The stabilizer acts unitarily on an imprimitivity fibre

The assignment $c(g,x):={\phi }_g(x)$ is a Borel $U(K)$-valued cocycle on $G\times G\mathrm{/}H$ satisfying the a.e. cocycle law and the local-measure continuity of [F1]; hence [F2] applies and produces a Borel unitary field $B$ and a strongly continuous unitary $\sigma :H\to U(K)$ with ${\phi }_g(x)=B(gx)\sigma (h(g,x))B(x{)}^{-1}$ for every $g$ and a.e. $x$.

Normalization at $eH$: if $\{eH\}$ is $\mu$-null, redefine $B_{eH}=I$; this changes $B$ on a null set and the factorization remains valid a.e. If $\{eH\}$ is an atom, replace $B_x$ by $B_x{B}_{eH}^{-1}$ and $\sigma$ by $B_{eH}\sigma (\cdot )B_{eH}^{-1}$, which is a unitary equivalence of representations and makes the new field equal to $I$ at $eH$. In both cases the factorization holds for every $g$ and a.e. $x$, and $B_{eH}=I$.

Uniqueness and gauge: if $(B_1,{\sigma }_1)$ and $(B_2,{\sigma }_2)$ both factorize the same cocycle, the uniqueness clause of [F2] gives a single unitary $T$ with ${\sigma }_2(h)T=T{\sigma }_1(h)$ for all $h$; a Borel gauge change $B\mapsto AB$ multiplies the lifted trivializations on the left and does not change the equivalence class. A change of section changes $B$ by the corresponding $\sigma$-factor and leaves the class of $\sigma$ fixed.

Place the formula in strict form: define ${\widehat{\phi }}_g(x):=B(gx)\sigma (h(g,x))B(x{)}^{-1}$; by [step 1.1] ${\widehat{\phi }}_g={\phi }_g$ a.e. for every $g$, and the right-hand side is jointly Borel in $(g,x)$; the strict section identity of [F3] makes $\widehat{\phi }$ an exact cocycle on all of $G\times G\mathrm{/}H$, so replacing the original fields by $\widehat{\phi }$ changes nothing in the a.e. class and gives the displayed formula for every pair.

Evaluating the strict formula at $x=eH$: for $h^{\mathrm{\prime }}\in H$ one has $h(h^{\mathrm{\prime }},eH)=s(eH{)}^{-1}{h}^{\mathrm{\prime }}s(eH)=h^{\mathrm{\prime }}$, so ${\phi }_{h^{\mathrm{\prime }}}(eH)=B(eH)\sigma (h^{\mathrm{\prime }})B(eH{)}^{-1}=\sigma (h^{\mathrm{\prime }})$ because $B_{eH}=I$; and for $g=s(x)$, $h(s(x),eH)=s(x{)}^{-1}s(x)s(eH)=e$ gives ${\phi }_{s(x)}(eH)=B(x)\sigma (e)B_{eH}^{-1}=B(x)$. Thus the recovered data are exactly $\sigma (h^{\mathrm{\prime }})={\phi }_{h^{\mathrm{\prime }}}(eH)$ and $B_x={\phi }_{s(x)}(eH)$.

Boundary cases: if $H=G$ then $G\mathrm{/}H$ is a point, $s(eH)=e$, and the factorization collapses to ${\phi }_g=B\sigma (g)B^{-1}$, so $\sigma$ is unitarily equivalent to the original representation carried by the fields. If $H=\{e\}$ then $H$ is the trivial group and $\sigma$ is a strongly continuous unitary representation of the trivial group, hence the identity representation on its given fibre $K$; nothing more is asserted.

Steps 1.1, 2.1 and 3.1 give existence of $B,\sigma$ with the strict factorization and the two evaluation identities; [step 1.3] gives uniqueness up to unitary equivalence and gauge; [step 4.1] gives the two boundary cases. This proves the statement.