A system of imprimitivity integrates to a nondegenerate representation of the transformation algebra

For fixed $g$ put $M_{f(g,\cdot )}={\int }_{X}f(g,\cdot )dP$; by [F1] this is a bounded operator and $\mathrm{\parallel }{M}_{f(g,\cdot )}\mathrm{\parallel }\le \mathrm{\parallel }f(g,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}$. The map $g\mapsto M_{f(g,\cdot )}$ is norm continuous: if $K_G$ is the compact group projection of $\mathrm{supp}f$, then for $g,g_0\in K_G$ and $\epsilon >0$ uniform continuity of $f$ on the compact set $K_G\times (\text{compact }x\text{-support})$ gives $\mathrm{\mid }f(g,x)-f(g_0,x)\mathrm{\mid }\le \epsilon$ for all $x$ once $g$ is close to $g_0$, whence $\mathrm{\parallel }{M}_{f(g,\cdot )}-M_{f(g_0,\cdot )}\mathrm{\parallel }\le \mathrm{\parallel }f(g,\cdot )-f(g_0,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$; and $M_{f(g,\cdot )}=0$ for $g\mathrm{\notin }{K}_G$. Consequently $g\mapsto M_{f(g,\cdot )}{U}_g\xi$ is strongly continuous for every $\xi \in H$ (product of a norm-continuous and a strongly continuous factor) and supported in the compact set $K_G$.

Covariance in operator form: conjugating $M_b=\int bdP$ by the unitary $U_h$ and using $U_{h}P(E)U_h^{-1}=P(hE)$ gives $U_h{M}_b{U}_h^{-1}=\int b(h^{-1}x)dP(x)=M_{b\circ h^{-1}}$, that is $U_h{M}_b=M_{b\circ h^{-1}}{U}_h$ for every bounded Borel $b$ and $h\in G$.

Nondegeneracy, first factor: for every $\xi \in H$, $\mathrm{\parallel }U(e_U)\xi -\xi \mathrm{\parallel }\to 0$ along the approximate identity of [F7], where $U(a):={\int }_{G}a(g)U_{g}dg$ is the Bochner integral in $H$; indeed $\mathrm{\parallel }U(e_U)\xi -\xi \mathrm{\parallel }\le {\int }_G{e}_U(g)\mathrm{\parallel }{U}_g\xi -\xi \mathrm{\parallel }dg$ and, given $\epsilon >0$, strong continuity gives an identity neighbourhood $V$ with $\mathrm{\parallel }{U}_g\xi -\xi \mathrm{\parallel }<\epsilon$ for $g\in V$, while for $U\subseteq V$ the support condition and $\mathrm{\parallel }{e}_U{\mathrm{\parallel }}_1=1$ make the last integral at most $\epsilon$.

Define $\pi (f)\xi :={\int }_G{M}_{f(g,\cdot )}{U}_g\xi dg$ as a Bochner integral: strong measurability follows from [step 1.1], and ${\int }_G\mathrm{\parallel }{M}_{f(g,\cdot )}{U}_g\xi \mathrm{\parallel }dg\le \mathrm{\parallel }f{\mathrm{\parallel }}_{1,\mathrm{\infty }}\mathrm{\parallel }\xi \mathrm{\parallel }$ with $C(f):={\int }_G\mathrm{\parallel }f(g,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}dg<\mathrm{\infty }$ because the integrand vanishes off $K_G$ and is bounded there by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ on a compact set of finite Haar measure. Hence $\pi (f)$ is a well-defined bounded operator with $\mathrm{\parallel }\pi (f)\xi \mathrm{\parallel }\le C(f)\mathrm{\parallel }\xi \mathrm{\parallel }$ for every $\xi$, so $\mathrm{\parallel }\pi (f)\mathrm{\parallel }\le C(f)$. Pairing with $\eta$ and commuting the bounded functional $⟨\cdot ,\eta ⟩$ through the Bochner integral gives exactly the displayed identity $⟨\pi (f)\xi ,\eta ⟩={\int }_G⟨M_{f(g,\cdot )}{U}_g\xi ,\eta ⟩dg={\int }_G{\int }_{X}f(g,x)d{E}_{U_g\xi ,\eta }(x)dg$. The map $f\mapsto \pi (f)$ is complex-linear because the integrand is bilinear in $(f,\xi )$ and the Bochner integral is linear.

Nondegeneracy, second factor: $M_{b_n}\to I$ strongly for the sequence of [F8], by the pointwise dominated convergence of [F1], since $b_n\to 1$ pointwise on $X$ and $0\le b_n\le 1$. For the product function $(g,x)\mapsto e_U(g)b_n(x)$ one has $\pi (e_U\otimes b_n)=M_{b_n}U(e_U)$, because the bounded operator $M_{b_n}$ commutes with the Bochner integral ${\int }_G{e}_U(g)M_{b_n}{U}_{g}dg=M_{b_n}{\int }_G{e}_U(g)U_{g}dg$. Given $\xi$ and $\epsilon >0$ choose $n$ with $\mathrm{\parallel }{M}_{b_n}\xi -\xi \mathrm{\parallel }<\epsilon \mathrm{/}2$ and then $U$ small enough that $\mathrm{\parallel }U(e_U)\xi -\xi \mathrm{\parallel }<\epsilon \mathrm{/}2$; then $\mathrm{\parallel }\pi (e_U\otimes b_n)\xi -\xi \mathrm{\parallel }<\epsilon$. Hence the closed span of $\pi (C_c(G\times X))H$ contains every $\xi$, so $\pi$ is nondegenerate.

Multiplicativity: for $\xi ,\eta \in H$, using [step 2.1] twice, [step 1.2] with $b=f_2(r,\cdot )$ and $h$, and the left-Haar substitution $r=h^{-1}g$ (so $hr=g$, $dr=dg$) one computes $⟨\pi (f_1)\pi (f_2)\xi ,\eta ⟩={\int }_G{\int }_G⟨M_{f_1(h,\cdot )}{M}_{f_2(r,h^{-1}\cdot )}{U}_{hr}\xi ,\eta ⟩drdh={\int }_G{\int }_G⟨M_{f_1(h,\cdot )f_2(h^{-1}g,h^{-1}\cdot )}{U}_g\xi ,\eta ⟩dgdh$; the scalar kernel is integrable on the compact support, so Fubini's theorem turns the iterated integral into ${\int }_G⟨M_{{\int }_G{f}_1(h,\cdot )f_2(h^{-1}g,h^{-1}\cdot )dh}{U}_g\xi ,\eta ⟩dg=⟨\pi (f_1\ast f_2)\xi ,\eta ⟩$, using the identification of the inner Bochner integral of multiplication operators through its pairings and the definition of the twisted product in [F3]. As $\eta$ is arbitrary this gives $\pi (f_1)\pi (f_2)=\pi (f_1\ast f_2)$.

Adjoint: taking adjoints in the defining Bochner integral and using $U_g^{\ast }=U_{g^{-1}}$ and $M_b^{\ast }=M_{\bar{b}}$, $\pi (f{)}^{\ast }={\int }_G{U}_g^{\ast }{M}_{\overline{f(g,\cdot )}}dg={\int }_G{M}_{\overline{f(g,g\cdot )}}{U}_{g^{-1}}dg$, where the second equality is [step 1.2] with $b=\overline{f(g,\cdot )}$ and $h=g^{-1}$. Substituting $h=g^{-1}$ in the Haar integral and using [F4] in the form ${\int }_{G}w(g)dg={\int }_G{\mathrm{\Delta }}_G(h{)}^{-1}w(h^{-1})dh$ gives $\pi (f{)}^{\ast }={\int }_G{M}_{{\mathrm{\Delta }}_G(h{)}^{-1}\overline{f(h^{-1},h^{-1}\cdot )}}{U}_{h}dh=\pi (f^{\ast })$, with $f^{\ast }$ the modular involution of [F3].

Steps 2.1, 3.1 and 3.2 show that $f\mapsto \pi (f)$ is a bounded $\ast$-representation of the transformation algebra with the stated norm bound, and step 2.2 shows it is nondegenerate. The homogeneous-space case $X=G\mathrm{/}H$ of the pair satisfies the added topological hypotheses, since $G\mathrm{/}H$ is second-countable LCH with continuous left action (Second-countable locally compact Hausdorff spaces are Polish, and homogeneous quotients are standard Borel).