Transforming an inhomogeneous Markov chain to a homogeneous one

I fail to understand Cinlar's transformation of an inhomogeneous Markov chain to a homogeneous one. It appears to me that $\hat{P}$ is not fully specified. Generally speaking, given a $\sigma$-algebra $\mathcal A$, a measure can be specified either explicitly over the entire $\sigma$-algebra, or implicitly by specifying it over a generating ring and appealing to Caratheodory's extension theorem. However, Cinlar specifies $\hat{P}$ over a proper subset of $\hat{\mathcal{E}}$ that is not a ring.

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We give the condition that $\widehat P$ is a Markow kernel, and we have that $$\widehat P((n,x),\{n+1\}\times E)=P_{n+1}(x,E)=1,$$ hence the measure $\widehat P((n,x),\cdot)$ is concentrated on $\{n+1\}\times E\}$? Therefore, we have $\widehat P((n,x),I\times A)=0$ for any $A\subset E$ and $I\subset \Bbb N$ which doesn't contain the integer $n+1$.
Thank you, Davide. What if $\left\{n+1\right\}\subsetneq I$? –  Evan Aad Aug 8 '12 at 13:01
In this case, the set $I\times E$ has necessarily measure $0$ by $\hat P((n,x),)$, since $\{n+1\}\times E$ concentrated all the measure. –  Davide Giraudo Aug 8 '12 at 13:49
But $A:=\left\{n+1\right\}\times E \subseteq B:=I \times E$, so we should have $a:=\hat{P}\left(\left(n,x\right),A\right)\leq b:=\hat{P}\left(\left(n,x\right),B\right)$, since $\hat{P}$ is a probability transition kernel, whereas you claim that $a=1,\space b=0$. –  Evan Aad Aug 8 '12 at 13:58
Sorry, I misread your comment. I believed you wrote "what if $\{n+1\}$ is not included in $I$", whereas you wrote "what if $\{n+1\}$ is contained in $I$ but not equal". In this case, we have $\widehat P((n,x),I\times A)=P_{n+1}(A)$. –  Davide Giraudo Aug 8 '12 at 14:00
OK. I think i get it: $\hat{P}\left(\left(n,x\right), I\times A\right) = P_{n+1}\left(x,A\right)\cdot \mathbb 1_I\left(n+1\right)$. –  Evan Aad Aug 8 '12 at 14:25