# Why $\mathbb P\{X_v\in A\mid \mathcal F_s\}=\int P_{s,t}(X_s,dy)P_{t,v}(y,A)$?

Let $$X$$ a stochastic process and $$\mathcal F_s=\sigma (X_u\mid u\leq t)$$. In the book "Continuous Martingale and Brownian motion" (third edition) of Yor and Revuz, page 80 : Let $$P_{s,t}(X_s,A)=\mathbb P\{X_t\in A\mid \mathcal F_s\}.$$

Q1) They say that $$\mathbb E[f(X_t)\mid \mathcal F_s]=P_{s,t}f(X_s)$$ a.s. where $$f$$ is a bounded and continuous function. Is it a definition or a proposition ? Because, they say that it can be proved using classical argument on conditional expectation, but I don't see how I can prove it.

Q2) After, they say that for $$s, $$\mathbb P\{X_v\in A\mid \mathcal F_s\}=\mathbb P\{X_v\in A\mid \mathcal F_t\mid \mathcal F_s\}=\mathbb E[P_{t,v}(X_t,A)\mid \mathcal F_s]=\int P_{s,t}(X_s,dy)P_{t,v}(y,A).$$

1) I'm not sure what mean $$\mathbb P\{X_v\in A\mid \mathcal F_t|\mathcal F_s\}$$. Is it $$\mathbb P \{\mathbb P\{X_v\in A\mid \mathcal F_t\}|\mathcal F_s\}= \mathbb E[\mathbb E[\boldsymbol 1_{X_v\in A}\mid \mathcal F_t]\mid \mathcal F_s] \ \ ?$$

2) How can I prove that $$\mathbb E[P_{t,v}(X_t,A)\mid \mathcal F_s]=\int P_{s,t}(X_s,dy)P_{t,v}(y,A),$$ I by the way, didn't even know that we have a formula for $$\mathbb E[P_{t,v}(X_t,A)\mid \mathcal F_s]$$