By Proposition

9.A.2, for every

\(n\ge 1\), there exist a bimeasurable function

\(p^0_n:\mathsf {X}^2\rightarrow \mathbb {R}_+\) and a kernel

\(S_n\) on

\(\mathsf {X}\times \mathscr {X}\) such that for all

\(x\in \mathsf {X}\) and

\(A\in \mathscr {X}\),

$$\begin{aligned} P^n(x, A) = \int _\mathsf {X}p_n^0(x,y) \phi (\mathrm {d}x) + S_n(x, A) \;. \end{aligned}$$

Define inductively the sequence of positive measurable functions

Open image in new window on

\(\mathsf {X}^2\) in the following way: set

\(p_1 = p^0_1\), and for all

\(n >1\) and

\(x, y \in \mathsf {X}\), set

$$\begin{aligned} p_n(x, y) = p^0_n(x, y) \vee \sup _{1\le k< n} \int _\mathsf {X}P^{n-k}(x,\mathrm {d}z) p_{k}(z, y) \;. \end{aligned}$$

(9.A.6)

By construction,

\(p_n\) satisfies the first inequality in (

9.A.4). We now show by induction on

\(n\ge 1\) that for every

\(x\in \mathsf {X}\),

\(p_n(x, y)=p^0_n(x, y)\), for

\(\phi -\mathrm {almost\ all}\) *y*. Indeed, this is true for

\(n=1\) by definition of

\(p_1\). For

\(n\ge 2\), assume that the induction assumption is true for

\(n-1\), i.e., for all

\(k=1,\dots , n-1\),

\(p_k(x, y)=p^0_k(x, y)\),

\(\phi -\mathrm {almost\ all}\) *y*. Then we have, for all

\(k=1,\dots , n-1\) and

\((x, A)\in \mathsf {X}\times \mathscr {X}\),

$$\begin{aligned} P^{n}(x,A)&= \int P^{n-k}(x,\mathrm {d}z) P^k(z, A) \\&\ge \int P^{n-k}(x,\mathrm {d}z) \int _A p^0_k(z,y) \phi (\mathrm {d}y) = \int _A \int P^{n-k}(x, dz) p^0_k(z, y) \phi (\mathrm {d}y) \;. \end{aligned}$$

Let the set

\(B^x_n\in \mathscr {X}\) be such that

\(\phi (B^x_n)=0\) and

\(S_n(x, B^x_n)=S_n(s,\mathsf {X})\). Then applying the previous inequality and the induction assumption, we obtain

$$\begin{aligned} \int _A p^0_n(x,y) \phi (\mathrm {d}y)&= P^n(x, A\setminus B^x_n) \ge \int _{A\setminus B^x_n} \left( \int P^{n-k}(x,\mathrm {d}z) p^0_k(z, y)\right) \phi (\mathrm {d}y) \\&= \int _A\left( \int P^{n-k}(x,\mathrm {d}z)p^0_k(z,y)\right) \phi (\mathrm {d}y) \\&= \int _A\left( \int P^{n-k}(x,\mathrm {d}z) p_k(z, y)\right) \phi (\mathrm {d}y) \;. \end{aligned}$$

This implies that for all

\(1\le k\le n\) and

\(x \in \mathsf {X}\),

$$\begin{aligned} p^0_n(x,\cdot ) \ge \int P^{n-k}(x,\mathrm {d}z)p_k(z,\cdot ) \quad \quad \phi -\text {a.e.}\end{aligned}$$

Since the set

*A* is arbitrary, this proves that the induction assumption is true for

*n*. Therefore, for all

\(x\in \mathsf {X}\), (

9.A.3) and the first inequality in (

9.A.4) hold. This in turn proves the second inequality in (

9.A.4).

We now prove the last statement. Fix one particular

\(x_0\in \mathsf {X}\). Set

\(F = \cap _{n\ge 1} (B_n^{x_0})^c\), where

\(B_n^{x_0}\) was defined above. Then

\(\phi (F^c) \le \sum _{n\ge 1} \phi (B_n^{x_0})=0\), and for every

\(n\ge 1\),

\(S_n(x_0,F)=0\). Since

\(\phi \) is an irreducibility measure for

*P*, for

\(B\subset F\) such that

\(\phi (B)>0\), there exists

*m* such that

\(P^m(x_0,B) = \int _B p_m(x_0,y) \phi (\mathrm {d}y)>0\). This implies

$$\begin{aligned} \int _B \sum _{n\ge 1} p_n(x_0,y) \phi (\mathrm {d}y) \ge \int _B p_m(x_0,y) \phi (\mathrm {d}y) > 0 \;. \end{aligned}$$

Since

*B* is an arbitrary subset of

*F* and

\(\phi (F^c)=0\), this implies

\(\sum _{n\ge 1} p_n(x_0,\cdot )>0\) \(\phi \)-

*a.e.* This proves (

9.A.5).

\(\Box \)