Measurable Lebesgue decomposition of a kernel on a countably generated space

Choose a disjoint measurable partition $(E_j{)}_{j\ge 1}$ with $\mu (E_j)<\mathrm{\infty }$ and positive total measure. Set $c_j=2^{-j}\mathrm{/}(1+\mu (E_j))$ and $c={\sum }_j{c}_j\mu (E_j)$. Then $0<c<\mathrm{\infty }$, and $\bar{\mu }(A)=c^{-1}{\sum }_j{c}_j\mu (A\cap E_j)$ is an equivalent probability measure. Its density $w=c_j\mathrm{/}c$ on $E_j$ is positive. For each fixed $x$, choose $K(x,\cdot )=f_x\bar{\mu }+S_x$ from [F3]. Let $({\mathcal{P}}_n)$ be increasing finite partitions generating $\mathcal{X}$, and set $f_n(x,y)={\sum }_{B\in {\mathcal{P}}_n:\bar{\mu }(B)>0}\frac{K(x,B)}{\bar{\mu }(B)}{1}_B(y),$ with value zero on zero-measure cells. For a positive-measure cell $B\in {\mathcal{P}}_n$, its average at level $n+1$ integrates over $B$ to the sum of $K(x,C)$ over the positive-measure children $C\subseteq B$, which is at most $K(x,B)$; zero-measure parents and all their children have value zero. Thus $(f_n(x,\cdot ))$ is a nonnegative $\bar{\mu }$-supermartingale. Also $\int f_n(x,\cdot )d\bar{\mu }\le K(x,\mathsf{X})=1$, so $-f_n(x,\cdot )$ has uniformly bounded positive-part expectations and [F4] gives a finite almost-sure limit for $f_n(x,\cdot )$.

Define $f_{\mathrm{\infty }}(x,y)={\mathrm{lim\hspace{0.17em}sup}}_n{f}_n(x,y)$, a jointly measurable extended nonnegative function; for each fixed $x$ it agrees almost everywhere with the finite limit from step 1.1. The conditional-expectation averages $g_n^x$ of the integrable density $f_x$ over ${\mathcal{P}}_n$ are defined by $g_n^x(y)=\bar{\mu }(B{)}^{-1}{\int }_B{f}_{x}d\bar{\mu }$ on positive-measure cells $B$, and by zero on zero-measure cells. Since $\int f_{x}d\bar{\mu }\le 1$, [F4] and the AC hypothesis [A1] give $g_n^x\to f_x$ almost everywhere: ${\mathcal{P}}_n$ generates $\mathcal{X}$. Cellwise $g_n^x\le f_n(x,\cdot )$ because ${\int }_B{f}_{x}d\bar{\mu }\le K(x,B)$, with both averages zero on null cells. Taking limits proves $f_x\le f_{\mathrm{\infty }}(x,\cdot )$ almost everywhere.

For a cell $B$ at partition level $N$, every $n\ge N$ has ${\int }_B{f}_n(x,y)\bar{\mu }(dy)\le K(x,B)$, since the positive-measure level-$n$ cells inside $B$ contribute only a sub-sum of the kernel masses of a partition of $B$. Fatou therefore gives ${\int }_B{f}_{\mathrm{\infty }}(x,y)\bar{\mu }(dy)\le K(x,B)$. By [F8], the left side defines a measure; Fatou at the whole space gives its total mass at most one, so both sides are finite measures. Their sum is a finite measure, and [F6] approximates any measurable set in symmetric difference under that sum; the inequality extends from the generating partition algebra to every measurable set. Choose a singular support $N_x$ with $\bar{\mu }(N_x)=0$ and $S_x(N_x^c)=0$. Since the left measure has density $f_{\mathrm{\infty }}$ with respect to $\bar{\mu }$, it assigns zero mass to $N_x$. Applying the inequality to $A\cap N_x^c$ therefore gives ${\int }_A{f}_{\mathrm{\infty }}d\bar{\mu }\le {\int }_A{f}_{x}d\bar{\mu }$. Together with step 2.1, this proves $f_{\mathrm{\infty }}=f_x$ almost everywhere.

Since $\bar{\mu }(dy)=w(y)\mu (dy)$, set $p(x,y)=f_{\mathrm{\infty }}(x,y)w(y)$ and $S(x,A)=K(x,A)-{\int }_{A}p(x,y)\mu (dy)$. Step 3.1 shows sectionwise that $S(x,\cdot )=S_x$, a nonnegative subprobability measure singular to $\mu$. The integral is measurable in $x$ by [F5]; hence $S(\cdot ,A)$ is measurable for every $A$, and its sections are countably additive. Thus $S$ is a subprobability kernel and the decomposition holds.