Let (left(T_{t} ight)_{t geqslant 0}) be a Markov semigroup given by (T_{t} u(x)=int_{mathbb{R}^{d}} u(y) p_{t}(x, d y))

Let \(\left(T_{t}\right)_{t \geqslant 0}\) be a Markov semigroup given by \(T_{t} u(x)=\int_{\mathbb{R}^{d}} u(y) p_{t}(x, d y)\) where \(p_{t}(x, C)\) is a kernel in the sense of Remark 7.6. Show that the semigroup property of \(\left(T_{t}\right)_{t \geqslant 0}\) entails the Chapman-Kolmogorov equations

\[p_{t+s}(x, C)=\int p_{t}(y, C) p_{s}(x, d y) \quad \text { for all } \quad x \in \mathbb{R}^{d} \quad \text { and } \quad C \in \mathscr{B}\left(\mathbb{R}^{d}\right)\]

Data From 7.6 Remark

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7.6 Remark (From semigroups to processes). So far we have considered semigroups given by a Brownian motion. It is an interesting question whether we can construct a Markov process starting from a semigroup of operators. If (T+)tao is a Markov semigroup where pt(x, C) := T+1c(x) is an everywhere defined kernel such that Cpt(x, C) is a probability measure for all x Rd, (t, x)p(x, C) is measurable for all CB(Rd), this is indeed possible. We can use the formula from Theorem 7.5 to define for every X Rd P(Cxx Cn) := T [1c Tt-ti [1cz... Tt-tn-11C]](x),