Iterated kernel densities retain convolution lower bounds

Apply [F3] to each $P^n$, obtaining preliminary densities $p_n^0$ and singular residuals $S_n$. Inductively set $p_1=p_1^0$ and, for $n>1$, $p_n(x,y)=\max (p_n^0(x,y),{\sup }_{1\le k<n}\int P^{n-k}(x,dz)p_k(z,y))\mathrm{.}$ For each $k<n$, the lifted kernel $((x,y),A)\mapsto P^{n-k}(x,A)$ is a probability kernel by [F2]. Apply [F4] to this kernel and the nonnegative product-measurable integrand $((x,y),z)\mapsto p_k(z,y)$; finite maxima preserve measurability. Thus the right side is jointly measurable. For each fixed $x$, composition of kernels gives $P^n(x,A)\ge {\int }_A\int P^{n-k}(x,dz)p_k(z,y)\varphi (dy)$. Since $S_n(x,\cdot )$ is singular, this domination forces each candidate density to be at most $p_n^0(x,\cdot )$ $\varphi$-almost everywhere. Thus $p_n=p_n^0$ $\varphi$-a.e. for each $x$, so it is still the density part and has both displayed convolution lower bounds pointwise by construction.

Fix $x$. For each $n$, choose a $\varphi$-null support $N_n$ of $S_n(x,\cdot )$ and let $F={\bigcap }_{n\ge 1}{N}_n^c$. Then $\varphi (F^c)=0$ and every residual assigns zero mass to $F$. If $B\subseteq F$ has $\varphi (B)>0$, [F1] gives an $m$ with $P^m(x,B)>0$; because the residual gives $B$ zero mass, ${\int }_B{p}_m(x,y)\varphi (dy)>0$.

Therefore every positive-$\varphi$ subset of $F$ intersects $\{{\sum }_n{p}_n(x,\cdot )>0\}$ in positive measure. Its complement in $F$ must be $\varphi$-null; otherwise take that complement as $B$ in step 1.2, which would make every integral zero. Hence the density sum is positive $\varphi$-almost everywhere.