A deterministic kernel from a measurable map

For fixed s, $A\mapsto {\mathbf{1}}_A(g(s))$ vanishes on the empty set and equals one on T. If $A_j$ are disjoint, at most one contains g(s), so ${\mathbf{1}}_{{\bigcup }_j{A}_j}(g(s))={\sum }_j{\mathbf{1}}_{A_j}(g(s))$. This is countable additivity, so the section is the Dirac probability at g(s). For a measurable A, the evaluation is ${\mathbf{1}}_{g^{-1}(A)}(s)$; its set $g^{-1}(A)$ is measurable by hypothesis. These are exactly [F1].

By [F2] and [F3], for a measurable $C\subseteq U$, $(K_g{K}_h)(s,C)={\int }_T{\mathbf{1}}_C(h(t))K_g(s,dt)=K_g(s,h^{-1}(C))={\mathbf{1}}_C(h(g(s)))\mathrm{.}$ The middle equality is the integral of an indicator of the measurable set $h^{-1}(C)$. The composite h after g is measurable because $(h\circ g{)}^{-1}(C)=g^{-1}(h^{-1}(C))$. This proves the formula at every s, with no reference to a null set. For example on real Borel spaces take $g(s)=s^2$ and $h(t)=t+1$. Then the composite section is ${\delta }_{s^2+1}$; at s=2 its mass on (4,6) is one and on (0,4] is zero. If S is empty the assertions are vacuous; an empty T with nonempty S cannot support the assumed map g.