Question:
If $\mu(\cdot,\cdot)$ is a probability kernel and satisfies: $$\mu(x,B)=\int\mu(x,dy)\mu(y,B)$$ for any $x$ and measurable set $B$. Then $\mu(x,B)=\delta_x(B)$.
Motivation: I want to know if we can get the transition kernel $\mu_{s,s}(x,B)=\delta_x(B)$ just from Chapman-Kolmogorov equations $\mu_{s,t}=\mu_{s,u}\mu_{u,t}$.
My attempt: We can easily get for each fixed $x,B$, there is a measurable set $\Omega_B$ with $\mu(x,\Omega_B)=1$ s.t. $\mu(y,B)=\delta_y(B)$ for any $y\in\Omega_B$. But I don't know how to go on.