Proof.

The condition in the right-hand side of (3.1) states that $h=1$ on $\supp(f)$ and $h\perp g$, so the implication “$\Leftarrow$” is immediate. The implication “$\Rightarrow$” (i.e., the existence of such $h$ provided $f\perpp g$) follows from Urysohn's Lemma and path-connectedness of $S_X$.

Proof.

By Lemma 3.1, $T$ is a $\perpp$-isomorphism, so let $\phi$ be the $T$-homeomorphism. We prove that $T$ is $\phi$-basic in a few simple steps.

1. If $f,h\in C_c(X,\mathbb{R})$, then $f=1$ on $\supp(h)$ (i.e., $fh=h$) if and only if $Tf=1$ on $\supp(Th)$ (i.e., $(Tf)(Th)=(Th)$). This implies that

2. If $f,h\in C_c(X,\mathbb{R})$, then $f\neq 0$ on $\supp(h)$ if and only if $Tf\neq 0$ on $\supp(Th)$.

   Indeed, if $f\neq 0$ on $\supp(h)$, we can find $g\in C_c(X,\mathbb{R})$ such that $g=1/f$ on $\supp(h)$. Item 1. implies that $TfTg=1$ on $\supp(Th)$, and in particular $Tf\neq 0$ on $\supp(Th)$. As the sets of the form $\sigma(h)$ form a basis of $X$, we conclude that

3. If $f\in C_c(X,\mathbb{R})$ and $y\in Y$, then $f(\phi(y))\neq 0$ if and only if $Tf(y)\neq 0$.

   A similar argument to that of item 2. also proves that

4. If $f,g,h\in C_c(X,\mathbb{R})$, then $f$ and $g$ coincide and are nonzero on $\supp(h)$ if and only if $Tf$ and $Tg$ coincide and are nonzero on $\supp(h)$.

5. In particular, from 3., $f(\phi(y))=0$ if and only if $Tf(y)=0$.

6. If $f\in C_c(X,\mathbb{R})$ and $y\in Y$, then $f(\phi(y))=1$ if and only if $Tf(y)=1$:

   Suppose this was not the case, say $f(\phi(y))=1$ but $Tf(y)\neq 1$. From item 1. above we know that $y$ is not isolated. If necessary, as $T$ is a $\perp$-isomorphism, Theorem 1.16 allows us to change $f$ by a function which equals $1/f$ on some neighbourhood of $\phi(y)$ and assume that $Tf(y)\greaterthan 1$. The rest of the argument is similar to that of [MR0029476, Lemma 4.1]: We first take a sequence of distinct points $y_n$, all contained in some compact neighbourhood of $y$, satisfying

   • $f(\phi(y_n))^n\to 1=f(\phi(y))$; and

   • $Tf(y_n)\to Tf(y)$, so in particular $Tf(y_n)^n\to\infty$.

   Using Propositions 2.9 and 2.10(b), we may find $g\in C_c(X,\mathbb{R})$ which coincides with $f^n$ on some neighbourhood of $\phi(y_n)$. But then $Tg$ coincides with $(Tf)^n$ on a neighbourhood $y_n$ (because $T$ is a $\perp$-isomorphism), contradicting the fact that $Tg$ is bounded.

We conclude, from 5. and 6., that $T$ is $\phi$-basic. Let $\chi\colon Y\times\mathbb{R}\to\mathbb{R}$ be the $T$-transform.

Let $F=\left\{y\in Y:\chi(y,\cdot)\text{ is discontinuous}\right\}$. Let us prove that $F$ is closed and discrete, or equivalently that $F\cap K$ is finite for all compact $K\subseteq Y$. Otherwise, by Proposition 3.3, there would be distinct $y_1,y_2,\ldots\in K$ and a strictly decreasing sequence $t_n\to 0$ such that $\chi(y_n,t_n)\greaterthan n$. Going to a subsequence if necessary, we can construct, by Proposition 2.10(b), $f\in C_c(X,\mathbb{R})$ with $f(\phi(y_n))=t_n$, so $Tf(y_n)\greaterthan n$ for all $n$, a contradiction to the boundedness of $Tf$.

To prove that $F$ is isolated we prove that it is open: If $z\in F$, then there is $t\in(0,1)$ with $\chi(z,t)\greaterthan 1$. Take $f\in C_c(X,\mathbb{R})$ such that $f=t$ on a neighbourhood $U$ of $\phi(z)$. In particular $Tf(z)\greaterthan 1$, so there is a neighbourhood $W$ of $z$ such that $Tf\greaterthan 1$ on $W$. Then for all $y\in\phi^{-1}(U)\cap W$, $\chi(y,t)=Tf(y)\greaterthan 1$, so $y\in F$.

Therefore, $F$ consists of isolated points, since $Y$ is locally compact. For $y\not\in F$, Proposition 2.7 and Proposition 3.3 imply that $\chi(y,\cdot)$ has the form $\chi(y,t)=\operatorname{sgn}(t)|t|^{p(y)}$ for some $p(y)\greaterthan 0$. As for the continuity of $p$, Let $U$ be any relatively compact open subset of $Y$ not intersecting $F$, and $f\in C_c(X,\mathbb{R})$ with $f=2$ on $\phi(U)$. Then $Tf(y)=f(y)^{p(y)}=2^p(y)$, so $p=\log_{2}\circ Tf$ on $U$. This proves that $T$ is continuous on $Y\setminus F$.

Proof.

By Theorem 2.14 and Proposition 2.8, there is a homeomorphism $\phi\colon Y\to X$ such that $T$ is $\phi$-basic, and the sections $\chi(y,\cdot)\colon G\to H$ of the $(\phi,T)$-transform $\chi$ are group morphisms by Proposition 2.7. Similar facts hold for $T^{-1}$, so Proposition 2.6 implies that each section $\chi(y,\cdot)$ is bijective. Letting $w(y)=\chi(y,\cdot)$ we are done with the first part.

Now note that for all $\alpha\in G$ and $y\in Y$,

which implies that every $w(y)$ is continuous if and only if $T$ is continuous on the constant functions (because the map $\alpha\mapsto\overline{\alpha}$ from $G$ to $C(X,G)$ is a homeomorphism onto its image). In this case, from the equality we can readily see that $T$ is continuous. The last part, assuming also that $T^{-1}$ is continuous on the constant functions, is similar, using $T^{-1}$ and $w(y)^{-1}$ in place of $T$ and $w(y)$.

Proof.

Weak regularity is immediate from (L2) and the fact that $R$ does not have a supremum, and item (a) is trivial. Let us prove (b). First suppose $f\Subset g$ and $\mathcal{H}\subseteq\mathcal{A}$ is a bounded subset such that $h\subseteq f$ for all $h\in \mathcal{H}$. Let $\alpha\in R$ be an upper bound of $\bigcup_{h\in\mathcal{H}}h(X)$. From weak regularity and compactness of $\supp(f)$, we can take finitely many functions $k_1,\ldots,k_n$ such that $k_i\Subset g$, and for every $x\in\supp(f)$ there is some $i$ with $k_i(x)\greaterthan \alpha$. Letting $k=\bigvee_{i=1}^n k_i$ we obtain the desired properties.

Conversely, suppose that condition (K) holds. Let $\alpha$ be any upper bound of $f_0(X)$ and again take $\beta\greaterthan \alpha$. Let $\mathscr{H}=\left\{h\in\mathcal{A}_{\geq f_0}:h\leq \beta, h\subseteq f\right\}$. Let $k$ be an upper bound of $\mathcal{H}$ with $k\subseteq g$. By Property (L2), we have $\sigma(f)=\bigcup_{h\in\mathscr{H}}\sigma(h)$, so $k\geq\beta$ on $\sigma(f)$ and thus also on $\supp(f)$, which implies $f\Subset k$. Since $k\subseteq g$ then $f\Subset g$.

Proof.

We will use the superscript “$f_0$” as in Definition 1.1, that is, for all $f\geq f_0$.

and similarly with $Tf_0$ in place of $f_0$. Then $\phi$ is the only homeomorphism satisfying $\sigma^{f_0}(f)=\phi(\sigma^{Tf_0}(Tf))$, or equivalently $Z^{f_0}(f)=\phi(Z^{Tf_0}(Tf))$, for all $f\geq f_0$.

First, assume that $f\leq g$, and let $U=\operatorname{int}([f=g])$. For all $x\in[f\smallerthan g]$, choose a function $k_x\in\mathcal{A}(X)_{\geq f_0}$ such that

• $\sigma^{f_0}(k_x)\cap [f=g]=\varnothing$;

• $k_x(x)=g(x)$;

• $k_x\leq g$.

Then $g=\sup\left\{f,k_x:x\in[f\smallerthan g]\right\}$, so $Tg=\sup\left\{Tf,Tk_x:x\in[f\smallerthan g]\right\}$..

Let us prove that $Tf(y)=Tg(y)$ for all $y\in\phi(U)$. Since $U\subseteq\bigcap_{x\in[f\smallerthan g]}Z^{f_0}(k_x)$, then $\phi^{-1}(U)\subseteq\bigcap_{x\in[x\smallerthan g]}Z^{Tf_0}(Tk_x)$.

Given $y\in\phi(U)$, use property (L2) to find $h\in\mathcal{A}(Y)$ such that $h(y)=Tf(y)$ and $h=Tg$ outside of $\phi(U)$. Then $h'=(Tf\lor h)\land Tg$ is an upper bound of $\left\{Tf,Tk_x:x\in[f\smallerthan g]\right\}$, so it is also an upper bound of $Tg$, and in particular

so $Tg(y)=Tf(y)$.

In the general case, if $f=g$ on an open set $U$ then $f=f\land g=g$ on $U$, so the previous case implies that $Tf$ and $Tg$ both coincide with $T(f\land g)$ on $\phi^{-1}(U)$.

Proof.

For each $f_0\in \mathcal{A}(X)$, let $\phi^{f_0}\colon Y\to X$ be the $T|_{\mathcal{A}(X)_{\geq f_0}}$-homeomorphism. Given $f_0,g_0\in\mathcal{A}(X)$, Lemma 3.10 implies that $\phi^{f_0\land g_0}$ satisfies the property of both the $T|_{\mathcal{A}(X)_{\geq f_0}}$ and the $T|_{\mathcal{A}(X)_{\geq g_0}}$-homeomorphisms, so $\phi^{f_0}=\phi^{f_0\land g_0}=\phi^{g_0}$. We are done by letting $\phi=\phi^{f_0}$ for some arbitrary $f_0\in\mathcal{A}(X)$.

Proof.

Let $\phi\colon Y\to X$ be the $T$-homeomorphism. We just need to prove that $T$ is $\phi$-basic, so assume $y\in Y$ and $f(\phi(y))=g(\phi(y))$. In order to prove that $Tf(y)=Tg(y)$, we may assume that $f\leq g$, by considering the auxiliary function $f\land g$.

If $y$ is isolated in $Y$, then $f$ and $g$ coincide on the open set $\left\{\phi(y)\right\}$, so $Tf$ and $Tg$ coincide on the open set $\left\{y\right\}$.

Assume then that $y$ is not isolated. Since $Y$ is first-countable, let $(y_n)_n$ be an injective sequence in $Y$ converging to $y$. By Proposition 3.12, there is $h\in C_c(X,\theta)$ such that

• If $n$ is even, $h=f$ on a neighbourhood of $\phi(y_n)$;

• If $n$ is odd, $h=g$ on a neighbourhood of $\phi(y_n)$.

Then $\phi(y)\in\overline{\operatorname{int}[f=h]}$, so $t\in\overline{\operatorname{int}[Tf=Th]}$ and so $Tf(y)=Th(y)$. Similarly, $Tg(y)=Th(y)=Tf(y)$. This proves that $T$ is $\phi$-basic. Let $\chi$ be the $(\phi,T)$-transform.

Proposition 2.7, applied to the signature of lattices (with the binary symbol “$\lor$” interpreted as “join”) implies that the sections $\chi(y,\cdot)$ are lattice isomorphisms of $R$ for all $n$, and in particular homeomorphisms. The proof that $\chi$ is continuous is similar to that of implication (ii)$\Rightarrow$(i) of Theorem 2.11 (using Proposition 3.12 instead of 2.10).

Proof.

First note that for all $f\in C_c(X)$, $|f|=(f\lor 0)-(f\land 0)$, so $T|f|=|Tf|$. Let $\phi\colon Y\to X$ be the $T$-homeomorphism, given by Theorem 3.11.

The proof that $T$ is $\phi$-basic is similar to that of item 6. of the proof of Theorem 3.4. Let $\chi$ be the $T$-transform. Each section $\chi(y,\cdot)$ is an additive order-preserving bijection (Propositions 2.7 and 2.6) and hence has the form $\chi(y,t)=p(y)t$ for some $p(y)\greaterthan 0$. If $Tf(y)\neq 0$, then $f(\phi(y))\neq 0$ as well and $p=Tf/(f\circ \phi)$ on a neighbourhood of $y$, thus $p$ is continuous.

Proof of Theorem 3.15.

We will use Theorem 3.11. First we need to modify $T$ to obtain a lattice isomorphism. Since $\mathcal{A}(X)$ is a sublattice, then for all $f\in\mathcal{A}(X)$,

As $\mathcal{A}(X)$ is regular and $X$ is compact, we can take finitely many functions $f_1,\ldots,f_n\in\mathcal{A}(X)$ such that for all $x\in X$, $f_i(x)\neq 0$ for some $i$, and therefore the function $F=\sum_{i=1}^n|f_i|$ belongs to $\mathcal{A}(X)$ and is non-vanishing, so $TF$ is also non-vanishing. We define new classes of functions

It is immediate to see that $\mathcal{B}(X)$ and $\mathcal{B}(Y)$ are regular, and contain the constant functions of $X$ and $Y$, respectively. Define a linear isomorphism $S\colon \mathcal{B}(X)\to\mathcal{B}(Y)$, $S(f)=T(fF)/TF$, which preserves non-vanishing functions and satisfies $S(1)=1$. Given a scalar $\lambda$, linearity and the non-vanishing property of $S$ imply that, for all $f\in\mathcal{B}(X)$, so $f(X)=Sf(Y)$, i.e., $S$ preserves images of functions.

However, in order to apply Theorem 3.11 we also need to make sure that $\mathcal{B}(X)$ and $\mathcal{B}(Y)$ are lattices. As $F\greaterthan 0$, it readily follows that $\mathcal{B}(X)$ is a (self-adjoint) sublattice of $C(X,\mathbb{K})$, however this is not so immediate for $\mathcal{B}(Y)$.

As $S$ preserves images of functions, it preserves real functions. If $f\in\mathcal{B}(X)$, then $S(\operatorname{Re}(f))$ and $S(\operatorname{Im}(f))$ are real functions such that $Sf=S(\operatorname{Re}(f))+iS(\operatorname{Im}(f))$. As $T$ is a $\perp$-isomorphism then $S$ is also a $\perp$-isomorphism, so we also obtain $S(\operatorname{Re}(f))\perp S(\operatorname{Im}(f))$. This is enough to conclude that $S$ preserves real and imaginary parts of functions, from which it follows that $\mathcal{B}(Y)$ is self-adjoint. Similarly, $S$ preserves positive and negative parts of functions. In particular, if $f\in\mathcal{B}(Y)$ then $f^+\in\mathcal{B}(Y)$, and this is enough to conclude that $\mathcal{B}(Y)$ is a sublattice of $C(Y,\mathbb{K})$, and that $S$ is an order-preserving isomorphism.

We may then consider only real-valued functions, and the complex case will follow by linearity (and since $S$ preserves real and imaginary parts). By Kaplansky's Theorem (3.11), we can construct the $S$-homeomorphism $\phi\colon Y\to X$. Now we need to prove that $S$ is $\phi$-basic.

Suppose $f(x)\neq 0$ for a given $x\in X$, and let us assume, without loss of generality, that $f(x)\greaterthan 0$. Then $f\greaterthan 0$ on some neighbourhood $U$ of $x$. Again using compactness of $X\setminus U$ and regularity of the sublattice $\mathcal{B}(X)$ we can construct a function $g\in\mathcal{B}(X)$ such that $g=0$ on some neighbourhood of $x$ and $g\greaterthan 0$ on $X\setminus U$. Letting $\widetilde{f}=f\lor g$, we have $\widetilde{f}=f$ on some neighbourhood of $x$, so $S\widetilde{f}=Sf$ on some neighbourhood of $\phi^{-1}(x)$. But $\widetilde{f}$ is non-vanishing, so $S\widetilde{f}$ is also non vanishing and in particular $Sf(\phi^{-1}(x))\neq 0$. This proves that $S$ is basic with respect to $\phi$.

Letting $\chi\colon Y\times\mathbb{R}\to\mathbb{R}$ be the $S$-transform, we have that all sections $\chi(y,\cdot)$ are linear and increasing (Theorem 2.7), hence of the form $\chi(y,t)=P(y)t$ for a certain $P(y)\greaterthan 0$. Since $S(1)=1$ then $P=1$, that is, $\chi(y,t)=t$ for all $t\in\mathbb{R}$.

Finally, for all $f\in\mathcal{A}(X)$ and $y\in Y$,

as we wanted.

Proof.

First assume that $X$ and $Y$ are compact, and let us show that $f\neq 0$ everywhere if and only if $Tf\neq 0$ everywhere. Suppose otherwise, say $f(x)=0$, and we have two cases: first, if $f$ is constant on a neighbourhood of $x$, this means that $Z(f)\neq \varnothing$, and Theorem 1.16 implies that $Z(Tf)\neq \varnothing$, and in particular $Tf(y)=0$ for any $y\in Z(Tf)$.

In the second case, if $f$ is not constant on any neighbourhood of $x$, an argument similar to the one in the proof of Theorem 3.4 yields a contradiction to $Tf$ being bounded, so $Tf(y)=0$ for some $y\in Y$.

The result follows in this case from the Li‒Wong Theorem (Theorem 3.15).

Now let $X$ and $Y$ be arbitrary locally compact Hausdorff. Given $b\in C_c(X)$, set $T_b\colon C(\supp(b))\to C(\supp(Tb))$ as $T_bf=(Tf')|_{\supp(Tb)}$, where $f'$ is any element of $C_c(X)$ extending $f$. Note that $T_bf$ does not depend on the choice of $f'$, because $T$ is an additive $\perp$-isomorphism (and Theorem 1.16).

Since $f\perpp g$ if and only if $f|_{\supp(b)}\perpp g|_{\supp(b)}$ for all $b$, the previous case allows us to obtain functions $p^b$ and $\phi^b$ such that $Tf(y)=p^b(y)f(\phi^b(y))$ for all $y\in\supp(Tb)$. Clearly, if $b\subseteq b'$ then $p^{b'}|_{\supp(b)}=p^b$ and $\phi^{b'}|_{\supp(b)}=\phi^{b'}$. Thus defining $p$ and $\phi$ as $p(y)=p^b(y)$ and $\phi(y)=\phi^b(y)$ where $b\in C_c(X)$ is such that $y\in\supp(b)$ we obtain the desired maps.

Proof.

By the previous lemma, $T$ is a $\perp$-isomorphism, so Jarosz' Theorem (3.17) implies that there are a homeomorphism $\phi\colon Y\to X$ and a non-vanishing continuous function $P\colon Y\to\mathbb{C}$ such that $T(f)(y)=P(y)f(\phi(y))$ for all $f\in C_c(X)$ and $y\in Y$.

Now using the fact that $T$ is isometric, we have, for every $f\in C_c(X)$,

which means that $|P\circ\phi^{-1}|$ is a continuous instance of the Radon‒Nikodym derivative $d\mu_X/d(\phi_*\mu_Y)$. Since $p=P/|P|\colon Y\to\mathbb{S}^1$ is continuous, we obtain the result.

Proof.

From the second condition one can construct two nets $(a_i)_i$ and $(b_i)_i$ (over the same ordered set) converging to $a$ and $b$, respectively, such that $\so(a_i)=\ra(b_i)$, and so $\so(a)=\ra(b)$ because $\mathcal{G}^{(0)}$ is Hausdorff. The reverse implication is trivial.

Proof.

Using invariance of $\mu$ and $\lambda$, we have, for every $f\in C_c(\mathcal{G}^{\so(a)})$,

Thus $t\mapsto D^{\ra(a)}(at)$ satisfies the property of the Radon‒Nikodym derivative $d\lambda^{\so(a)}/d\mu^{\so(a)}=D^{\so(a)}$, hence these functions coincide $\mu^{\so(a)}$-a.e.

Proof.

Again applying Lemma 3.19 and Jarosz' Theorem (3.17), we can find a homeomorphism $\phi\colon\mathcal{H}\to\mathcal{G}$ and a continuous non-vanishing scalar function $P$ such that

Let us check that $\phi$ is a groupoid morphism. Suppose $(a,b)\in\mathcal{H}^{(2)}$, and consider neighbourhoods $U$ and $V$ of $\phi(a)$ and $\phi(b)$, respectively.

Choose non-negative functions $f_U,f_V\in C_c(\mathcal{H})$ such that

Then $ab\in\supp(f_af_b)$, because $\lambda_{\mathcal{H}}$ has full support, and so $\phi(ab)\in\supp(T^{-1}(f_af_b))$. As $\phi$ is the $T$-homeomorphism and $T$ is an isomorphism, the inclusion in (3.4) implies $\phi(ab)\subseteq UV$. By Lemma 3.23, the product $\phi(a)\phi(b)$ is defined, and moreover, continuity of the product implies that every neighbourhood of $\phi(a)\phi(b)$ contains $\phi(ab)$. Since $\mathcal{G}$ is Hausdorff then $\phi(ab)=\phi(a)\phi(b)$. Therefore $\phi$ is a morphism and a homeomorphism, thus a topological groupoid isomorphism.

We now need to rewrite the function $P$ as $P(h)=p(h)D(\phi(h))$ as in the statement of the theorem, and to this end we use multiplicativity of $T$ and compare integrals. If $f,g\in C_c(\mathcal{G})$ and $c\in\mathcal{H}$, then on one hand

and on the other

Now let $f\in C_c(\mathcal{G}^{\phi(\ra(c))})$ be an arbitrary non-negative function. Define $g\in C_c(\mathcal{G}_{\phi(\so(c))})$ by $g(t)=f(\phi(c)t^{-1})$. Extending $f$ and $g$ arbitrarily to elements of $C_c(\mathcal{G})$, Equations (3.5) and (3.6) become

for all non-negative $f\in C_c(\mathcal{G}^{\phi(\ra(c))})$ and all $c\in\mathcal{H}$. Define $D\colon\mathcal{G}\to\mathbb{C}$ (or $\mathbb{R}$ in the real case) by Using Equation (3.7) with $y=\ra(c)$ in place of $c$, we obtain for all $f\in C_c(\mathcal{G}^{\phi(y)})$, thus $D$ is a continuous instance of the Radon‒Nikodym derivative Now applying this to Equation (3.7), and using regularity of all measures involved, we conclude that for $\lambda_{\mathcal{G}}^{\phi(\ra(c))}$-a.e. $s\in\mathcal{G}^{\phi(\ra(c))}$ and since all functions involved are continuous, and $\lambda_{\mathcal{G}}^{\phi(\ra(c))}$ has full support, the same equality is actually valid for all $s\in\mathcal{G}^{\phi(\ra(c))}$. Equivalently, for all $c\in\mathcal{H}$ and all $t\in\mathcal{H}^{\ra(c)}$, $P(t)P(t^{-1}c)=D(\phi(t))P(c)$. Together with Lemma 3.24, this implies that the map $p=P/(D\circ\phi)$ is a continuous groupoid morphism from $\mathcal{H}$ to the group of non-zero scalars.

Now let us verify that $\mu_{\mathcal{G}}$ and $\phi_*\mu_{\mathcal{H}}$ are equivalent measures. By regularity of the measures, we may extend the formula $T(f)=P\cdot (f\circ\phi)$ to obtain a linear isomorphism between the classes of measurable functions on $\mathcal{G}$ and on $\mathcal{H}$, and which restricts to an isometry from $L^1(\lambda_{\mathcal{G}}\circ\mu_{\mathcal{G}})$ to $L^1(\lambda_{\mathcal{H}}\circ\mu_{\mathcal{H}})$. Suppose that $K\subseteq\mathcal{G}^{(0)}$ has positive $\mu_{\mathcal{G}}$-measure, and let $f=1_{\ra^{-1}(K)}$ be the characteristic function of $\ra^{-1}(K)$. Then $Tf=P\cdot 1_{\phi^{-1}(\ra(K))}$. As $\Vert f\Vert_{\mathcal{G}}\greaterthan 0$, then $\Vert Tf\Vert_{\mathcal{H}}\greaterthan 0$ as well, so the support of $Tf$ has positive $(\lambda_{\mathcal{H}}\circ\mu_{\mathcal{H}})$-measure, and this implies that $\ra(\supp(Tf))=\phi^{-1}(K)$ has positive $\mu_{\mathcal{H}}$-measure. Therefore $\mu_{\mathcal{G}}$ is absolutely continuous with respect to $\phi_*\mu_{\mathcal{H}}$. The reverse implication is similar.

To prove that $p$ takes value in $\mathbb{S}^1$, let us denote by $\Vert\cdot\Vert_Z$ the $L^1$-norm with respect to $(\lambda_Z\circ\mu_Z)$ when $Z\in\left\{\mathcal{G},\mathcal{H}\right\}$. For all $f\in C_c(\mathcal{G})$ we have

and since $\Vert Tf\Vert_{\mathcal{H}}=\Vert f\Vert_{\mathcal{G}}=\int_{\mathcal{G}}|f|d(\lambda_{\mathcal{G}}\circ\mu_{\mathcal{G}})$, we obtain Since $p$ is a morphism then $p(\mathcal{H}^{(0)})=\left\{1\right\}$, which, along with Equation (3.8) and continuity of $p$, yields $|p|=|p\circ\ra|=1$ on $\mathcal{H}$. The same Equation (3.8) then also implies $\mu_{\mathcal{G}}=\phi_*\mu_{\mathcal{H}}$.

Proof.

By the Banach‒Stone Theorem (3.18), there is a homeomorphism $\phi\colon\mathcal{H}^{(0)}\to\mathcal{G}^{(0)}$ and a continuous function $P\colon\mathcal{H}^{(0)}\to\mathbb{S}^1$ such that $Tf(y)=P(y)f(\phi(y))$ for all $f\in C_c(\mathcal{G}^{(0)})$ and $y\in\mathcal{H}^{(0)}$. Since $T$ is multiplicative we obtain $P=1$. (The same conclusion can be obtained in a similar manner by Milgram's or Jarosz' Theorem.)

For each $x\in\mathcal{G}^{(0)}$, let $\left\{a^x_i:i\in I_x\right\}$ be a net of functions in $C_c(\mathcal{G}^{(0)})$ satisfying:

1. $0\leq a^x_i\leq 1$, and $a^x_i(x)=1$;

2. $\bigcap_i\supp(a^x_i)=\left\{x\right\}$;

3. If $j\geq i$ then $[a^x_j\neq 0]\subseteq[a^x_i\neq 0]$.

Items (ii)-(iii) and compactness of each $\supp(a^x_i)$ imply that $\left\{[a^x_i\neq 0]:i\in I_x\right\}$ is a neighbourhood basis at $x$. For $y\in\mathcal{H}^{(0)}$, let $a^y_i=T(a^{\phi(y)}_i)=a^{\phi(y)}_i\circ\phi$, so that the net $\left\{a^y_i:i\in I_{\phi(y)}\right\}$ satisfies (i)-(iii) as well.

Continuity of $\lambda_G$ implies that for all $x\in \mathcal{G}^{(0)}$ and $f\in C_c(\mathcal{G})$, $\lim_{i\in I_x}\Vert a^x_i f\Vert_{I,\ra}=\int |f(\gamma)|d\lambda^x(\gamma)$, and similarly on $\mathcal{H}$.

Given $f,g\in C_c(\mathcal{G})$, we use Lemma 3.19 to obtain

and the last condition is preserved by $T$, so by Jarosz' Theorem $T$ is of the form $Tf(\alpha)=\widetilde{P}(\alpha)f(\widetilde{\phi}(\alpha))$ for a certain homeomorphism $\widetilde{\phi}\colon\mathcal{H}\to\mathcal{G}$ and a non-vanishing continuous scalar function $\widetilde{P}$. We can readily see that $\widetilde{\phi}$ and $\widetilde{P}$ are extensions of $\phi$ and $P$, respectively, so instead let us simply denote $\widetilde{\phi}=\phi$ and $\widetilde{P}=P$.

The proof that $\phi$ is a groupoid isomorphism, and that $P$ can be decomposed as $P=(D\circ\phi)p$ for the (continuous) Radon‒Nikodym derivative $D$ and some continuous morphism $p\colon\mathcal{H}\to\mathbb{C}\setminus\left\{0\right\}$ is the same as in Theorem 3.25, but the verification that $|p|=1$ is different.

Given $y\in\mathcal{H}^{(0)}$ and $f\in C_c(\mathcal{G})$, using the definition of $D$ as a Radon‒Nikodym derivative,

Considering again the functions $a^{\phi(y)}_i$ and $a^y_i$, and the fact that $T$ is isometric we obtain for all $f\in C_c(\mathcal{G})$, which implies that $|p|=1$ $\lambda_{\mathcal{H}}^y$-a.e. Since $p$ is continuous and $\lambda_{\mathcal{H}}$ is fully supported, we conclude that $|p|=1$ on $\mathcal{H}$.

Proof.

Let $g$ be an inverse of $f$ relatively to $D_R(\mathcal{G})$. First note that $fg=f1_{\so(\supp(f))}g\in D_R(\mathcal{G})$.

Let $a\in\supp(f)$. Since $fg$ is an idempotent in $D_R(\mathcal{G})$, the product in $D_R(\mathcal{G})$ is pointwise and $R$ is indecomposable, then $fg(\ra(a))\in\left\{0,1\right\}$. Moreover, as $\supp(f)$ is a bisection we have $f(a)=fgf(a)=(fg)(\ra(a))f(a)$, so $(fg)(\ra(a))=1$.

Again using that $\supp(f)$ is bisection, we obtain

so $f(a)$ is invertible in $R$.

Proof.

The cocycle condition states that $\chi(ab)(rs)=\chi(a)(r)\chi(b)(s)$ for all $a,b,r,s$. Taking $a=b=x$ yields (a). Taking $b=a^{-1}$, $r=u$ and $s=u^{-1}$ yields (b), and for item (c) we use commutativity of $S$:

Proof.

Since $T$ preserves the respective diagonal algebras, it also preserves their normalizers, i.e., $T(N_R(\mathcal{G}))=N_S(\mathcal{H})$. Let us describe disjointness first for elements in $N_R(\mathcal{G})$. The local bisection hypothesis implies, by Lemma 3.32, that an element $f$ of $N_R(\mathcal{G})$ has the form

where $\lambda_1,\ldots,\lambda_n$ are invertible elements in $R$ and $U_1,\ldots,U_n$ are disjoint compact-open bisections of $\mathcal{G}$ such that $\bigcup_{i=1}^n U_i=\supp(f)$ is also a compact-open bisection. A similar statement holds for $N_S(\mathcal{H})$.

If $f,g\in N_R(\mathcal{G})$, then $f\subseteq g$ if and only if $f=g p$ for some $p\in D_R(\mathcal{G})$: Indeed,

• If $f=gp$ then $\supp(f)\subseteq\supp(g)\supp(p)\subseteq\supp(g)$;

• Conversely, if $\supp(f)\subseteq\supp(g)$ take $p=g^*f$. Then

  \begin{equation*} \supp(p)\subseteq\supp(g^*)\supp(f)\subseteq(\supp(g))^{-1}\supp(g)\subseteq\mathcal{G}^{(0)} \end{equation*}
  where the last inclusion follows from $\supp(g)$ being a bisection. The equality $f=gp$ follows from the definition of $p$ and since $\ra(\supp(f))\subseteq\ra(\supp(g))$.

Therefore $T$ preserves inclusion of normalizers. Since $N_R(\mathcal{G})$ contains $\left\{1_U:U\in\KB(\mathcal{G})\right\}$ then it is regular (Definition 1.5), because $\KB(\mathcal{G})$ is a basis for the topology of $\mathcal{G}$. Hence $T$ also preserver disjointness of normalizers (Proposition 1.8).

To prove that $T$ preserves disjointness in all of $A_R(\mathcal{G})$, we decompose elements of $A_R(\mathcal{G})$ in terms of elements of $N_R(\mathcal{G})$ and $D_R(\mathcal{G})$: if $f,g\in A_R(\mathcal{G})$, then $f\perp g$ if and only if there are finite collections of normalizers $f_i,g_j\in N_R(\mathcal{G})$ and elements $\widetilde{f_i},\widetilde{g_j}\in D_R(\mathcal{G})$ ($1\leq i\leq n$, $1\leq j\leq m$) such that

Indeed, if there are such $f_i,g_j,\widetilde{f_i},\widetilde{g_j}$ then $\supp(f)\subseteq\bigcup_i\supp(f_i)$ and $\supp(g)\subseteq\bigcup_j\supp(g_j)$, and the latter sets are disjoint.

Conversely, we write $f=\sum_i\lambda_i1_{A_i}$, where the $A_i$ are pairwise disjoint compact-open bisections and $\lambda_i\neq 0$, and take $f_i=1_{A_i}$ and $\widetilde{f_i}=\lambda_i1_{\so(A_i)}$, so that $\supp(f)=\bigcup_{i=1}^n\supp(f_i)$. Similarly, writing $g=\sum_j\widetilde{g_j}g_j$ where $g_j\in N_R(\mathcal{G})$ and $\supp(g)=\bigcup_j\supp(g_j)$, then $f\perp g$ implies $f_i\perp g_j$ for all $i$ and $j$.

Therefore, $T$ is a $\perp$-isomorphism. Note that $\perpp$ and $\perp$ coincide in $A_R(\mathcal{G})$, since its elements are locally constant (similarly to Example 1.4). Then $T$ is a $\perpp$-isomorphism, so let $\phi\colon\mathcal{H}\to\mathcal{G}$ be the $T$-homeomorphism. The verification that $\phi$ is a groupoid isomorphism is similar to that of Theorem 3.25.

Since elements of $A_R(\mathcal{G})$ (and $A_S(\mathcal{H})$) are locally constant, then for all $f\in A_R(\mathcal{G})$,

and therefore $T$ is basic (by additivity of $T$ and Proposition 2.8). Let $\chi$ be the $T$-transform. Since $T$ is additive with the pointwise operations, each section $\chi(\alpha)=\chi(\alpha,\cdot)$ is additive (by Proposition 2.7). This yields a map $\chi\colon\mathcal{G}\to\operatorname{Iso}_+(R,S)$, and we need now to verify that $\chi$ is a cocycle.

If $(a,b)\in\mathcal{H}^{(2)}$ and $r,s\in R$, choose compact-open bisections $U,V$ of $\mathcal{G}$ containing $\phi(a)$ and $\phi(b)$, respectively. Then using multiplicativity of $T$ we obtain

If $cd=ab$ is such that the last term above is nonzero, then $\ra(c)=\ra(a)$ and $\phi(c)\in U$, so since $U$ is a bisection we obtain $a=c$. Similarly, $d=b$, therefore $\chi(ab)(rs)=\chi(a)(r)\chi(b)(s)$, and $\chi$ is a cocycle.

It remains only to prove that $\chi$ is continuous: Let $r\in R$ be fixed, $a\in\mathcal{H}$ and $U$ any compact-open bisection containing $\phi(a)$. For all $b\in\phi^{-1}(U)$,

which means that the map $b\mapsto\chi(b)(r)$ coincides with $T(r1_U)$ on $\phi^{-1}(U)$ and thus it is continuous.

Proof.

For every point $y\in F$, choose a clopen set $U_y$ containing $y$ such that $f(U_y)\cup g(U_y)$ does not contain $1$. For every other point $x\in X':=X\setminus\bigcup_{y\in F}U_y$, there is a clopen set $U\subseteq X'$ such that $f(U)\cup g(U)\neq\mathbb{S}^1$. Using compactness of $X'$ and taking complements and intersections if necessary we can find a clopen partition $U_1,\ldots,U_n$ of $X'$ such that $f(U_i)\cup g(U_i)\neq\mathbb{S}^1$ for all $i$. Simply choose $z_i\in \mathbb{S}^1\setminus(f(U_i)\cup g(U_i))$ and define $h=z_i$ on $U_i$, and $h=1$ on $\bigcup_{f\in F}U_f$.

Proof of Theorem 3.43.

For the notion of support we will use (Definition 1.1), we take $\theta=1$, the constant function at $1$, so regularity of $C(X,\mathbb{S}^1)$ is immediate.

Suppose that $f\perp g$ but that $Tf$ and $Tg$ are not disjoint. By Lemma 3.44, there exists $H\in C(Y,\mathbb{S}^1)$ such that $H\neq Tf,Tg$ everywhere, but that $1\in H(Y)$. Let $h=T^{-1}H$. Then $T(f^{-1}h)$ and $T(g^{-1}h)$ do not attain $1$, which implies that $f^{-1}h$ and $g^{-1}h$ do not attain $1$ as well. Thus

But $(Th)^{-1}(1)=H^{-1}(1)$ is nonempty, contradicting the given property of $T$.

Therefore $f\perp g$ implies $Tf\perp Tg$, and the same argument yields the opposite implication, so $T$ is a $\perp$-isomorphism. Let $\mathcal{A}(X)$ and $\mathcal{A}(Y)$ be the subgroups of order-$2$ elements of $C(X,\mathbb{S}^1)$ and $C(Y,\mathbb{S}^1)$, respectively (i.e., the groups of continuous functions with values in $\left\{-1,1\right\}$).

$\mathcal{A}(X)$ and $\mathcal{A}(Y)$ are also regular, since $X$ and $Y$ are zero-dimensional, and the restriction $T|_{\mathcal{A}(X)}\colon\mathcal{A}(X)\to \mathcal{A}(Y)$ is a $\perpp$-isomorphism, because $\perp$ and $\perpp$ coincide on $\mathcal{A}(X)$ and $\mathcal{A}(Y)$. Let $\phi\colon Y\to X$ be the corresponding $T|_{\mathcal{A}(X)}$-homeomorphism.

Let $h\in C(X,\mathbb{S}^1)$ be arbitrary. Since $\sigma(h)=\bigcup_{a\in \mathcal{A}(X),a\subseteq h}\sigma(a)$ and $T$ is a $\perp$-isomorphism, we obtain by Theorem 1.16 that, for all $h\in C(X,\mathbb{S}^1)$,

Since $\phi$ is a homeomorphism it preserves closures, from which it follows that $T$ is also a $\perpp$-isomorphism, and $\phi$ is also the $T$-homeomorphism.

Claim: $f(\phi(y))=1\iff Tf(y)=1$.

Therefore $T$ is basic. Let $\chi$ be the $T$-transform, so that each section $\chi(y,\cdot)$ is an automorphism of the circle. If $\chi(y,\cdot)$ is continuous then it has the form $\chi(y,z)=z^{p(y)}$ where $p(y)\in\left\{\pm1\right\}$. Let us prove that for all except finitely many $y\in Y$, the section $\chi(y,\cdot)$ is continuous.

Let $F=\left\{y\in Y:\chi(y,\cdot)\text{ is discontinuous}\right\}$, and suppose that $F$ were infinite. By Proposition 2.9, there are countably infinitely many distinct points $y_n\in F$ ($n\in\mathbb{N}$), such that no $y_n$ lies in the closure of the other ones. We can choose a sequence $z_n\to 1$ such that $\chi(y_n,z_n)$ lies in the second quadrant of the circle. Define $f(\phi(y_n))=z_n$, $f=1$ on the boundary of $\{\phi(y_n):n\in\mathbb{N}\}$ and extend $f$ continuously to all of $X$. Let $y$ be a cluster point of $\{y_n\}_n$, so that in particular $f(\phi(y))=1$. Then

But $y$ is an accumulation point of the $y_n$, and $Tf(y_n)$ lies in the second quadrant while $Tf(y)=1$, a contradiction to the continuity of $Tf$.

Therefore $F$ is finite, so now we show that it is open in order to conclude that its points are isolated in $Y$. Let $y\in Y$ and choose $z_0\in\mathbb{S}^1$ of the form $z_0=e^{it}$ where $-\pi/4\leq t\leq \pi/4$, but such that $\chi(y,z_0)$ is in the second or third quadrant, so in particular it is not $z_0$ nor $z_0^{-1}$. Denote by $z_0$ the constant function at $z_0$, we that

Since $T(z_0)$ is continuous, there is a neighbourhood $U$ of $y$ such that $\chi(x,z_0)\neq z_0,z_0^{-1}$ for all $x\in U$, so $x\in F$.

Therefore $Y'=Y\setminus F$ is also compact, and we already constructed the function $p\colon Y'\to\{\pm 1\}$ with the desired property. To see that $p$ is continuous, denote by $i$ the constant function $x\mapsto i$ and note that

and similarly $p^{-1}(-1)=T(i)^{-1}(-i)\cap Y'$, so these two sets, which are complementary in $Y'$, are closed and hence clopen.

Proof.

We identify each $\lambda\in\mathbb{S}^1$ with the corresponding constant map on $X$ or $Y$. The constant function $-1$ is characterized by the following two properties:

• $(-1)^2=1$;

• If $g^3=1$, then $d_\infty(-1,g)\in\left\{1,2\right\}$.

Thus $T(-1)=-1$. A function $f$ does not attain $1$ if and only if $d_\infty(-1,f)\smallerthan 1$, so $T$ preserves functions not attaining $1$, and we apply Theorem 3.43 (or more precisely its proof) in order to obtain a homeomorphism $\phi\colon Y\to X$, a function $\chi\colon Y\times\mathbb{S}^1\to\mathbb{S}^1$ and a continuous function $p\colon Y'\to\left\{-1,1\right\}$, where $Y'=\left\{y\in Y:\chi(y,\cdot)\text{ is continuous}\right\}$, such that

for all $y\in Y$, $y'\in Y'$ and $f\in C(X,\mathbb{S}^1)$. It remains only to prove that $Y'=Y$, i.e., every section $\chi(y,\cdot)$ is continuous.

If $\lambda_i\to\lambda$ in $\mathbb{S}^1$ then we also have uniform convergence of the corresponding constant functions, so

thus $\chi(y,\cdot)$ is continuous for all $y$.