that having the functional calculus always on your mind, this is a natural sequence to consider). To see the result for general operators, find the relationship between $R(T)$ and $R(TT^*)$.)

If $\{P_\alpha\}$ is a family of projections on $H$, we define $\vee_\alpha P_\alpha$ to be the projection on the closed subspace spanned by the subspaces $P_\alpha H$, and we define $\wedge_\alpha P_\alpha$ to be the projection onto the intersection $\cap_\alpha P_\alpha$. The projections $\vee P_\alpha$ and $\wedge P_\alpha$ are called the sum and intersection of the family. Sometimes, one writes $\inf P_\alpha$ for $\wedge P_\alpha$ and $\sup P_\alpha$ for $\vee P_\alpha$.

Proposition 3: If all the projections in the family $\{P_\alpha\}$ are contained in a von Neumann algebra $A$, then the sum and intersection are also in $A$

Proof: It is enough to prove the claim for the sum (why?). Recall that a projection $p$ belongs to $A'$ if and only if the range of $p$ is invariant for $A$ (see the lemma used in the proof of the double commutant theorem, in Lecture 2). Reversing the role of $A$ and $A'$, we see that if every $P_\alpha$ is in $A$, then every space $P_\alpha H$ is invariant for $A'$. It follows that $\vee_\alpha P_\alpha H$ is invariant for $A'$, thus $\vee_\alpha P_\alpha \in A'' = A$.

The following two exercises give an alternative way of proving the above proposition.

Exercise C: If $P,Q$ are projections, then $P \vee Q = R(P+Q)$ and $P \wedge Q = P - R(P(I-Q))$ ( if this is difficult, consult Kadison-Ringrose, Vol. I, Section 2.5).

Exercise D: Prove that if $A$ is a von Neumann algebra, and if $\{P_\alpha\} \subseteq A$, then $\vee_\alpha P_\alpha \in A$ and $\wedge_\alpha P_\alpha \in A$, by making use of the previous exercise, and applying the theorem on monotone nets of operators.

#### 3. The center of a von Neumann algebra, factors

Definition 4: The center of a von Neumann algebra $A \subseteq B(H)$ is defined to be $Z(A) = A \cap A'$. A projection $p \in A$ is said to be a central projection if it is contained in the center of $A$.

The center of a von Neumann algebra is an commutative von Neumann algebra. Note that the although the commutant of a von Neumann algebra depends on the particular representation of the von Neumann algebra and not on the *-algebraic structure, the center depends only on the *-algebraic structure.

Definition 5: A von Neumann algebra $A \subseteq B(H)$ is said to be a factor if $Z(A) = \mathbb{C} I$.

Clearly, a von Neumann algebra is a factor if and only it has no central projections.

Examples:

1. The algebra $B(H)$ is a factor (we have essentially seen already that $B(H)' = \mathbb{C}I$ so $Z(B(H)) = \mathbb{C} I$). At the other extreme, every commutative von Neumann algebra is its own center. The only commutative von Neumann algebra that is a factor is $\mathbb{C}$.
2. We will see in a later lecture that if $G$ is a countable group and if for every $g \neq e$, the conjugacy class $C(g) := \{xgx^{-1} : x \in G\}$ is infinite, then the group von Neumann algebra $L(G)$ is a factor. (A group for which $|C(g)| = \infty$ for every $g \neq e$ is called an ICC group; examples of countable ICC groups are the free groups $F_n$, and the group $S_\infty$ consisting of all permutations of the natural numbers that fix all but finitely many elements.)

In a way that can be made precise (but will probably not be made precise in this course) factors form the “building blocks” of von Neumann algebras. We will now see that $A$ is a factor if and only if it has no weakly closed ideals, so that the factor are in a sense the “simple” von Neumann algebras.

If $c \in Z(A)$, then it is easy to see that $cA = cAc$ is a *-subalgebra of $A$, and in fact, it is WOT closed. To see that it is WOT closed, note that every element $ca = cac$ in $cA$ can be identified with the operator $ca\big|_{cH} \subseteq B(cH)$. The map $a \mapsto ca$ can therefore be considered as a (contractive) *-homomorphism of $A$ onto $cAc \subseteq B(cH)$, which maps $A_1$ onto $(cAc)_1$ (onto, because this map is “the identity” on $cA$). Since this map is WOT continuous and $A_1$ is WOT compact, we obtain that $(cA)_1$ is WOT compact. Invoking Corollary 15 of the previous lecture, we see that $cA$ can be considered to be a von Neumann algebra in $B(cH)$. It follows that $cA$ is a WOT closed, *-closed, two sided ideal in $cA$. Note that in this case we have the decomposition $A = cA \oplus (1-c)A$, i.e.,

$A = \begin{pmatrix}cA & \\ & (1-c)A \end{pmatrix}$.

It turns out that all WOT closed (two sided) ideals have this form.

Theorem 6: Let $A$ be a von Neumann algebra, and let $I \subseteq A$ be a WOT closed two sided ideal. Then $I^* = I$, and, moreover, there exists a central projection $c \in Z(A)$ such that $I = cA$.

Exercise E: Prove Theorem 4. (Hint: To prove that $I$ is selfadjoint, use the the polar decomposition. To prove the existence of the form $I = cA$, note that the projection $c$, if it exists, must be equal to the sup of all projections in $I$, thus one can define $c$ as the supremum and prove that such a supremum does what we want. To show that there are sufficiently many projections in $I$ use the Borel functional calculus. The notion of range projection (see the next section) will also be useful for proving that $I = cA$.)

Corollary 7: A von Neumann algebra $A$ is a factor if and only if $A$ has no non-trivial WOT closed ideals.

Proof: Indeed, $A$ is a factor precisely when $Z(A)$ has no non-trivial projections, and this corresponds to the situation where there are no non-trivial WOT closed ideals.

Definition 8: Let $A$ be a von Neumann algebra. For every operator $T \in B(H)$, the central cover of $T$ (also called the central support or central carrier) is the projection

$C_T := \inf \{P \in Z(A) : TC = T\}$.

Exercise F: Suppose that $P$ is a projection in a von Neumann algebra $A \subseteq B(H)$. Prove that $C_P$ is the orthogonal projection onto the subspace

$[A P H] := \{ah : a \in A , h \in PH\}$.

#### 4. More projection constructs

Suppose that $A \subseteq B(H)$ is a von Neumann algebra, an let $p$ be a projection. We can define a set $A_p \subseteq B(pH)$ by

$A_p = \{pa \big|_{pH} : a \in A\}$.

If $p$ is either in $A$ or in $A'$, then $A_p$ is a *-algebra, called the reduced or the induced von Neumann algebra, respectively. In the case $p \in A$, then $A_p$ can be identified with the *-subalgebra $pAp \subseteq A$. In the case that $p \in A'$, then the compression map $a \mapsto pa\big|_{pH}$ is a *-homomorphism (make sure you understand why), and can be identified with the map $a \mapsto ap$ or $a \mapsto a\big|_{pH}$.

Proposition 9: Let $A \subseteq B(H)$ be a von Neumann algebra, and let $p \in A'$. Then $A_p$ and $(A')_p$ are both von Neumann algebras on $pH$, and are mutual commutants: $(A_p)' = (A')_p$

Proof: To prove that that $(A')_p$ is a von Neumann algebra, one runs an arguent similar to the case where $p \in Z(A)$, which we treated above. We will show that $A_p$ is a von Neumann algebra (this will also follow from $(A_p)' = (A')_p$, but the proof is interesting in itself).

To see that $A_p$ is a von Neumann on $pH$, we consider the map $\varphi : a \mapsto ap$. $\varphi$ is a WOT continuous *-homomorphism. Its kernel $\ker \varphi$ is therefore a WOT closed ideal, hence, by Theorem 4, $\ker \varphi = cA$ for some $c \in Z(A)$. On $(1-c)A$ the map $\varphi$ must be injective. Therefore, it is isometric (Theorem 1). So $\varphi(A_1) = (A_p)_1$, so the latter – as the WOT continuous image of a WOT compact set – is WOT compact, and therefore – by Corollary 15 in the previous lecture – $A_p$ is a von Neumann algebra.

Let $b \in A'$. Then for all $a \in A$,

$pb\big|_{pH} p a \big|_{pH} = p b a \big|_{pH} = pa b \big|_{pH} =p a \big|_{pH} p b \big|_{pH}$,

and this shows that $(A')_p \subseteq (A_p)'$.

For the reverse inclusion (which we will prove as $((A')_p)' \subseteq (A_p)'' = A_p$, using the fact that we already established that $A_p$ is a von Neumann algebra), it suffices to show that every unitary $u \in ((A')_p)'$ is the compression $vp$ of some $v \in A$ (recall Exercise D in the first lecture, which shows that a C*-algebra is spanned by its unitary elements). For such a $u$, we define an operator $v : [A' pH] \to [A'pH]$ by

$v b g = b u g$,

for $b \in A'$ and $g \in pH$, and extending linearly. We see that

$\langle b_1 u g_1, b_2 u g_2 \rangle = \langle u^* p b_2^* b_1 pu g_1, g_2 \rangle =$

$= \langle p b_2^* b_1 u^* u g_1, g_2 \rangle = \langle b_1 g_1 , b_2 g_2 \rangle$,

for $b_1, b_2 \in A'$, and $g_1, g_2 \in pH$.

So, $v$ is an isometry on $[A' pH]$. We extend $v$ to be a partial isometry on $H$ by defining $v \equiv 0$ on $[A' pH]^\perp$. (The partial isometry $v$ satisfies that $v^*v$ is the orthogonal projection onto $[A'pH]$, that is, it is $c = C_p$ (computed in $A'$)). For every $b \in A'$ we have (1) for all $g \in pH$ and $b' \in A'$

$b v b' g = b b' u g = v b b' g$

and (2) for every $h \perp [A'pH]$ we have $bh \perp [A'pH]$, so

$b v h = 0 = v b h$.

Therefore, $vb = bv$, so $v \in A'' = A$. But $u = vp \in A_p$, and the proof is complete.

Exercise G: Prove that also in the case that $p \in A$, it also holds that $A_p$ is a von Neumann algebra and that $(A_p)' = (A')_p$. Moreover, show that if $p \in A$ or $p \in A'$, then $Z(A_p) = Z(A)_p$.

#### 5. Every countably generated commutative von Neumann algebra is singly generated

Theorem 10: Let $H$ be a separable Hilbert space and $A \subseteq B(H)$ a commutative von Neumann algebra. Then there is a selfadjoint operator $a \in A$ such that $A = W^*(a)$

Proof: By Corollary 15 in Lecture 2, the unit ball of $A$ is WOT compact, and by Exercise E in that lecture, it is metrizable. Thus, $A_1$ is separable (as every compact metric space is). Since every selfadjoint operator can be approximated by its spectral projections associated with intervals with rational endpoints, we see that there is a sequence $\{e_n\}_{n=1}^\infty$ that generates $A$ as a von Neumann algebra.

We now claim that the operator $a = \sum_{n=1}^\infty 3^{-n} e_n$ generates $A$ as a von Neumann algebra. To establish this, it is enough to show that $e_n \in C^*(a)$ for all $n$. And really, it suffices to concentrate on showing that $e_1 \in C^*(a)$, because if that’s true then $b:= a - \frac{1}{3} e_1 = \sum_{n\geq 2} 3^{-n} e_n$ will be in $C^*(a)$ and one can proceed inductively.

We have the direct sum $a = e_1 a \oplus (1-e_1)a$ (recall that $A$ is a commutative algebra). Let $b = \sum_{n\geq 2} 3^{-n} e_n$ as above. Since $0 \leq b \leq \frac{1}{6}$,

$0 \leq (1-e_1)a = (1-e_1)b (1-e_1) \leq \frac{1}{6}$.

It follows that $\sigma((1-e_1)a) \subseteq [0,\frac{1}{6}]$. Likewise, $e_1 a = \frac{1}{3}e_1 + e_1 b e_1$, so $\frac{1}{3}e_1 \leq e_1 a \leq \frac{1}{2}e_1$, therefore $\sigma(e_1 a) \subseteq \{0\} \cup [\frac{1}{3},\frac{1}{2}]$.

We see that $\sigma(a) \subseteq [0,\frac{1}{6}] \cup [\frac{1}{3}, \frac{1}{2}]$. Therefore, $f = \chi_{[\frac{1}{3}, \frac{1}{2}]}$ is continuous on $\sigma(a)$, and $e_1 = f(a) \in C^*(a)$, as required.

#### 6. Separating vectors and cyclic vectors

Definition 11: Let $A \subseteq B(H)$ be a *-algebra. A vector $h \in H$ is said to be cyclic if $[Ah] = H$. It is said to be separating for $A$ if for all $a \in A$, $ah = 0$ implies that $a = 0$.

Lemma 12: If $h \in H$ is cyclic for a *-algebra $A$, then $h$ is separating for $A'$. The converse is also true when $A$ is non-degenerate.

Proof: Exercise (easy).

Proposition 13: Every commutative von Neumann algebra on a separable Hilbert space has a separating vector.

Proof: We will prove that for every von Neumann algebra $A \subseteq B(H)$ $A'$ has a cyclic vector; the result then follows from the lemma above.

As in Exercise I of Lecture 1, we may write $H = \oplus_{n} H_{n}$ as a direct sum of “cyclic subspaces”, where $H_n = [A' h_n]$ for some $h_n \in H$. Since every $H_n$ is invariant for $A'$, the projection $p_n$ onto $H_n$ belongs to $A'' = A \subseteq A'$ (the inclusion following from commutativity). It follows that the vector $\sum 2^{-n}h_n$ is cyclic for $A'$. Indeed, for every $n$, $h_n = 2^n p_n h \in A' h$, so $H_n \subseteq [A' h]$, and this completes the proof.

#### 7. Classification of commutative von Neumann algebra

We now describe what commutative von Neumann algebras “look like”, and this will lead to a classification of all commutative von Neumann algebras.

Recall that the support of a Borel measure $\mu$ on $\mathbb{R}$, denoted $supp(\mu)$, is the closure of the set of all points $x \in \mathbb{R}$, that satisfy

for every neighborhood $U \ni x$, $\mu(U) >0$.

A measure is said to be compactly supported if its support is compact.

Theorem 14: Let $A$ be a commutative von Neumann algebra on a separable Hilbert space. Then there exists a regular, compactly supported, Borel probablity measure $\mu$ on $\mathbb{R}$, such that $A$ is *-isomorphic to $L^\infty(\mu)$

Proof: Let $h$ be a separating vector for $A$, given by Proposition 13. Let $M = [Ah]$, and $p = P_M$. Then $p \in A'$, so $b \mapsto pb\big|_{M}$ is a *-homomorphism from $A$ onto $A_p$. In fact, it is a *-isomorphism, because $h$ is a separating vector. Will show that $A_p$ is unitarily equivalent to $L^\infty(\mu)$ for some $\mu$.

Let $a \in A_{sa}$ be an operator – the existence of which is guaranteed by Theorem 10 – for which $A = W^*(a)$. As the map $b \mapsto pb\big|_{M}$ is a WOT continuous *-isomorphism, we have that $A_p$ is generated as a von Neumann algebra by $a_p:= pa \big|_{M}$. The vector $h$ is cyclic for $a_p$, so by a previous result (see Exercise I in Lecture 2) we have that $A_p$ is unitarily to $L^\infty(\mu)$, where $\mu$ is as we require. That completes the proof.

Now it is our goal to understand what kind of C*-algebras arise as $L^\infty(\mu)$ for a regular, compactly supported, Borel probablity measure $\mu$ on the real line.

Given such a measure, let $E = \{x \in \mathbb{R} : \mu(\{x\}) \neq 0\}$. Then the cardinality of $E$ is at most $\aleph_0$. Define $\mu_d$ by

$\mu_d (F) = \mu(F \cap E)$

and $\mu_c$ by

$\mu_c(F) = \mu(F \cap E^c)$.

Generally, a measure $\nu$ is said to be discrete if $\nu = \sum_i c_i \delta_{x_i}$, and continuous if $\nu(\{x\}) = 0$ for every point $x$ in the space. The measures $\mu_d$ and $\mu_c$ are then seen to be discrete and continuous, respectively. We have the decomposition $\mu = \mu_c + \mu_d$. The measures $\mu_d$ and $\mu_c$ are called the discrete and continuous parts of $\mu$, and the decomposition above is unique up to sets of measure zero. Both measures are also regular Borel measures if $\mu$ is.

Exercise H: Prove that $L^\infty(\mu)$ is *-isomorphic to $L^\infty(\mu_c) \oplus L^\infty(\mu_d)$.

Exercise I: Prove that if $\mu$ is a discrete probability measure, then there exists a countable set $\Omega$ such that $L^\infty(\mu)$ is *-isomorphic to $\ell^\infty(\Omega)$.

It remains to understand $L^\infty(\mu)$ for finite continuous measures. It turns out that the only von Neumann algebra that arises this way, is $L^\infty[0,1]$ (where the measure is the Lebesgue measure).

Theorem 15: Let $\mu$ be a compactly supported and continuous regular Borel probability measure on the real line. Then $L^\infty(\mu)$ is *-isomorphic to $L^\infty[0,1]$

Proof: Assume that the support of $\mu$ is contained in $[a,b]$, and assume also that $a,b \in supp(\mu)$. Thus, we can think of $\mu$ as a measure on $[a,b]$. It will be convenient also to set $X:= supp(\mu)$, and to think of $\mu$ as a Borel probability measure on $X$. It is straightforward to show that $L^\infty([a,b],\mu) = L^\infty(X,\mu)$, and we switch back and forth between the two viewpoints as convenient.

We will construct a map $\phi : X \to [0,1]$ such that $f \mapsto f \circ \phi$ is a *-isomorphism of $L^\infty[0,1]$ onto $L^\infty(X,\mu)$.

The map $\phi$ is defined by

$\phi(t) = \mu([a,t])$.

(In order for this to have meaning it is convenient to think of $\mu$ as a measure on $[a,b]$. In fact we can also think of $\phi$ as a map defined $[a,b] \to [0,1]$). Then $\phi$ is a non-decreasing function, so it is Borel measureable. In fact, the map $\phi$ is strictly increasing on $X$, with the exception of a countable number of pairs $x_i < y_i$ for which $\phi(x_i) = \phi(y_i)$ (there at most $\aleph_0$ such pairs, because for all i, the interval $(x_i,y_i)$ is disjoint from $X$ but contained in $[a,b]$, thus the total length of these intervals is finite). Moreover, $\phi$ is onto, because $\mu$ is continuous. We can therefore invert $\phi$, missing at most countably many points in $X$. The map $\phi^{-1}$ is also increasing, so it Borel measurable.

Now, $\phi$ is measure preserving. By this, we mean that if $E \subseteq [0,1]$, then $\mu(\phi^{-1}(E))$ is equal to the Lebesgue measure of $E$. Since both measures are regular, and since $\phi$ maps intervals to intervals, it is enough to consider the case of $E = [0,t]$. Let $x \in X$ such that $\phi(x) = t$. Then

$\mu (\phi^{-1}([0,t])) = \mu ([a,x] \cap X) = \mu([a,x]) = \phi(x) = t$,

but the latter is the Lebesgue measure of $[0,t]$, as required. The converse is also true, so $\phi$ preserves the measure in both directions.

Now one has to show that the map $f \mapsto f \circ \phi$ is a well defined *-isomorphism of $L^\infty[0,1]$ onto $L^\infty(X,\mu)$. The final details are left to the reader.

Conclusion: Every commutative von Neumann algebra on a separable Hilbert space has one of the following forms (up to *-isomorphism):

1. $L^\infty[0,1]$,
2. $\ell^\infty(\Omega)$, for a countable set $\Omega$,
3. $L^\infty[0,1] \oplus \ell^\infty(\Omega)$, for a countable set $\Omega$.