Conditional independence under kernel extension and conditioning

Fix bounded measurable $g(R)$ and set $r=E[g(R)\mid \sigma (C)]$. Conditional independence implies $E[h(L)b(C)g(R)]=E[h(L)b(C)r]$ for every bounded measurable $h,b$: test its conditional-product identity against $b(C)$ and then the defining conditional expectation of $h(L)$. Rectangle indicators generate $\sigma (L,C)$, so [F3] gives $E[Hg(R)]=E[Hr]$ for every bounded $\sigma (L,C)$-measurable $H$. Equivalently $E[g(R)\mid \sigma (L,C)]=r$. All extensions can be made for each fixed $g$: the class of testing events is a lambda-system by complements and disjoint countable unions; bounded versions and monotone convergence justify the sums. No common null set for all functions is needed.

For bounded measurable $h(l,v)$ define $H(l,c)=\int h(l,v)K(l,c,dv)$, bounded and measurable by [F2]. Kernel integration gives $E[b(C)h(L,V)g(R)]=E[b(C)H(L,C)g(R)]=E[b(C)H(L,C)r]=E[b(C)h(L,V)r]$, where the middle identity is step 1.1. Thus $E[h(L,V)g(R)\mid \sigma (C)]=E[h(L,V)\mid \sigma (C)]r$. In taking out bounded measurable conditioning factors here, the event-integral definition first proves the rule for simple factors and [F3] proves it for bounded factors. This proves the first claimed independence. Testing the same calculation with rectangle factors in $(L,V,C)$ and using [F3] also gives $E[g(R)\mid \sigma (L,V,C)]=r$.

Since $r$ is $\sigma (C)$-measurable, [F4] yields $E[g(R)\mid \sigma (C,V)]=r$. For bounded $f(L)$, conditioning first on $\sigma (L,V,C)$ in step 2.1 and then on $\sigma (C,V)$ gives $E[f(L)g(R)\mid \sigma (C,V)]=E[f(L)\mid \sigma (C,V)]r$. The bounded-factor rule justified in step 2.1 applies at both stages. This is the second claimed independence by [F1]. Singleton state spaces and constant kernels obey the same identities.