Fuzzy Observables: from Weak Markov Kernels to Markov Kernels

Abstract

We provide a proof based on transfinite induction that every weak Markov kernel is equivalent to a Markov kernel. We only assume the space where the weak Markov kernel is defined to be second countable and metrizable. That generalizes some previous results where the kernel is required to be defined on a standard Borel space (which is second countable and completely metrizable) and the framework is the theory of stochastic operators. This property of weak Markov kernels is at the root of the characterization of a commutative POVM as the fuzzification of a spectral measure through a Markov kernel. As a result, the characterization of commutative POVMs is also generalized. We then revisit the relationships between weak Markov kernels, Markov kernels, commutative POVMs and fuzzy observables.

1 Introduction

Fuzzy observables play a very important role in quantum physics and in the analysis of its foundations. Moreover, they are connected to fuzzy sets since they can be interpreted as fuzzification of sharp observables. From the mathematical viewpoint, they are described by commutative positive operator valued measures (POVMs). A POVM is a map $F:\mathscr {B}(X)\rightarrow \mathscr {L}^+_s(\mathscr {H})$ from the Borel $\sigma$-algebra of a topological space X to the space of linear, self-adjoint, positive operators in the Hilbert space $\mathscr {H}$ satisfying the following properties: 1) $F(X)=\textbf{1}$ where $\textbf{1}$ is the identity operator, 2) for every countable family $\{\varDelta _n\}$ of disjoint sets in $\mathscr {B}(X)$,

$$\begin{aligned} F\big (\bigcup _{n=1}^{\infty }\varDelta _n\big )=\sum _{n=1}^{\infty }F(\varDelta _n). \end{aligned}$$

where the series converges in the weak operator topology. It is said to be commutative if $\big [F(\varDelta _1),F(\varDelta _2)\big ]=\textbf{0}$ for all $\varDelta _1\,,\varDelta _2\in \mathscr {B}(X)$. A POVM such that $F(\varDelta )$ is a projection operator for every $\varDelta \in \mathscr {B}(X)$ is called a projection valued measure (PVM). In particular, a real PVM ($X=\mathbb {R}$) is called a spectral measure. As it is well known, spectral measures are in a one-to-one correspondence with self-adjoint operators, the latter representing standard (or sharp) quantum observables. By analyzing the process of measurement in quantum physics, it can be shown [14, 19, 23] that POVMs provide the right mathematical representation for quantum observable; spectral measures being a too restrictive mathematical tool. Quantum observables represented by POVMs that are not sharp are called generalised or unsharp observables.

We recall that $\langle \psi ,F(\varDelta )\psi \rangle$ (where $\langle \cdot ,\cdot \rangle$ denotes the inner product in $\mathscr {H}$) is interpreted as the probability that a measurement of the observable represented by F gives a result in $\varDelta$.

Before we proceed with the proof of the main results it is helpful to illustrate the connections between Markov kernels, POVMs and fuzzy observables. We can consider, for example, the joint measurement of position and momentum observables. If the position and momentum observables are represented by the spectral measures $E^Q$ and $E^P$ respectively, the mathematical formalism does not allow us to describe their joint measurement which should be represented by a joint POVM of which $E^Q$ and $E^P$ are the marginals. Such a POVM exists if and only if the spectral measures $E^Q$ and $E^P$ commute and it is well known that they do not. On the contrary, there are couples of non commuting POVMs that are the marginals of a joint POVM. An example is provided by the fuzzification of $E^Q$ and $E^P$ by means of two Markov kernels $\mu ^Q$ and $\mu ^P$,

$$\begin{aligned} F^Q(\varDelta )&=\int _{\mathbb {R}}\mu _\varDelta ^Q(\lambda )\,dE^Q_\lambda =\mu ^Q_\varDelta (Q),\\ F^P(\varDelta )&=\int _{\mathbb {R}}\mu _\varDelta ^P(\lambda )\,dE^P_\lambda =\mu ^P_\varDelta (P).\nonumber \end{aligned}$$

(1)

where $\mu _{(\cdot )}^Q(\lambda )$ is a probability measure for every $\lambda \in \mathbb {R}$ and $\mu _\varDelta ^Q(\cdot )$ is measurable for every $\varDelta$ and the same is true for $\mu _{(\cdot )}^P(\lambda )$ and $\mu _{\varDelta }^P(\cdot )$. Those are the conditions defining a Markov kernel of which we recall the definition below.

Definition 1

Let $(\varLambda ,\mathscr {A})$ be a measurable space and $\mathscr {B}(X)$ the Borel $\sigma$-algebra of a topological space X. A Markov kernel is a map $\mu :\varLambda \times \mathscr {B}(X)\rightarrow [0,1]$ such that,

• 1. $\mu _{\varDelta }(\cdot )$ is a measurable function for each $\varDelta \in \mathscr {B}(X)$,

• 2. $\mu _{(\cdot )}(\lambda )$ is a probability measure for each $\lambda \in \varLambda$.

Going back to the position and momentum POVMs $F^Q$ and $F^P$ we remark that they can be interpreted [12, 24] as fuzzification of $E^Q$ and $E^P$ respectively. Indeed, for every $\varDelta \in \mathscr {B}(\mathbb {R})$, $(\mathbb {R},\mu _\varDelta ^Q(\cdot ))$ defines a fuzzy set [29] (see below). A very relevant feature of $F^Q$ and $F^P$ is that there is a third POVM $F(\varDelta _1\times \varDelta _2)$ of which $F^Q$ and $F^P$ are the marginals, i.e., $F^Q(\varDelta _1)=F(\varDelta _1\times \mathbb {R})$, $F^P(\varDelta _2)=F(\mathbb {R}\times \varDelta _2)$. The POVM F represents the joint measurement of position and momentum [3, 15, 25]. The property that, in some cases, two not commuting POVMs can have a joint POVM is at the root of the formulation of quantum mechanics on phase space [10, 11, 25, 26].

Note that $F^Q$ and $F^P$ are commutative POVMs. Indeed, every operator $F^Q(\varDelta )$ is a function of the self-adjoint operator Q. As a consequence the family of operators $\{F^Q(\varDelta )\}_{\varDelta \in \mathscr {B}(\mathbb {R})}$ is commutative. The same is true for $F^P$. That is true in general, i.e., a POVM which is the fuzzification of a spectral measure is commutative.

The fundamental result about commutative POVMs is that all of them are the fuzzification of a spectral measure [2, 4, 18, 20]: let X be Hausdorff, second countable, and locally compact. Every commutative POVM $F:\mathscr {B}(X)\rightarrow \mathscr {L}^+_s(\mathscr {H})$ is the fuzzy version of a spectral measure $E^F$ (the sharp version of F) with the fuzzification represented by a Markov kernel $\mu$,

$$\begin{aligned} \langle \psi ,F(\varDelta )\psi \rangle&:=\int \mu _{\varDelta }(\lambda )\,d\langle \psi ,E^F_\lambda \psi \rangle ,\quad \varDelta \in \mathscr {B}(X),\quad \psi \in \mathscr {H}. \end{aligned}$$

(2)

The quantity $\langle \psi ,E^F(\varDelta )\psi \rangle$ can be interpreted as the probability that a perfectly accurate measurement (sharp measurement) of the observable represented by the spectral measure E gives a result in $\varDelta$. A possible interpretation of (2) is that [1, 2, 4, 25], due to measurement imprecision¹, the outcomes of the measurement of $E^F$ are randomized: if the sharp value of the outcome of the measurement of $E^F$ is $\lambda$ then the apparatus produces with probability $\mu _{\varDelta }(\lambda )$ a reading in $\varDelta$. As a result, the probability of an outcome in $\varDelta$ is given by $\langle \psi ,F(\varDelta )\psi \rangle$ so that F represents an unsharp measurement of E.

In the framework of fuzzy sets theory, (2) can be interpreted as follows (see [12] for more details). The Markov kernel $\mu$ provides a family of fuzzy events $\{(\mathbb {R},\mu _\varDelta )\}_{\varDelta \in \mathscr {B}(\mathbb {R})}$. For every $\psi \in \mathscr {H}$, the expression

$$\begin{aligned} \langle \psi ,F(\varDelta )\psi \rangle =\int _{\mathbb {R}}\mu _{\varDelta }(\lambda )\,d\langle \psi ,E^F_\lambda \psi \rangle \end{aligned}$$

can then be interpreted as the probability of the fuzzy event $(\mathbb {R},\mu _{\varDelta })$ with respect to the probability measure $\langle \psi ,E^F(\cdot )\psi \rangle$. In other words, the unsharp observable F gives the probabilities of the fuzzy events $(\mathbb {R},\mu _\varDelta )$ with respect to the probability measures corresponding to $E^F$ (they are $\langle \psi ,E^F(\cdot )\psi \rangle$, $\psi \in \mathscr {H}$).

It is worth remarking that starting from the fuzzy observable F it is possible to obtain the sharp observable $E^F$ of which F is a fuzzy version (see Ref.s [5,6,7,8]).

Several proofs of (2) have been provided [2, 4, 18, 20, 21]. All of them are based on the existence, for every weak Markov kernel $\gamma$ (see Definition 2 below), of a Markov kernel $\mu$ which is equivalent to $\gamma$. In order to illustrate this point, let us consider the von Neumann algebra $\mathscr {A}(F)$ generated by $\{F(\varDelta )\}_{\varDelta \in \mathscr {B}(X)}$. It is commutative and then singly generated by a self-adjoint operator $A^F$ with spectral measure $E^F$. Therefore, the commutative POVM $F:\mathscr {B}(X)\rightarrow \mathscr {L}_s^+(\mathscr {H})$ admits an integral representation

$$\begin{aligned} \langle \psi ,F(\varDelta )\psi \rangle =\int _{\mathbb {R}}\gamma _\varDelta (\lambda )\,d\langle \psi ,E^F_\lambda \psi \rangle \end{aligned}$$

(3)

where $\{\gamma _\varDelta (\cdot )\}_{\varDelta \in \mathscr {B}(X)}$ is a family of measurable functions. Moreover, it is straightforward to show that the family of functions $\{\gamma _{\varDelta }(\cdot )\}_{\varDelta \in \mathscr {B}(X)}$ defines a weak Markov kernel (see Definition 2 below and the proof of Theorem 5) with respect to the measure $\nu (\cdot ):=\langle \psi _0, E^F(\cdot )\psi _0\rangle$ where $\psi _0$ is a separating vector for $\mathcal {A}^W(F)$.

Definition 2

Let $\nu$ be a measure on $\varLambda$. A map $\mu :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ is a weak Markov kernel with respect to $\nu$ if:

• 1. $\mu _{\varDelta }(\cdot )$ is a measurable function for each $\varDelta \in \mathscr {B}(X)$,

• 2. for every $\varDelta \in \mathscr {B}(X)$,  $0\le \mu _{\varDelta }(\lambda )\le 1$,   $\nu -a.e.$,

• 3. $\mu _{\emptyset }(\lambda )=0$, $\mu _{X}(\lambda )=I$   $\nu -a.e.$,

• 4. for any sequence $\{\varDelta _i\}_{i\in \mathbb {N}}$, $\varDelta _i\cap \varDelta _j=\emptyset$,

  $$\begin{aligned} \sum _i\mu _{\varDelta _i}(\lambda )=\mu _{(\cup _i\varDelta _i)}(\lambda ),\quad \nu -a.e. \end{aligned}$$

  (4)

In order to prove the integral representation (2) (where $\mu$ is a Markov kernel), it is necessary to show that the $\gamma$ in (3) can be replaced by a Markov kernel $\mu$. In other words, it is necessary to show that the weak Markov kernel $\gamma$ is equivalent to a Markov kernel $\mu$ according to the following equivalence definition.

Definition 3

Two weak Markov kernels $\gamma :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ and $\beta :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ are said to be equivalent if, for every $\varDelta \in \mathscr {B}(X)$, $\gamma _\varDelta (\lambda )=\beta _\varDelta (\lambda )$, $\nu -a.e.$.

As a final remark on the differences between weak Markov kernels and Markov kernels, we observe that every Markov kernel is a weak Markov kernel with respect to every measure $\nu$. Note moreover that, in the case of a weak Markov kernel, for every partition $\varDelta =\cup _i\varDelta _i$ (which is a disjoint union), there is a subset $\varLambda _{\{\varDelta ,\varDelta _i\}}\subset \varLambda$ of measure one such that $\mu _{\varDelta }(\lambda )=\sum _i\mu _{\varDelta _i}(\lambda )$, $\lambda \in \varLambda _{\{\varDelta ,\varDelta _i\}}$. That does not ensure that $\mu _{(\cdot )}(\lambda )$ is a probability measure because of the dependence of $\varLambda _{\{\varDelta ,\varDelta _i\}}$ on both $\varDelta$ and its partitions $\{\varDelta _i\}_{i\in \mathbb {N}}$.

Going back to the connections between weak Markov kernels and Markov kernels, an indirect proof that for every weak Markov kernel there is an equivalent Markov kernel is provided in [20, 21, 28] where the space X is required to be second countable and completely metrizable (e.g., a standard Borel space) and the proof is given in the framework of stochastic operators (see Section 2 below).

In Section 2 we generalize this result to the case of a second countable, metrizable space (not necessarily completely metrizable) by giving an independent and direct proof which is based on transfinite recursion. Then, in Section 3, we use this result in order to generalize the integral representation (2) and some previous results about commutative POVMs.

2 From Weak Markov Kernels to Markov Kernels

In the present section we prove that if X is second countable and metrizable, a weak Markov kernel is equivalent to a Markov kernel. In [21, 28] the equivalence is proved in the more restrictive case of a standard Borel space X (which is second countable and completely metrizable). The proof is indirect in the sense that it is derived as a consequence of the fact that every stochastic operator can be represented by means of a Markov kernel and of the observation that every weak Markov kernel defines a stochastic operator (see Theorem 6.3 in [21]) for example.

In order to prove the theorem in the more general case of a metrizable space and in order to give a direct proof, a more fundamental approach is required. As we pointed out in the comment after Definition 2, in the case of a weak Markov kernel $\mu$, $\mu _{(\cdot )}(\lambda )$ is not, in general, a probability measure because $\varLambda _{\{\varDelta ,\varDelta _i\}}$ depends both on $\varDelta$ and $\{\varDelta _i\}$. Nevertheless, it is worth observing that: 1) being X second countable, there is a countable basis $\mathscr {L}$ for his topology which generates a countable ring $\mathscr {R}(\mathscr {L})$, 2) given a set $\varDelta \in \mathscr {B}(X)$, the family of all its partitions, $\varDelta =\cup _i\varDelta _i$, has the power of the continuum. All of that rises the following question: given a weak Markov kernel $\gamma :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$, is it possible to use transfinite recursion in order to show the existence of a subset $\varLambda _1\subset \varLambda$, $\nu (\varLambda _1)=1$, such that the restriction $\gamma :(\varLambda _1,\nu )\times \mathscr {R}(\mathscr {L})\rightarrow [0,1]$ is a Markov kernel? Theorem 2 below answers in the positive. The proof is based on Lemma 1, Corollary 2 and Theorem 1 (that in its turn is based on transfinite recursion). Moreover, it is proved (Theorem 3) that $\gamma :(\varLambda _1,\nu )\times \mathscr {R}(\mathscr {L})\rightarrow [0,1]$ is the restriction to $\varLambda _1$ of a Markov kernel $\alpha :(\varLambda ,\nu )\times \mathscr {R}(\mathscr {L})\rightarrow [0,1]$ that can be extended to a Markov kernel $\mu :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ which is equivalent to the weak Markov kernel $\gamma :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$.

In what follows, $\mathscr {B}(X)$ denotes the Borel $\sigma$-algebra of a topological space X, $\varLambda$ a compact subsets of [0, 1], $\nu$ a probability measure on $\varLambda$ and $L^\infty (\varLambda ,\nu )$ the space of essentially bounded measurable functions (with two functions identified if they coincide up to $\nu$-null sets). Finally I denotes a closed subset of [0, 1].

Lemma 1

Let X be a second countable metrizable topological space, $\mathscr {S}$ a basis for its topology and $\mathscr {R}(\mathscr {S})$ the ring generated by $\mathscr {S}$. Let $\gamma :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ be a weak Markov kernel. Then, there is a weak Markov kernel $\beta :(I,\nu _0)\times \mathscr {B}(X)\rightarrow [0,1]$ with $\beta _\varDelta$ continuous for every $\varDelta \in \mathscr {R}(\mathscr {S})$, a function $g:\varLambda \rightarrow I$ and a set $N\subset \varLambda$, $\nu (N)=1$, such that, for every $\varDelta \in \mathscr {R}(\mathscr {S})$ and for all $x\in N$, $\beta _\varDelta (g(x))=\gamma _\varDelta (x)$.

Proof

Without loss of generality, we can assume $\varLambda$ to be the support of $\nu$. Let $\mathscr {A}_\nu$ be the von Neumann algebra of multiplication operators in $\mathscr {H}=L^2(\varLambda ,\nu )$ which is isometrically $^*$-isomorphic [16] to $L^\infty (\varLambda ,\nu )$. In particular, for every function $f\in L^\infty (\varLambda ,\nu )$ there is a multiplication operator

$$\begin{aligned} M_f:L^2(\varLambda ,\nu )\rightarrow L^2(\varLambda ,\nu ) \end{aligned}$$

$$\begin{aligned}{}[M_f(h)](x)=f(x)h(x),\quad h\in L^2(\varLambda ,\nu ). \end{aligned}$$

The self-adjoint operator $B:=M_x$, $[Bh](x)=[M_x(h)](x)=xh(x)$, $x\in \varLambda$ generates $\mathscr {A}_\nu$. The spectrum of B, $\sigma (B)$, coincides with the support, $\varLambda$, of $\nu$ and the spectral measure corresponding to B is $E^B(\varDelta )=M_{\chi _{\varDelta }}$. Moreover, $\nu$ is a scalar-valued spectral measure for B, i.e., $\nu$ and $E^B$ are mutually absolutely continuous (see [16], page 133).

Now, we define the commutative POVM,

$$\begin{aligned} F(\varDelta )=M_{\gamma _\varDelta }=\int \gamma _\varDelta (x)\,M_{\chi _{dx}},\quad \varDelta \in \mathscr {B}(X). \end{aligned}$$

(5)

Let us consider the von Neumann algebra $\mathscr {A}^W(F)$ generated by $\{F(\varDelta )\}_{\varDelta \in \mathscr {B}(X)}$. It coincides with the von Neumann algebra generated by the set $O_2:=\{F(\varDelta )\}_{\varDelta \in \mathscr {R}(\mathscr {S})}$ (see Proposition 1 in Appendix A). We recall that both $\mathscr {S}$ and $\mathscr {R}(\mathscr {S})$ are countable. It can be proved that there is a generator A of $\mathscr {A}^W(F)$ with spectrum $I\subset [0,1]$ and scalar valued spectral measure $\nu _0$. Moreover, there is a weak Markov kernel $\beta :(I,\nu _0)\times \mathscr {B}(X)\rightarrow [0,1]$ such that, for every $\varDelta \in \mathscr {R}(\mathscr {S})$, $M_{\gamma _\varDelta }=\beta _\varDelta (A)$ with $\beta _\varDelta$ continuous (see Theorem 7 in the appendix for the details). Since $A\in \mathscr {A}^W(F)\subset \mathscr {A}_\nu$, there must be a measurable function $g:\varLambda \rightarrow I$ such that $A=M_g=\int _\varLambda g(x)\,M_{\chi _{dx}}=g(B)$. Then, the spectral measure $E^A$ corresponding to A is such that $E^A(D)=E^B(g^{-1}(D))=M_{\chi _{g^{-1}(D)}}$, $D\in \mathscr {B}(I)$. Hence, $\forall \varDelta \in \mathscr {B}(X)$,

$$\begin{aligned} \int _\varLambda \gamma _\varDelta (\lambda )\,M_{\chi _{d\lambda }}&=M_{\gamma _\varDelta }=\int _I \beta _\varDelta (x)\,E^A_{dx}\\&=\int _I \beta _\varDelta (x)\,M_{\chi _{g^{-1}(dx)}}=\int _\varLambda \beta _\varDelta (g(\lambda ))\,M_{\chi _{d\lambda }} \end{aligned}$$

so that, $\beta _\varDelta (g(\lambda ))=\gamma _\varDelta (\lambda )$, $\nu$-a.e.. In particular, being $\mathscr {R}(\mathscr {S})$ countable, there is a set $N\subset \varLambda$, $\nu (N)=1$, such that, for every $\varDelta \in \mathscr {R}(\mathscr {S})$ and for all $\lambda \in N$, $\beta _\varDelta (g(\lambda ))=\gamma _\varDelta (\lambda )$. $\square$

Theorem 1

Let X be a second countable topological space, $\mathscr {S}$ a basis for its topology and $\mathscr {R}(\mathscr {S})$ the ring generated by $\mathscr {S}$. Let $\beta :(I,\nu _0)\times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ be a weak Markov kernel such that $\beta _\varDelta$ is continuous. Then there is a subset $M\subset I$, $\nu _0(M)=1$, such that $\beta :M\times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ is a Markov kernel.

Proof

Let $\varDelta \in \mathscr {R}(\mathscr {S})$. A partition, $\mathfrak {p}^\varDelta =\{\varDelta _i\}_{i\in J}$, $J\subset \mathbb {N}$, of $\varDelta$ is a collection of disjoint sets $\varDelta _i\in \mathscr {R}(\mathscr {S})$ such that $\varDelta =\cup _{i\in J}\varDelta _i$.

Since $\beta$ is a weak Markov kernel, to every partition $\mathfrak {p}^\varDelta =\{\varDelta _i\}_{i\in \mathbb {N}}$ there corresponds a set $N_{\mathfrak {p}^\varDelta }\subset I$ such that $\nu _0(N_{\mathfrak {p}^\varDelta })=1$ and

$$\begin{aligned} \beta _{\varDelta }(\lambda )-\sum _{i} \beta _{\varDelta _i}(\lambda )=0,\quad \lambda \in N_{\mathfrak {p}^\varDelta }. \end{aligned}$$

(6)

Consider the collection $\mathfrak {P}_\varDelta$ of all the partitions of $\varDelta$. The power of $\mathfrak {P}_\varDelta$ is at most $\mathfrak {c}$, the power of the continuum. Let Z be the collection of all countable ordinals. Let $\mathfrak {P}_\varDelta (Z)$ be a well ordering of $\mathfrak {P}_\varDelta$ through Z. Then, $\mathfrak {P}_\varDelta =\{\mathfrak {p}^\varDelta _\alpha \}_{\alpha \in Z}$, $\mathfrak {p}^\varDelta _\alpha =\{\varDelta _i^\alpha \}_{i\in J^\alpha }$, $J^\alpha \subset \mathbb {N}$. Now we use transfinite recursion in order to define a new family of sets. Let $N^\varDelta _\alpha :=N_{\mathfrak {p}^\varDelta _\alpha }$ denote the set corresponding to the partition $\mathfrak {p}^\varDelta _\alpha$ as in (6). Let $M^\varDelta _1=N^\varDelta _1$, $M^\varDelta _{\alpha +1}=M^\varDelta _\alpha \cap N^\varDelta _{\alpha +1}$ and $M^\varDelta _\eta :=\cap _{\alpha <\eta }M^\varDelta _{\alpha }$ if $\eta$ is a limit ordinal.

Since every $\eta \in Z$ is an enumerable ordinal, $\nu (M^\varDelta _\eta )=1$ for every $\eta \in Z$. Moreover, $\{M^\varDelta _\alpha \}_{\alpha \in Z}$ is non increasing. Now, take the closure $\overline{M^\varDelta _\alpha }$ of $M^\varDelta _\alpha$ for every $\alpha \in Z$. By the continuity of $\beta _\varDelta$,

$$\begin{aligned} \beta _{\varDelta }(\lambda )-\sum _{i} \beta _{\varDelta _i^\alpha }(\lambda )=0,\quad \lambda \in \overline{M^\varDelta _\alpha }. \end{aligned}$$

(7)

Since $\{\overline{M^\varDelta _\alpha }\}_{\alpha \in Z}$, is a non-increasing family of closed sets, there must be an index $\eta <\omega _1$ ($\omega _1$ denotes the first uncountable ordinal) such that $\cap _{\alpha<\omega _1}\overline{M^\varDelta _\alpha }=\cap _{\alpha <\eta }\overline{M^\varDelta _\alpha }=M_\varDelta$. Since $\eta$ is countable, $\nu _0(M_\varDelta )=\nu _0(\cap _{\alpha <\eta }\overline{M^\varDelta _\alpha })=1$. Then, for every partition $p^\varDelta _\alpha =\{\varDelta ^\alpha _i\}_{i\in J^\alpha }\in \mathfrak {P}_\varDelta$.,

$$\begin{aligned} \beta _{\varDelta }(\lambda )-\sum _{i} \beta _{\varDelta _i^\alpha }(\lambda )=0,\quad \forall \lambda \in M_\varDelta . \end{aligned}$$

(8)

Since $\mathscr {R}(\mathscr {S})$ is countable, $M:=\cap _{\varDelta \in \mathscr {R}(\mathscr {S})} M_\varDelta$ is such that $\nu _0(M)=1$. Then, for every $\varDelta \in \mathscr {R}(\mathscr {S})$ and $p^\varDelta _\alpha =\{\varDelta ^\alpha _i\}_{i\in I^\alpha }\in \mathfrak {P}_\varDelta$,

$$\begin{aligned} \beta _{\varDelta }(\lambda )-\sum _{i} \beta _{\varDelta ^\alpha _i}(\lambda )=0,\quad \lambda \in M. \end{aligned}$$

(9)

Define

$$\begin{aligned} h_\varDelta (\lambda )= {\left\{ \begin{array}{ll} \beta _\varDelta (\lambda ) &{} \lambda \in M\\ \varphi (\varDelta ) &{} \lambda \in I/M \end{array}\right. } \end{aligned}$$

(10)

where $\varphi$ is an arbitrary probability measure on $\mathscr {R}(\mathscr {S})$. The map $h:\varLambda \times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ is a Markov kernel and coincides with $\beta$ on M. $\square$

The proof of the previous theorem contains the following Corollary (see equation (10)).

Corollary 1

Let X be a second countable topological space, $\mathscr {S}$ a basis for its topology and $\mathscr {R}(\mathscr {S})$ the ring generated by $\mathscr {S}$. Let $\beta :(I,\nu _0)\times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ be a weak Markov kernel such that $\beta _\varDelta$ is continuous. Then, there is a set $M\subset I$, $\nu _0(M)=1$, and a Markov kernel $h:I\times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ such that $\beta _\varDelta (\lambda )=h_\varDelta (\lambda )$ for every $\lambda \in M$. In particular, $\beta$ and h are $\nu _0$-equivalent.

Corollary 2

Let X be a second countable metrizable topological space, $\mathscr {S}$ a basis for its topology and $\mathscr {R}(\mathscr {S})$ the ring generated by $\mathscr {S}$. Then, there is a Markov kernel $\alpha :\varLambda \times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ and a set $\varLambda _1\subset \varLambda$, $\nu (\varLambda _1)=1$, such that $\alpha _\varDelta (\lambda )=\gamma _\varDelta (\lambda )$, $\lambda \in \varLambda _1$, $\varDelta \in \mathscr {R}(\mathscr {S})$.

Proof

Let $\gamma$, g and h be as in Lemma 1 and Theorem 1. Since $h:I\times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ is a Markov kernel, $h\circ g:\varLambda \times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ is a Markov kernel as well. Let $\varLambda _0:=g^{-1}(M)$ where M has been defined in Theorem 1. Then, with the same notation of Lemma 1 and Theorem 1, $E^B(\varLambda _0)=E^B[g^{-1}(M)]=E^A(M)=\textbf{1}$, where $\textbf{1}$ denotes the identity operator. Therefore, $\nu (\varLambda _0)=1$. Let $\varLambda _1:=\varLambda _0\cap N$ where N has been defined in Lemma 1. Note that $\nu (\varLambda _1)=1$. By Corollary 1 and Lemma 1, $\alpha (\lambda ):=h_\varDelta (g(\lambda ))=\beta _\varDelta (g(\lambda ))=\gamma _\varDelta (\lambda )$ for every $\varDelta \in \mathscr {R}(\mathscr {S})$ and $\lambda \in \varLambda _1$. $\square$

The following theorem is a consequence of Lemma 1, Theorem 1 and Corollary 2

Theorem 2

Let X be a second countable metrizable topological space, $\mathscr {S}$ a basis for its topology and $\mathscr {R}(\mathscr {S})$ the ring generated by $\mathscr {S}$. Let $\gamma :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ be a weak Markov kernel. Then, there is a subset $\varLambda _1\subset \varLambda$, $\nu (\varLambda _1)=1$, such that $\gamma :(\varLambda _1,\nu )\times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ is a Markov kernel.

Proof

By Corollary 2, $\gamma _\varDelta (\lambda )=\alpha _\varDelta (\lambda )$, $\lambda \in \varLambda _1$, $\nu (\varLambda _1)=1$, $\varDelta \in \mathscr {R}(\mathscr {S})$, where $\alpha :\varLambda \times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ is a Markov kernel. $\square$

The proof of the following theorem is based on the use of the Borel hierarchy on X, which requires the space X to be metrizable since in such case every open set is a $F_\delta$ set (see section 3.6 in [27]) or section 30 in [22]).

Theorem 3

The Markov kernel $\alpha :\varLambda \times \mathscr {R}(\mathscr {S})\rightarrow [0,1]$ in Corollary 2 can be extended to a Markov kernel $\mu :\varLambda \times \mathscr {B}(X)\rightarrow [0,1]$ such that, for every $\varDelta \in \mathscr {B}(X)$, $\mu _\varDelta (\lambda )=\gamma _\varDelta (\lambda )$, $\nu$-a.e..

Proof

For every $\lambda \in \varLambda$, the measure $\alpha _{(\cdot )}(\lambda ):\mathscr {R}(\mathscr {S})\rightarrow [0,1]$ can be extended to the Borel $\sigma$-algebra $\mathscr {B}(X)$. Let $\mu _{(\cdot )}(\lambda ):\mathscr {B}(X)\rightarrow [0,1]$ denotes such an extension. We want to show that, for each $\varDelta \in \mathscr {B}(X)$, $\mu _\varDelta :\varLambda \rightarrow [0,1]$ is measurable and $\mu _\varDelta =\gamma _\varDelta$, $\nu$-a.e.. That can be proved by using transfinite induction. We start by recalling the definition of Borel Hierarchy in a second countable metrizable space [22, 27]. Let $\omega _1$ be the first uncountable ordinal. Let $\mathscr {K}$ be a family of subsets of X. The Borel classes, $\mathscr {B}(\mathscr {K})$, generated by $\mathscr {K}$ are defined inductively as follows. $\Sigma _0^0=\emptyset$, $\Sigma _1^0=\mathscr {K}$, $\Sigma _2^0$ is the class of countable unions of sets in $\Pi _1^0=\{X-\varDelta ,\,\varDelta \in \mathscr {K}\}$, $\Sigma _\alpha ^0$, $2<\alpha <\omega _1$, is the class of countable unions of sets in $\cup _{\beta <\alpha }\Pi _\beta ^0$ where $\Pi _\beta ^0=\{X-\varDelta \,|\,\varDelta \in \Sigma _\beta ^0\}$. Then $\mathscr {B}(\mathscr {K})=\cup _{\alpha <\omega _1}\Sigma _\alpha ^0$. If $\mathscr {K}=\mathscr {R}(\mathscr {S})$ then $\cup _{\alpha <\omega _1}\Sigma _\alpha ^0=\mathscr {B}(X)$ (see [27], Proposition 3.6.1, page 116).

Let $G\in \mathscr {R}(\mathscr {S})$. Then $\mu _G=\alpha _G$ is measurable and, by Corollary 2, it is such that $\mu _G(\lambda )=\gamma _G(\lambda )$, $\lambda \in \varLambda _1$. Suppose that for every $\varDelta \in \Sigma _\beta ^0$, $\beta <\alpha$, the function $\mu _\varDelta$ is measurable and such that $\mu _\varDelta =\gamma _\varDelta$, $\nu$-a.e.. As a consequence, $\mu _{\varDelta }$ is measurable and such that $\mu _{\varDelta }=\gamma _{\varDelta }$, $\nu$-a.e. for every $\varDelta \in \Pi ^0_\beta$. Let $\varDelta \in \Sigma _\alpha ^0$. Then, $\varDelta =\cup _{i=1}^\infty \varDelta _i$, $\varDelta _i\in \Pi _\beta ^0$. Setting $\widetilde{\varDelta }_n=\sum _{i=1}^n\varDelta _i$, we obtain a non-decreasing family of sets such that $\widetilde{\varDelta }_n\uparrow \varDelta$. Note that, due to the fact that the pi-classes $\Pi ^0_\beta$ are closed with respect to finite unions, for every $n\in \mathbb {N}$ there is an index $\beta <\alpha$ such that $\widetilde{\varDelta }_n\in \Pi ^0_\beta$. Hence, $\mu _{\widetilde{\varDelta }_n}$ is measurable and, for every $\lambda \in \varLambda$,

$$\begin{aligned} \mu _\varDelta (\lambda )=\lim _{n\rightarrow \infty }\mu _{\widetilde{\varDelta }_{n}}(\lambda ) \end{aligned}$$

which prove the measurability of $\mu _\varDelta$. Hence, $\mu :\varLambda \times \mathscr {B}(X)\rightarrow [0,1]$ is a Markov kernel. Moreover, by the inductive hypothesis, $\mu _{\widetilde{\varDelta }_n}(\lambda )=\gamma _{\widetilde{\varDelta }_n}$, $\nu$-a.e.. Then,

$$\begin{aligned} \int \gamma _{\varDelta }(\lambda )\,dE^B_\lambda =F(\varDelta )&=\lim _{n\rightarrow \infty } F(\widetilde{\varDelta }_n)\\&=\lim _{n\rightarrow \infty }\int _\varLambda \gamma _{\widetilde{\varDelta }_n}(\lambda )\,dE^B_\lambda =\lim _{n\rightarrow \infty }\int _\varLambda \mu _{\widetilde{\varDelta }_n}(\lambda )\,dE^B_\lambda \!=\!\int _\varLambda \mu _{\varDelta }(\lambda )\,dE^B_\lambda \end{aligned}$$

so that $\mu _{\varDelta }(\lambda )=\gamma _{\varDelta }(\lambda )$, $\nu$-a.e. $\square$

The following theorem is a straightforward consequence of Lemma 1, Theorem 1, Corollary 2 and Theorem 3.

Theorem 4

Let X be a second countable metrizable topological space. Let $\gamma :(\varLambda ,\nu )\times \mathscr {B}(X)\rightarrow [0,1]$ be a weak Markov kernel. Then, there is a Markov kernel $\mu :\varLambda \times \mathscr {B}(X)\rightarrow [0,1]$ such that, for every $\varDelta \in \mathscr {B}(X)$, $\mu _\varDelta =\gamma _\varDelta$, $\nu$-a.e..

3 Fuzzy Observables

In the introduction we recalled that the connection between commutative POVMs and Markov kernels (see equation (2)) is at the root of the interpretation of the former as the fuzzification of a spectral measure. The result obtained in the previous section can be used to generalize such a connection to the case of POVMs defined on a second countable metrizable space.

The following theorem gives a characterization of commutative POVMs as fuzzification of spectral measures with the fuzzification realized by means of Markov kernels and generalizes some previous results [2, 4, 20, 21] where complete metrizability is required. The possible extension of the results connecting Naimark’s operators and sharp reconstructions [5,6,7, 9, 13] to the case of a metrizable second countable space can be analyzed as well.

In the following, the symbol $\mu$ is used to denote both Markov kernels and weak Markov kernels. The symbol $\mathscr {A}^W(F)$ denotes the von Neumann algebra generated by the POVM F, i.e., the von Neumann algebra generated by the set $\{F(\varDelta )\}_{\mathscr {B}(X)}$. Analogously $\mathscr {A}(B)$ denotes the von Neumann algebra generated by the self-adjoint operator B.

Definition 4

Whenever F, A, and $\mu$ are such that $F(\varDelta )=\mu _{\varDelta }(A)$, $\varDelta \in \mathscr {B}(X)$, we say that $(F,A,\mu )$ is a von Neumann triplet.

Theorem 5

Let X be a second countable, metrizable space. A POVM $F:\mathscr {B}(X)\rightarrow \mathscr {L}^+_s(\mathscr {H})$ is commutative if and only if, there exists a bounded self-adjoint operator $A=\int \lambda \,dE_\lambda$ with spectrum $\sigma (A)\subset [0,1]$ and a Markov Kernel $\mu$ such that:

$$\begin{aligned} F(\varDelta )=\int _{\sigma (A)}\mu _{\varDelta }(\lambda )\,dE_{\lambda }=\mu _\varDelta (A),\quad \varDelta \in \mathscr {B}(X). \end{aligned}$$

(11)

Proof

If there is an operator A satisfying (11), the operators $F(\varDelta )=\mu _\varDelta (A)$, $\varDelta \in \mathscr {B}(X)$, commute since they are functions of the same operator A. In order to prove that commutativity implies the existence of A, let us consider the commutative von Neumann algebra $\mathscr {A}^W(F)$. It is singly generated by a self-adjoint operator A with compact spectrum $\sigma (A)\subset [0,1]$. Hence, for every $\varDelta \in \mathscr {B}(X)$, there is a measurable function $\gamma _\varDelta :\sigma (A)\rightarrow [0,1]$ such that

$$\begin{aligned} F(\varDelta )=\int _{\sigma (A)}\gamma _\varDelta (\lambda )\,dE^A_\lambda \end{aligned}$$

(12)

where $E^A$ is the spectral measure corresponding to A. Let $\varDelta$ be the disjoint union of the sets $\{\varDelta _i\}_{i\in \mathbb {N}}$. Then,

$$\begin{aligned} \int _{\sigma (A)}\gamma _\varDelta (\lambda )\,dE^A_\lambda =F(\varDelta )=\sum _{i=1}^\infty F(\varDelta _i)=\sum _{i=1}^\infty \int _{\sigma (A)}\gamma _{\varDelta _i}(\lambda )\,dE^A_\lambda \end{aligned}$$

(13)

For each $\lambda \in \sigma (A)$, $\sum _{i=1}^n\gamma _{\varDelta _i}(\lambda )$ is an nondecreasing family of measurable functions. Then, it converges to a measurable function $f_\varDelta$ and, by the Lebesgue convergence theorem,

$$\begin{aligned} \int _{\sigma (A)}\gamma _\varDelta (\lambda )\,dE^A_\lambda =\sum _{i=1}^\infty \int _{\sigma (A)}\gamma _{\varDelta _i}(\lambda )\,dE^A_\lambda =\int _{\sigma (A)} f_{\varDelta }(\lambda )\,dE^A_\lambda \end{aligned}$$

(14)

Hence,

$$\begin{aligned} \gamma _\varDelta (\lambda )= f_{\varDelta }(\lambda )=\sum _{i=1}^\infty \gamma _{\varDelta _i}(\lambda ), \quad E^A-a.e. \end{aligned}$$

In other words, $\gamma :(\sigma (A),\nu ^A)\times \mathscr {B}(X)\rightarrow [0,1]$ is a weak Markov kernel. Here $\nu ^A(\varDelta ):=\langle \psi _0,E^A(\cdot )\psi _0\rangle$ where $\psi _0$ is a separating vector for $\mathscr {A}^W(F)=\mathscr {A}^W(A)$ and $E^A(\varDelta )=\textbf{0}$ if and only if $\nu ^A(\varDelta )=0$.

By Theorem 4 (which requires X to be metrizable) there is a Markov kernel $\mu :\sigma (A)\times \mathscr {B}(X)\rightarrow [0,1]$ which is equivalent to $\gamma$ with respect to $\nu ^A$. Therefore, $\mu$ is such that

$$\begin{aligned} F(\varDelta )=\int _{\sigma (A)} \mu _{\varDelta }(\lambda )\,dE^A_\lambda . \end{aligned}$$

(15)

$\square$

In Ref. [4] it has been proved that if X is Hausdorff, locally compact and second countable (and then completely metrizable), the Markov kernel $\mu$ in (15) can be replaced by a Feller Markov kernel (Theorem 4.3 in [4]). Thanks to Theorems 4 and to Proposition 1 in the appendix, the proof can be generalized to the case of an arbitrary POVM defined on a second countable metrizable space. We limit ourselves to restate the theorem since the only difference in the proof is in the use of transfinite induction (see Proposition 1 in the appendix) avoiding to require that F is regular and in the use of Theorem 4 in order to replace a weak Markov kernel by an equivalent Markov kernel.

Definition 5

Let $E:\mathscr {B}(\varLambda )\rightarrow \mathscr {L}_s^+(\mathscr {H})$ be a spectral measure. A map $\mu _{(\cdot )}(\cdot ):\varLambda \times \mathscr {B}(X)\rightarrow [0,1]$ is a strong Markov kernel with respect to E if it is a weak Markov kernel with respect to E and there exists a set $\varGamma \subset \varLambda$, $E(\varGamma )=\textbf{1}$, such that $\mu _{(\cdot )}(\cdot ):\varGamma \times \mathscr {B}(X)\rightarrow [0,1]$ is a Markov kernel. A strong Markov kernel is denoted by the symbol $(\mu , E,\varGamma \subset \varLambda )$.

Definition 6

A Feller Markov kernel is a Markov kernel $\mu _{(\cdot )}(\cdot ):\varLambda \times \mathscr {B}(X)\rightarrow [0,1]$ such that the function

$$\begin{aligned} G(\lambda )=\int _{X}f(x)\,\mu _{dx}(\lambda ),\quad \lambda \in \varLambda \end{aligned}$$

is continuous and bounded whenever f is continuous and bounded.

Theorem 6

Let X be a second countable, metrizable space. Let $F:\mathscr {B}(X)\rightarrow \mathscr {L}_s^+(\mathscr {H})$ be a POVM. Then, F is commutative if and only if, there exists a bounded self-adjoint operator $A=\int \lambda \,dE_\lambda$ with spectrum $\sigma (A)\subset [0,1]$ and a strong Markov Kernel $(\mu , E, \varGamma \subset \sigma (A))$ such that:

• 1) $\mu _{\varDelta }(\cdot ):\sigma (A)\rightarrow [0,1]$ is continuous for each $\varDelta \in \mathscr {R}(\mathscr {S})$,

• 2) $F(\varDelta )=\int _{\varGamma }\mu _{\varDelta }(\lambda )\,dE_{\lambda },\quad \varDelta \in \mathscr {B}(X)$.

• 3) $\mathscr {A}^W(F)=\mathscr {A}^W(A)$.

• 4) $\mu :\varGamma \times \mathscr {B}(X)\rightarrow [0,1]$ is a Feller Markov kernel.

Appendix

The following proposition has been proved in reference [12]. Both Proposition 1 and Theorem 7 below have been used in the proof of Lemma 1.

Proposition 1

([12]) Let X be second countable and metrizable. Let $\mathscr {S}$ be a basis for the topology of X. Let $\mathscr {R}(\mathscr {S})$ be the ring generated by $\mathscr {S}$. Let $F:\mathscr {B}(X)\rightarrow \mathscr {L}_s^+(\mathscr {H})$ be a POVM. Then, the von Neumann algebra $\mathscr {A}^W(\mathscr {R}(\mathscr {S}))$ generated by $\{F(\varDelta )\}_{\varDelta \in \mathscr {R}(\mathscr {S})}$ coincides with the von Neumann algebra $\mathscr {A}^W(F)$.

The following theorem has been proved in Ref. [4] where the POVM F was required to be normal. Such an assumption can be relaxed thanks to Proposition 1. The rest of the proof is unchanged and is repeated here for the readers convenience.

Theorem 7

Let X be second countable and metrizable. Let $F:\mathscr {B}(X)\rightarrow \mathscr {L}_s^+(\mathscr {H})$ be a POVM and $\mathscr {A}^W(F)$ the von Neumann algebra generated by F. Then, there is a generator A and a weak Markov kernel $\beta :\sigma (A)\times \mathscr {B}(X)\rightarrow [0,1]$ such that

$$\begin{aligned} F(\varDelta )=\int _{\sigma (A)}\beta _\varDelta (\lambda )\,dE^A_\lambda , \quad \varDelta \in \mathscr {B}(X), \end{aligned}$$

and $\beta _\varDelta$ is continuous for every $\varDelta \in \mathscr {R}(\mathscr {S})$.

Proof

By Proposition 1, the von Neumann algebra $\mathscr {A}^W(F)$ coincides with the von Neumann algebra generated by the set $O_2:=\{F(\varDelta )\}_{\varDelta \in \mathscr {R}(\mathscr {S})}$. We recall that both $\mathscr {S}$ and $\mathscr {R}(\mathscr {S})$ are countable. Now, let $\{\varDelta _i\}_{i\in \mathbb {N}}$ be an enumeration of the set $\mathscr {R}(\mathscr {S})$. Let $E^{(i)}$ denote the spectral measure corresponding to $F(\varDelta _i)\in O_2$. We have $F(\varDelta _i)=\int x\,d E^{(i)}_x$. Therefore, for each $i,k\in \mathbb {N}$ there exists a division $\{\varDelta _j^{(i,k)}\}_{j=1,\dots ,m_{i,k}}$ of [0, 1] such that

$$\begin{aligned} \big \Vert \sum _{j=1}^{m_{i,k}}x^{(i,k)}_j\,E^{(i)}(\varDelta _j^{(i,k)})-F(\varDelta _i)\big \Vert \le \frac{1}{k}. \end{aligned}$$

(16)

where, $x^{(i,k)}_j\in \varDelta _j^{(i,k)}$ for any $i,k\in \mathbb {N}$ and $j=1,\dots ,m_{i,k}$.

By the spectral theorem, $\{E^{(i)}(\varDelta _j^{i,k})\}_{j\le m_{i,k}}\subset \mathscr {A}^W(F)$ for any $i,k\in \mathbb {N}$. Therefore, the von Neumann algebra $\mathscr {A}^W(D)$ generated by the set $D:=\{E^{(i)}(\varDelta _j^{i,k}),\,j\le m_{i,k},\,i,k\in \mathbb {N}\}$ is contained in $\mathscr {A}^W(F)$

$$\begin{aligned} \mathscr {A}^W(D)\subset \mathscr {A}^W(F)=\mathscr {A}^W(O_2). \end{aligned}$$

(17)

Moreover, by (16)

$$\begin{aligned} \mathscr {A}^C(O_2)\subset \mathscr {A}^C(D)\subset \mathscr {A}^W(F). \end{aligned}$$

where $\mathscr {A}^C(O_2)$ and $\mathscr {A}^C(D)$ are the $C^*$-algebras generated by $O_2$ and D respectively.

By the double commutant theorem,

$$\begin{aligned} \mathscr {A}^W(F)&=[\mathscr {A}^C(O_2)]''\subset [\mathscr {A}^C(D)]''=\mathscr {A}^W(D) \end{aligned}$$

so that (see equation (17)),

$$\begin{aligned} \mathscr {A}^W(D)=\mathscr {A}^W(F). \end{aligned}$$

(18)

By Theorem 11, page 871 in Ref. [17], there is a homeomorphism $\pi :\varLambda \rightarrow \pi (\varLambda )\subset \prod _{i=1}^{\infty }\{0,1\}$ which identifies the spectrum $\varLambda$ of $\mathscr {A}^C(D)$ with a closed subset of $\prod _{i=1}^{\infty }\{0,1\}$. Moreover, the function $f:\varLambda \rightarrow [0,1]$,

$$\begin{aligned} f(\lambda ):=\sum _{i=1}^{\infty }\frac{x_i}{3^i};\quad (x_1,\dots ,x_n,\dots )=\pi (\lambda ) \end{aligned}$$

is continuous and injective and then it distinguishes the points of $\varLambda$. Since $\varLambda$ and [0, 1] are Hausdorff, $f:\varLambda \rightarrow f(\varLambda )$ is a homeomorphism.

By the Gelfand-Naimark theorem and the spectral theorem for representations of commutative $C^*$-algebras, there is an isometric $^*$-isomorphism between $\mathscr {A}^C(D)$ and $\mathscr {C}(\varLambda )$

$$\begin{aligned} T:\mathscr {C}(\varLambda )&\rightarrow \mathscr {A}^C(D)\subset B(\mathscr {H})\\ g&\mapsto T(g)=\int _{\varLambda }g(\lambda ) d\widetilde{E}_{\lambda }.\nonumber \end{aligned}$$

(19)

where $\widetilde{E}$ is the spectral measure from $\mathscr {B}(\varLambda )$ to $\mathscr {E}(\mathscr {H})$ corresponding to T.

Since f distinguishes the points of $\varLambda$, it generates $\mathscr {C}(\varLambda )$ and then

$$\begin{aligned} A=\int _{\varLambda } f(\lambda )\,d\widetilde{E}_{\lambda } \end{aligned}$$

generates both $\mathscr {A}^C(D)$ and $\mathscr {A}^W(F)$.

Now, we proceed to the proof of the existence of the weak Markov kernel $\beta$.

By (19), for each $\varDelta \in \mathscr {R}(\mathscr {S})$, there exists a continuous function $\gamma _{\varDelta }\in \mathscr {C}(\varLambda )$ such that

$$\begin{aligned} F(\varDelta )=\int _{\varLambda }\gamma _{\varDelta }(\lambda )\, d\widetilde{E}_{\lambda }. \end{aligned}$$

Let us consider the continuous function

$$\begin{aligned} \nu _{\varDelta }(t):=(\gamma _{\varDelta }\circ f^{-1})(t),\quad \varDelta \in \mathscr {R}(\mathscr {S}). \end{aligned}$$

By the change of measure principle, we have,

$$\begin{aligned} F(\varDelta )&=\int _{\varLambda }\gamma _{\varDelta }(\lambda )\,d\widetilde{E}_{\lambda }=\int _{\varLambda }\gamma _{\varDelta }(f^{-1}(f(\lambda )))\,d\widetilde{E}_{\lambda }\\&=\int _{I}\gamma _{\varDelta }(f^{-1}(t))\,d E^A_{t}=\int _{I}\nu _{\varDelta }(t)\,d E^A_{t}=\nu _{\varDelta }(A) \end{aligned}$$

where $I=f(\varLambda )$ and $E^A$ is the spectral measure corresponding to A and defined by $E^A(\varDelta )=\widetilde{E}(f^{-1}(\varDelta ))$, $\varDelta \in \mathscr {B}(I)$. Therefore, for each $\varDelta \in \mathscr {R}(\mathscr {S})$, $\nu _{\varDelta }(f(\lambda ))=\gamma _{\varDelta }(\lambda )$, $\lambda \in \varLambda$, and $F(\varDelta )=\nu _{\varDelta }(A)$.

Now, we extend $\nu$ to all $\mathscr {B}(X)$. Since A is the generator of $\mathscr {A}^W(F)$, for each $\varDelta \in \mathscr {B}(X)$, there exists a Borel function $\omega _{\varDelta }$ such that.

$$\begin{aligned} F(\varDelta )=\int _{I}\omega _{\varDelta }(t)\,d E^A_t \end{aligned}$$

Then, we can consider the map $\beta :\sigma (A)\times \mathscr {B}(X)\rightarrow [0,1]$ defined as follows

$$\begin{aligned} \beta _{\varDelta }(\lambda )= {\left\{ \begin{array}{ll} \nu _{\varDelta }(\lambda ) &{} if \quad \varDelta \in \mathscr {R}(\mathscr {S})\\ \omega _{\varDelta }(\lambda ) &{} if \quad \varDelta \notin \mathscr {R}(\mathscr {S}). \end{array}\right. } \end{aligned}$$

(20)

which coincides with $\nu$ on $\mathscr {R}(\mathscr {S})$ and is such that $\beta _\varDelta (A)=F(\varDelta )$. Now, let $\psi _0\in \mathscr {H}$ be a separating vector for $\mathscr {A}^W(A)$ and $\nu _0(\cdot ):=\langle \psi _0,E^A(\cdot )\psi _0\rangle$. In order to prove that $\beta :(I,\nu _0)\times \mathscr {B}(X)\rightarrow [0,1]$ is a weak Markov kernel, we proceed as in the proof of Theorem 5 (see equation (14)). Note that $\nu _0(\varDelta )=0$ if and only if $E^A(\varDelta )=\textbf{0}$. $\square$